Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Combinatorics
8
1 Permutations and Combinations . . . . . . . . . . . . . . . . . . . . . . . . .
8
2 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
4 Complementary Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
5 Principle of Inclusion and Exclusion . . . . . . . . . . . . . . . . . . . . . . .
67
6 Stars & Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
7 Combinatorial Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
3 Casework
8 Geometric Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
9 Geometric Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
10 Expected Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
11 Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
12 Probability States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
Algebra
151
13 Algebraic Manipulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
14 Vieta’s Formulas for Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 166
15 Polynomial Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
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Contents
16 Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189
17 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
18 Special Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
19 Mean, Median, Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
20 System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
21 Speed, Distance, and Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
Number Theory
239
22 Primes and Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
23 Divisibility & Legendre’s Formula . . . . . . . . . . . . . . . . . . . . . . . . 253
24 GCD & LCM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260
25 Modular Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268
26 Algebraic Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
27 Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
28 Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
29 Miscellaneous Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 310
Geometry
319
30 Angle Chasing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
31 Triangle Area and Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
32 Special Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
33 Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
34 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
35 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424
36 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
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37 3-D Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
38 Area and Length of Complex Shapes . . . . . . . . . . . . . . . . . . . . . . 514
39 Coordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538
Advanced Topics
550
40 Floor and Ceiling Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550
41 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
42 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 568
43 Algebraic Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582
44 Geometric Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595
45 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607
46 Additional Techniques and Strategies . . . . . . . . . . . . . . . . . . . . . . 625
3
Preface
Motivation
This book was created to provide a comprehensive overview of the most important concepts on
the AMC 10/12 examinations. The book includes video lectures for every chapter, formulas
for every topic, and hundreds of examples and practice problems with detailed video solutions.
This book covers the following topics:
• Combinatorics
• Algebra
• Number Theory
• Geometry
Discord Server
Join our Discord Server to discuss problems from the book and for help with math contest
preparation.
Feedback Form
If you have any feedback, find any errors, or think of any interesting problems that should be
added here, please fill out this Feedback Form or email us at omegalearn.info@gmail.org.
Book Updates
We appreciate your feedback and will update the book regularly by adding new topics and
problems. Please bookmark OmegaLearn.org to get the Latest Version of this book.
Video Content
The book includes video lectures for every chapter. All the lecture videos are available on
our YouTube Channel. These videos should provide a good foundation for anyone preparing
for math competitions.
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Preface
Information about math competitions
These videos provide some useful strategies for anyone just starting with math competitions
or working towards specific competitions like the AMC 10/12.
1. All you need to know about Math Competitions from Elementary to High School
2. How to prepare for AMC 10/12 and qualify for AIME and USA(J)MO
Free Mastering AMC 10/12 Course
Mastering AMC 10/12 Course
This is the video course accompanying this book, and includes video lectures for every
chapter. We explain all the important concepts from this book, along with many useful
examples showing the application of those concepts.
Free AMC 10/12 Fundamentals Course
AMC 10/12 Fundamentals Classes
This is an earlier course which covered the most fundamental topics on the AMC 10/12
exams. Here is the list of topics that were covered:
1. Quadratics, Polynomials, Vieta’s Formulas, Roots
Homework Solutions
2. Algebraic Manipulations, Factorizations, SFFT, Sophie Germain’s Identity
Homework solutions
3. Permutations, Combinations, Casework, Complementary Counting
Homework Solutions
4. Probability, Expected Value, Geometric Probability, States
Homework Solutions
5. Number Theory: Factors, Multiples, and Bases
Homework Solutions
6. GCD/LCM, Modular Arithmetic, Diophantine Equations
Homework Solutions
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Preface
7. Triangles, Polygons, Basic Trigonometry
Homework Solutions
8. Circular Geometry, Power of a Point, Cyclic Quadrilaterals
9. Logarithms
10. Meta-solving Techniques - how to find answers without solving the problem
How to Use This Book
To get the most out of this book, please give each problem your best effort before checking
out the solution! Even if you don’t solve it, the act of trying will help you gain useful insights
about the problem and will help you develop the intuition needed for the topic.
Each chapter includes relevant formulas for the topic along with instructive Example
Problems that show interesting applications of the concept.
Then, there is a Practice Problems section, which includes problems from AMC 10/12,
AIME and original problems. These problems have a detailed video solution.
There is also an Additional Problems section with more problems (on the harder side) for
extra practice.
Hopefully, the curated collection of examples and problems in this book will improve your
problem solving skills and help you perform better on the AMC 10/12 exams. Good Luck!
6
Combinatorics
7
Chapter 1
Permutations and Combinations
Video Lecture
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Chapter 1. Permutations and Combinations
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1.1
Chapter 1. Permutations and Combinations
Factorials
Definition 1.1.1. A factorial is the product of all positive integers less than or equal to a
given positive integer. In other words n! = n × (n − 1) × (n − 2) × · · · × 1.
1.2
Permutations Basics
Definition 1.2.1. A permutation is a possible arrangement of objects in a set where the
order of objects matter.
Theorem 1.2.2 (Number of Orderings of Objects)
The number of ways of arranging n objects in a line is n!
The number of ways of arranging n objects in a circle where rotations of the same
arrangement aren’t considered distinct is (n − 1)!
The number of ways of arranging n objects in a circle where rotations of the same
arrangement aren’t considered distinct and reflections of the same arrangement aren’t
considered distinct is (n−1)!
2
Remark 1.2.3
The reason that this is true is because we can simply fix 1 person to be at the top and
there are (n − 1)! ways to arrange the other people. This accounts for rotations since
rotating an arrangement will result in someone else on top. We divide by 2 for reflections
because of symmetry on both the left and right sides of the person chosen to be at the
top.
Theorem 1.2.4 (Permutations Formula)
The number of ways to arrange k objects out of n total objects is
n
Pk =
n!
(n − k)!
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1.3
Chapter 1. Permutations and Combinations
Permutations with Restrictions
Example 1.1
How many 4 digit numbers exist such that the first digit is odd and the other 3 digits
are even and all digits distinct?
Solution 1.1
The first digit has 5 choices (1, 3, 5, 7, or 9). The 2nd digit also has 5 choices (0, 2, 4, 6, 8).
Do we have to subtract any of the choices of the 2nd digit to make sure the
numbers are distinct?
Keep in mind that although the digits have to be distinct, the 2nd digit is even and
the first digit is odd, so there is no overlap.
The 3rd digit can be any even number except the one chosen for the 2nd digit, so it has
4 choices. The 4th digit can be any even number except those chosen for the 2nd and 3rd digit.
In total, the number of 4 digit numbers with these constraints is
5 × 5 × 4 × 3 = 300
Example 1.2 (Omega Learn)
Gauss, Pauli, Einstein, Newton, Edison, and Faraday are sitting at a circular table.
Einstein, Newton, and Faraday are enemies so they refuse to sit next to each other.
With this condition, how many ways are there for the 6 physicists to sit at the table if
rotations are not counted as distinct orientations?
Solution 1.2
Without the condition, the answer would just be (6 − 1)! = 5! = 120 as we can fix 1 of the 6
people at the top and permute the remaining 5 people.
Where must the enemies sit so that none of them are sitting next to each
other?
Since there are only 6 total people, there must be exactly one other person between each of
the enemies as seen in the diagram below where the red X’s represent the enemies.
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Chapter 1. Permutations and Combinations
Should we again try to fix someone at the top to deal with rotations?
Yes! By fixing Gauss at the top (arbitrarily), we can now simply order the remaining
5 people without having to worry about the rotation condition.
How many ways to permute the enemies?
Note that the 3 possible locations of the enemies is fixed so there are 3! ways to permute them.
How many ways to permute the people with no enemies?
We already fixed Gauss (someone without enemies) so there are 2! ways to permute the other
2 people who don’t have enemies.
In total, our answer is
3! × 2! = 12
Example 1.3 (Omega Learn)
How many 6 digit numbers exist such that all the digits are distinct, and it’s first 2 digits
are 6 or more, the last 2 digits are 5 or less, and the 3rd digit is a nonzero multiple of 7?
Video Solution
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Chapter 1. Permutations and Combinations
Example 1.4
Peter is packing his pencil case for the AMC 12. He must pack 3 distinct pencils, 2
distinct erasers, and 2 distinct markers in some order. However, he will not put an item
in his case if there are already more of that type of item than any other type of item.
(Eg. Peter will not put another pencil in his case if there are already more pencils than
markers or more pencils than erasers in the case). How many different orders are there
for Peter to put items in his case?
Video Solution
1.4
Word Rearrangements
Theorem 1.4.1 (Word Rearrangements)
The number of ways to order a word is
n!
d1 ! × d2 ! × d3 ! × . . .
where n is the number of letters and d1 , d2 , d3 , . . . are the number of times each of the
letters that occur more than 1 time appear in the word.
Example 1.5
How many ways are there to rearrange the letters in COMPUTER such that the C, O,
M, and P stay together (not necessarily in the same order)?
Solution 1.3
This problem is slightly different from the standard word rearrangement problems because
we now have a constraint.
How should we deal with the condition that the COMP must stay together?
Let’s treat COMP as 1 block. Since the letters anyways have to stay together, let call this
block a COM P . Now, instead, we can find the number of arrangments of COM P UTER.
How many ways are there to arrange COM P UTER?
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Chapter 1. Permutations and Combinations
Notice that this is just 5! since all the letters (or symbols) are different.
Now, are we done, or is there something we forgot to account for?
Remember that letters in the word COMP can appear in any order inside the box, so
we have to also account for rearrangements of these letters within the COM P .
For each unique arrangement of COM P UTER, how many ways are there to
arrange the letters in the word COM P ?
There are 4! ways to arrange the 4 letters C O M P.
Therefore, in total, there are 5! × 4! = 120 × 24 = 2880 ways to rearrange COMPUTER
such that the letters COMP remain together.
Example 1.6 (AIME)
In a shooting match, eight clay targets are arranged in two hanging columns of three
targets each and one column of two targets. A marksman is to break all the targets
according to the following rules:
1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest remaining target in the chosen column.
If the rules are followed, in how many different orders can the eight targets be broken?
Video Solution
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Chapter 1. Permutations and Combinations
Remark 1.4.2
This is not only true for words! The number of ways of arranging objects or anything
else is also the same.
Example 1.7 (AMC 10)
Using the letters A, M , O, S, and U , we can form five-letter ”words”. If these ”words”
are arranged in alphabetical order, then the ”word” U SAM O occupies position
Video Solution
Remark 1.4.3
Don’t forget to add 1 to the number of words before USAMO in alphabetical order!
1.5
Combinations
Definition 1.5.1. A combination is a possible arrangement in a collection of items where
the order of the selection does not matter.
Remark 1.5.2
Usually, the words ”permute”, ”order does matter”, etc. imply a permutation while the
words ”choose”, ”select”, ”order doesn’t matter”, etc. imply a combination.
Theorem 1.5.3 (Combinations Formula)
The number of ways to choose k objects out of a total of n objects is
n
n!
n(n − 1) . . . (n − k + 1)
=
=
k
k!(n − k)!
k!
!
This is typically spoken as ”n choose k”
Remark 1.5.4
Notice that
!
n
n
=
k
n−k
!
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Chapter 1. Permutations and Combinations
This is true because we can see choosing k objects on the left hand side is equivalent to
the n − k objects that will not be selected on the right hand side.
Example 1.8 (Omega Learn)
There are 6 experienced applicants and 7 inexperienced applicants applying for a job.
Out of the experienced applicants, 2 managers are selected. Out of all of the other
applicants who aren’t selected for a manager (both experienced and inexperienced), 4
other employees are selected. How many ways are there to do this?
Solution 1.4
To solve this problem, we will find the number of ways of choosing managers and other
employees separately.
First, how many ways are there to select 2 managers?
Since they must be from the 6 experienced applicants, there are
30
6
6×5
=
= 15
=
2
2!
2
!
Next, how many ways are there to select the 5 other employees?
Since 2 of the experienced applicants were already selected, there are 4 experienced applicants and 7 inexperienced applicants left.
Therefore, we must select 4 employees out of the 11 remaining people. We can do this
in
!
11 × 10 × 72
72
11
11 × 10 × 9 × 8
=
= 11 × 10 ×
= 11 × 10 × 3 = 330
=
4!
24
24
4
Therefore, in total, how many ways are there to select the 3 managers and
5 other employees?
Since there are 15 ways to select the managers and 330 ways to select the other employees,
there will be 15 × 330 = 4950 ways to select both the managers and the employees.
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Chapter 1. Permutations and Combinations
Example 1.9 (Omega Learn)
You have 12 different energy bars that you need to give to 3 people: Alice, Betty, and
Chase. Alice needs 3 bars, Betty needs 4 bars, and Chase needs 5 bars. How many ways
are there to distribute the 12 bars to satisfy their requirements?
Video Solution
Example 1.10 (EMCC)
How many ways are there to place 4 balls into a 4 × 6 grid such that no column or row
has more than one ball in it? (Rotations and reflections are considered distinct.)
Video Solution
Example 1.11
Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin
has an engraving of one face on one side, but not on the other. He wants to stack the
eight coins on a table into a single stack so that no two adjacent coins are face to face.
Find the number of possible distinguishable arrangements of the 8 coins.
Solution 1.5
First, let’s deal with the orderings of the coins.
How many ways to arrange the 4 gold and 4 silver coins?
8!
We can use the word rearrangement formula to get 4!×4!
= 70. Now, we will the number of ways to put the coins face up or face down in the stack
If a coin is face down, what must the coin above it be?
It can be anything and the faces will never meet.
If a coin is face up, what must the coin above it be?
It must also be face up as if it’s face down, then the faces will meet.
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Chapter 1. Permutations and Combinations
The moment the coin first becomes face up, all the remaining coins on top must be face up
as well! If none of the coins are face up, then there is 1 orientation.
How many orientations exist if at least 1 of the coins is face up?
We simply have to pick a coin to be the first face up coin and every other coin’s orientation is fixed. There are 8 ways to do this.
In total, we have 9 orientations for the faces of the coins and
the gold and silver coins giving us an answer of 70 × 9 = 630 .
1.6
8
4
= 70 ways to order
Subsets
Theorem 1.6.1
The number of subsets of a set of size n is 2n .
Remark 1.6.2
We have 2 choices for each element in the set: whether to include or not include the
element in our subset, and since there are n elements in the set, the total number of
subsets is 2 × 2 × 2 × · · · × 2 (n times).
Note: This means that one of our subsets is the empty subset, where we decide to
not include all of the elements. Remember to check the problem wording whether we
should count the empty subset as valid or not.
Example 1.12 (AMC 10)
Two subsets of the set S = {a, b, c, d, e} are to be chosen so that their union is S and
their intersection contains exactly two elements. In how many ways can this be done,
assuming that the order in which the subsets are chosen does not matter?
Video Solution
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Chapter 1. Permutations and Combinations
Example 1.13 (AIME)
For {1, 2, 3, . . . , n} and each of its non-empty subsets a unique alternating sum is defined
as follows. Arrange the numbers in the subset in decreasing order and then, beginning
with the largest, alternately add and subtract successive numbers. For example, the
alternating sum for {1, 2, 3, 6, 9} is 9 − 6 + 3 − 2 + 1 = 5 and for {5} it is simply 5. Find
the sum of all such alternating sums for n = 7.
Solution 1.6
It’s definitely possible to manually add them for all possible subsets. However, there is a
much more clever way of approaching this problem.
For n = 3, what is the sum of all alternating sums?
As seen in the table, the total sum is 12. In fact, we can pair up the subsets so that
the sum of each pair is 3!
Why can we pair up the subsets as shown?
For every alternating sum that includes 3 of the form 3 - x, by removing the 3 from
the alternating sum becomes x as all the numbers that were added in the original alternating
sum will be subtracted in the alternating sum without 3 and vice versa!. Therefore, everything
will cancel out but the 3!.
By extending this logic, what is the sum of alternating sums for n = 7?
Like earlier, we pair up all subsets that include and 7 and ones that don’t include a 7.
7
In total, we have 22 = 64 pairs each with a sum of 7 giving an answer of 64 × 7 = 448 .
1.7
Practice Problems
Problem 1.7.1 (AMC 10)
How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2
green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are
indistinguishable.)
Video Solution
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Chapter 1. Permutations and Combinations
Problem 1.7.2 (AMC 10)
A child builds towers using identically shaped cubes of different colors. How many
different towers with a height of 8 cubes can the child build with 2 red cubes, 3 blue
cubes, and 4 green cubes? (One cube will be left out.)
Video Solution
Problem 1.7.3 (AMC 10)
A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2
letters, also not necessarily distinct. These six characters may appear in any order, except
that the two letters must appear next to each other. How many distinct license plates
are possible?
Video Solution
Problem 1.7.4 (AMC 10)
At a gathering of 30 people, there are 20 people who all know each other and 10
people who know no one. People who know each other hug, and people who do not know
each other shake hands. How many handshakes occur within the group?
Video Solution
Problem 1.7.5 (AMC 10)
Henry’s Hamburger Haven offers its hamburgers with the following condiments: ketchup,
mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose
one, two, or three meat patties and any collection of condiments. How many different
kinds of hamburgers can be ordered?
Video Solution
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Chapter 1. Permutations and Combinations
Problem 1.7.6 (AMC 10)
Values for A, B, C, and D are to be selected from {1, 2, 3, 4, 5, 6} without replacement
(i.e. no two letters have the same value). How many ways are there to make such choices
so that the two curves y = Ax2 + B and y = Cx2 + D intersect? (The order in which the
curves are listed does not matter; for example, the choices A = 3, B = 2, C = 4, D = 1 is
considered the same as the choices A = 4, B = 1, C = 3, D = 2.)
Video Solution
Problem 1.7.7 (AIME)
A game uses a deck of n different cards, where n is an integer and n ≥ 6. The number
of possible sets of 6 cards that can be drawn from the deck is 6 times the number of
possible sets of 3 cards that can be drawn. Find n.
Video Solution
Problem 1.7.8 (AMC 10A)
What is the number of ways the numbers from 1 to 14 can be split into 7 pairs such that
for each pair, the greater number is at least 2 times the smaller number?
(A) 108
(B) 120
(C) 126
(D) 132
(E) 144
Video Solution
Problem 1.7.9 (AMC 10/12)
Values for A, B, C, and D are to be selected from 1, 2, 3, 4, 5, 6 without replacement
(i.e. no two letters have the same value). How many ways are there to make such choices
so that the two curves y = Ax2 + B and y = Cx2 + D intersect? (The order in which the
curves are listed does not matter; for example, the choices A = 3, B = 2, C = 4, D = 1 is
considered the same as the choices A = 4, B = 1, C = 3, D = 2.)
Video Solution
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Chapter 1. Permutations and Combinations
Problem 1.7.10 (AMC 12)
How many 15-letter arrangements of 5 A’s, 5 B’s, and 5 C’s have no A’s in the first 5
letters, no B’s in the next 5 letters, and no C’s in the last 5 letters?
(A)
5
X
k=0
5
k
!3
(B) 35 · 25
(C) 215
(D)
15!
(5!)3
(E) 315
Video Solution
Additional Problems
Problem 1.7.11 (AIME)
Ten points in the plane are given, with no three collinear. Four distinct segments
joining pairs of these points are chosen at random, all such segments being equally likely.
The probability that some three of the segments form a triangle whose vertices are among
the ten given points is m/n, where m and n are relatively prime positive integers. Find
m + n.
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Chapter 1. Permutations and Combinations
Problem 1.7.12 (AIME)
Three numbers, a1 , a2 , a3 , are drawn randomly and without replacement from the set
{1, 2, 3, . . . , 1000}. Three other numbers, b1 , b2 , b3 , are then drawn randomly and without
replacement from the remaining set of 997 numbers. Let p be the probability that, after
suitable rotation, a brick of dimensions a1 × a2 × a3 can be enclosed in a box of dimension
b1 × b2 × b3 , with the sides of the brick parallel to the sides of the box. If p is written as
a fraction in lowest terms, what is the sum of the numerator and denominator?
Answers
1.1 300
1.2 12
1.3 900
1.4 864
1.5 2880
1.6 560
1.7 115
1.8 4950
1.9 27720
1.10 360
1.11 630
1.12 40
1.13 448
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Chapter 1. Permutations and Combinations
1.7.1 420
1.7.2 1260
1.7.3 5 × 104 × 262
1.7.4 245
1.7.5 768
1.7.6 90
1.7.7 13
1.7.8 144
1.7.9 90
1.7.10 A
1.7.11 489
1.7.12 5
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Chapter 2
Probability
Video Lecture
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2.1
Chapter 2. Probability
Probability Basics
Definition 2.1.1. Probability is the chance something occurs.
Theorem 2.1.2 (Probability)
probability =
Total number of Desired Outcomes
Total Outcomes
Example 2.1
In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5
squares are colored blue. If one of all possible such colorings is chosen at random, the
probability that the chosen colored array appears the same when rotated 90 degrees
around the central square is n1 , where n is a positive integer. Find n.
Solution 2.1
The first step to any probability problem is finding the number of successful outcomes.
What must happen so that when the figure is rotated the same coloring appears?
The top L shape must look the same as the right L shape, which must look the same
as the bottom L shape, which must look the same as the left L shape. Essentially, all 4 L
shapes must be colored in the exact same way.
How can this be done with 5 blue and 8 red squares?
The number of total blue squares in all of the L’s is a multiple of 4 and since there are a
total of 5 blue squares, there must be 4 total blue squares amongst the L’s and the center
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Chapter 2. Probability
square must also be colored blue. Due to the identical coloration discovered earlier, we must
pick one of the squares for all the L’s and color them blue. There are 3 ways to do this.
How many total ways are there to color the array?
There are 13 squares and any 5 of them can be blue so
1
3
=
of 133 = 1287
.
(5)
429
2.2
13
5
. This gives a probability
Distinguishability
Example 2.2
I have cards numbered from 1 to 10. What is the probability I pick a pair of 2 different
cards that have an odd product?
Solution 2.2
This is a rather simple example however it demonstrates an important point about distinguishability in probability problems.
What must happen for the cards to have an odd product?
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Chapter 2. Probability
Both cards must be odd.
How many pairs of cards have both cards odd?
If order does not matter, then there are 52 = 10 combinations. However, if the order
does matter, then there are 5 × 4 = 20 permutations.
How many total pairs of cards can be chosen?
If order does not matter, then there are 10
= 45 combinations. However, if the or2
der does matter, then there are 10 × 9 = 90 permutations.
So does the order matter of the cards in the pair matter or not?
If the order of the cards in the pair does matter, then the probability is 20
=
90
=
the order of the cards in the pair does not matter, then the probability is 10
45
2
.
9
2
. If
9
Why are the probabilities are the same?
When dealing with ordered pairs, the number of successful and total pairs were multiplied by
2! so they would cancel out when dividing!
Remark 2.2.1
In probability problems, whether you decide to multiply for order or not is up to you as
both will give the same probability.
Example 2.3
The sum of 2 positive integers is 4. Find the probability that one of the integers is a 2.
Solution 2.3
There are 2 cases:
1. Both integers are 2
2. One integer is a 3 and the other is a 1
So is the probability just 12 ?
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Chapter 2. Probability
No! While there are only 2 cases, they both have different chances of happening.
How many ways are there for each case to happen?
There is only 1 way where both integers can be 2 however there are 2 ways one of the
integers is a 1 and the other is a 3 as we can flip them. The 2nd case is more likely.
What is the probability with this knowledge?
There are a total of 3 cases of which 1 satisfies the condition so the probability is
1
.
3
Remark 2.2.2
If you decide not to multiply for order when doing probability problems, make sure the
number of orderings is consistent across all cases. In the first problem, every pair had 2
orderings. The same would be true for groups of 5 for example as there are always 5!
ways to order the elements in the group. However, in the 2nd problem, the two cases
had different number of orderings so this doesn’t work.
Example 2.4 (AIME)
A deck of forty cards consists of four 1’s, four 2’s,..., and four 10’s. A matching pair (two
cards with the same number) is removed from the deck. Given that these cards are not
returned to the deck, let m/n be the probability that two randomly selected cards also
form a pair, where m and n are relatively prime positive integers. Find m + n.
Video Solution
Example 2.5 (AMC 10)
Each of the 20 balls is tossed independently and at random into one of the 5 bins. Let
p be the probability that some bin ends up with 3 balls, another with 5 balls, and the
other three with 4 balls each. Let q be the probability that every bin ends up with 4
balls. What is pq ?
Video Solution
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2.3
Chapter 2. Probability
Probability of Independent Events
Remark 2.3.1
Whenever we need to find the probabilities of 2 independent events happening (the
results of the events don’t depend on each other), we can simply find the probability of
each event individually and multiply them together.
Example 2.6 (Omega Learn)
Person A rolls a dice with the numbers 1, 1, 2, 6, 15, 30 and Person B rolls a dice
with the numbers 1, 2, 4, 18, 28, 44. What is the probability that the product of the 2
numbers is a multiple of 24?
Video Solution
Example 2.7
What is the probability that the product of 4 dice rolls is a multiple of 648?
Video Solution
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2.4
Chapter 2. Probability
Probability of Dependent Events
Remark 2.4.1
When the events are dependent, the probability of both events happening (say A and
B) is the probability of event A happening times the probability that event B happens
given that event A already happened.
Example 2.8 (AIME)
There is a 40% chance of rain on Saturday and a 30% chance of rain on Sunday. However,
it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on
Saturday. The probability that it rains at least one day this weekend is ab , where a and
b are relatively prime positive integers. Find a + b.
Video Solution
2.5
Dependent or Independent Events?
Example 2.9 (Omega Learn)
Mark must choose a seat from 5 x 5 grid of chairs in the classroom. His enemy Steve
does not like Mark so wants to choose a seat that is not in the same row or column
as Mark. However, to his dismay, the teacher randomly assigns seats. What is the
probability that Steve is not sitting in the same row or column as Mark?
Video Solution
2.6
Practice Problems
Problem 2.6.1
Two 6-sided dice are rolled. What is the probability that the sum of the two numbers is even?
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Chapter 2. Probability
Video Solution
Problem 2.6.2 (AMC 12)
A deck of cards has only red cards and black cards. The probability of a randomly
chosen card being red is 13 . When 4 black cards are added to the deck, the probability of
choosing red becomes 41 . How many cards were in the deck originally?
Video Solution
Problem 2.6.3 (AMC 10)
Bob and Alice each have a bag that contains one ball of each of the colors blue, green,
orange, red, and violet. Alice randomly selects one ball from her bag and puts it into
Bob’s bag. Bob then randomly selects one ball from his bag and puts it into Alice’s bag.
What is the probability that after this process the contents of the two bags are the same?
Video Solution
Problem 2.6.4 (AMC 10)
Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold
randomly selects 5 books from this list, and Betty does the same. What is the probability
that there are exactly 2 books that they both select?
Video Solution
Problem 2.6.5 (AMC 10)
Two different numbers are selected at random from {1, 2, 3, 4, 5} and multiplied together.
What is the probability that the product is even?
Video Solution
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Chapter 2. Probability
Problem 2.6.6 (AMC 10)
A fair 6-sided die is repeatedly rolled until an odd number appears. What is the
probability that every even number appears at least once before the first occurrence of
an odd number?
Video Solution
Problem 2.6.7 (AMC 12)
Two fair dice, each with at least 6 faces are rolled. On each face of each dice is printed
a distinct integer from 1 to the number of faces on that die, inclusive. The probability
of rolling a sum of 7 is 34 of the probability of rolling a sum of 10, and the probability
1
of rolling a sum of 12 is 12
. What is the least possible number of faces on the two dice
combined?
Video Solution
Problem 2.6.8 (AMC 10/12)
The numbers 1,2,. . . ,9 are randomly placed into the 9 squares of a 3×3 grid. Each
square gets one number, and each of the numbers is used once. What is the probability
that the sum of the numbers in each row and each column is odd?
Video Solution
Problem 2.6.9 (AMC 10)
Two cubical dice each have removable numbers 1 through 6. The twelve numbers
on the two dice are removed, put into a bag, then drawn one at a time and randomly
reattached to the faces of the cubes, one number to each face. The dice are then rolled
and the numbers on the two top faces are added. What is the probability that the sum
is 7?
Video Solution
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Chapter 2. Probability
Problem 2.6.10 (AMC 12)
Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio’s
number is larger than the sum of the two numbers chosen by Tina?
Video Solution
Problem 2.6.11 (AMC 12)
Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for which the first term is
not 1. A permutation is chosen randomly from S. The probability that the second term
is 2, in lowest terms, is a/b. What is a + b?
Video Solution
Problem 2.6.12 (AMC 10)
A box contains 5 chips, numbered 1, 2, 3, 4, and 5. Chips are drawn randomly one at
a time without replacement until the sum of the values drawn exceeds 4. What is the
probability that 3 draws are required?
Video Solution
Problem 2.6.13 (AMC 10)
Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9,
or 10, with each number entered on four slips. Four slips are drawn from the hat at
random and without replacement. Let p be the probability that all four slips bear the
same number. Let q be the probability that two of the slips bear a number a and the
other two bear a number b ̸= a. What is the value of q/p?
Video Solution
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Chapter 2. Probability
Problem 2.6.14 (AMC 10/12)
A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed
into bin k is 2−k for k = 1, 2, 3.... What is the probability that the red ball is tossed into
a higher-numbered bin than the green ball?
Video Solution
Problem 2.6.15 (AMC 12)
Two fair dice, each with at least 6 faces are rolled. On each face of each dice is printed
a distinct integer from 1 to the number of faces on that die, inclusive. The probability
of rolling a sum if 7 is 34 of the probability of rolling a sum of 10, and the probability
1
of rolling a sum of 12 is 12
. What is the least possible number of faces on the two dice
combined?
Video Solution
Problem 2.6.16 (AMC 10)
Ang, Ben, and Jasmin each have 5 blocks, colored red, blue, yellow, white, and green;
and there are 5 empty boxes. Each of the people randomly and independently of the
other two people places one of their blocks into each box. The probability that at least
one box receives 3 blocks all of the same color is m
, where m and n are relatively prime
n
positive integers. What is m + n?
Video Solution
Problem 2.6.17 (AMC 10)
A fair 6-sided die is repeatedly rolled until an odd number appears. What is the
probability that every even number appears at least once before the first occurrence of
an odd number?
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Chapter 2. Probability
Video Solution
Problem 2.6.18 (AMC 12)
Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin i is 2−i for
i = 1, 2, 3, .... More than one ball is allowed in each bin. The probability that the balls
end up evenly spaced in distinct bins is pq , where p and q are relatively prime positive
integers. (For example, the balls are evenly spaced if they are tossed into bins 3, 17, and
10.) What is p + q?
Video Solution
Additional Problems
Problem 2.6.19 (AMC 10/12)
An urn contains one red ball and one blue ball. A box of extra red and blue balls
lies nearby. George performs the following operation four times: he draws a ball from the
urn at random and then takes a ball of the same color from the box and returns those
two matching balls to the urn. After the four iterations the urn contains six balls. What
is the probability that the urn contains three balls of each color?
Answers
1
2.1 429
2.2 29
2.3 13
2.4 758
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Chapter 2. Probability
2.5 16
2.6 16
5
2.7 1296
2.8 107
2.9 23
2.6.1 12
2.6.2 12
2.6.3 13
2.6.4 25
63
2.6.5 0.7
1
2.6.6 20
2.6.7 17
1
2.6.8 14
2
2.6.9 11
2.6.10 25
3
2.6.11 16
2.6.12 15
2.6.13 162
2.6.14 13
2.6.15 17
2.6.16 471
1
2.6.17 20
2.6.18 55
2.6.19 15
37
Chapter 3
Casework
Video Lecture
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3.1
Chapter 3. Casework
Casework
Many counting or probability problems can be solved by dividing a problem into several cases
and calculating arrangements and probabilities for each case before summing them together.
Concept 3.1.1
When doing casework, always try to minimize the number of cases you have to deal with.
Example 3.1 (Omega Learn)
There are 3 identical blue boxes, 6 identical green boxes, and 5 distinct items. Each of
the items must be placed into either a green or blue box, and each box can contain a
maximum of 1 item. Assuming all of the boxes are different, how many ways are there
to do this?
Solution 3.1
There are 5 items to be put in some of the 9 green and blue boxes, so we must choose 5 of
these 9 boxes to put the items into.
How can we divide this problem into cases?
We can choose our cases based on the number of green and blue boxes we are selecting. The
cases are
1. 3 blue boxes, 2 green boxes
2. 2 blue boxes, 3 green boxes
3. 1 blue box, 4 green boxes
4. 0 blue boxes, 5 green boxes
Notice that this covers all the cases because it covers all possible number of blue boxes.
Clearly, 6 green boxes is impossible since we only have to select 5 of them to put items into.
Also, make sure to keep in mind that the boxes of the same color are identical.
Case 1: 3 blue boxes, 2 green boxes
For this case, how many ways are there to put 5 items into 3 blue and 2 green
boxes?
We can count this be seeing that we simply have to choose 3 of the items to place into the
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Chapter 3. Casework
blue boxes, and the other items will be put in green boxes. We don’t have to worry about
order of the items in the blue boxes
(or the green boxes) because the boxes of the same color
are identical. We can do this in 53 ways.
Case 2: 2 blue boxes, 3 green boxes Again, we do this by choosing 2 of the items to
put into
blue boxes, and the remaining 3 items will be put into green boxes. We can do this
in 52 ways.
Case 3: 1 blue boxes, 4 green boxes Out of the 5 items, we must choose
1 of them
for a blue box and the remaining 4 will go in green boxes. We can do this in 51 ways.
Case 4: 0 blue boxes, 5 green boxes Because all of the items must go in green boxes,
there is only 1 way to do this.
In total among all the cases, there are
5
5
5
5×4×3 5×4 5
60 20 5
+
+
+1 =
+
+ +1 =
+ + + 1 = 10 + 10 + 5 + 1 = 26
3
2
1
3!
2!
1!
6
2 1
!
!
!
Remark 3.1.2
For casework problems, pick one attribute to change and find the number of ways in each
of the cases. In the previous problem, this was the number of blue boxes. Sometimes,
the problem may require subcases.
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Chapter 3. Casework
Example 3.2 (AMC 8)
A △ or ⃝ is placed in each of the nine squares in a 3-by-3 grid. Shown below is a sample
configuration with three △s in a line. How many configurations will have three △s in a
line and three ⃝s in a line?
Video Solution
Example 3.3 (AIME)
A bug starts at a vertex of an equilateral triangle. On each move, it randomly selects
one of the two vertices where it is not currently located, and crawls along a side of the
triangle to that vertex. Given that the probability that the bug moves to its starting
vertex on its tenth move is m/n, where m and n are relatively prime positive integers,
find m + n.
Solution 3.2
For this problem, we will do casework on the number of clockwise and counterclockwise
moves. Say there are A clockwise moves and B counterclockwise moves.
What must happen for the bug to reach the starting vertex in 10 moves?
The number of clockwise and counterclockwise moves must differ by a multiple of 3 as
every 3 moves the bug will just cycle around the triangle and come back to the starting
vertex.
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Chapter 3. Casework
How many move orderings exist if the number of clockwise and counter-clockwise
moves differs by 0?
We have A + B = 10 and A − B = 0 which
implies A = 5, B = 5. There must be 5
clockwise and 5 counterclockwise moves so 10
= 252 orderings.
5
How many move orderings exist if the number of clockwise and counter-clockwise
moves differs by 3?
We have A + B = 10 and A − B = 3. Adding the equations, we get 2A = 13 which
has no integer solutions. There is no possible ways in this case. The same is true if B − A = 3
by symmetry.
How many move orderings exist if the number of clockwise and counter-clockwise
moves differs by 6?
We have A+B = 10 and A−B = 6. Adding the equations, we get 2A = 16 =⇒ A = 8, B = 2.
We have 8 clockwise moves and 2 counterclockwise moves or 8 counter-clockwise moves. By
symmetry,
2 counterclockwise moves and 8 clockwise moves also works. Both of these have
10
= 45 orderings.
2
In total, we have 2 × 45 + 252 = 342 sucessful ways and 21 0 = 1024 total ways so the
342
probability is 1024
= 171
giving 683 .
512
Example 3.4 (AMC 12)
A fancy bed and breakfast inn has 5 rooms, each with a distinctive color-coded decor.
One day 5 friends arrive to spend the night. There are no other guests that night. The
friends can room in any combination they wish, but with no more than 2 friends per
room. In how many ways can the innkeeper assign the guests to the rooms?
Video Solution
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Chapter 3. Casework
Example 3.5 (AMC 10)
A farmer’s rectangular field is partitioned into 2 by 2 grid of 4 rectangular sections. In
each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The
farmer does not want to grow corn and wheat in any two sections that share a border,
and the farmer does not want to grow soybeans and potatoes in any two sections that
share a border. Given these restrictions, in how many ways can the farmer choose crops
to plant in each of the four sections of the field?
Video Solution
Example 3.6 (AMC 12)
Azar and Carl play a game of tic-tac-toe. Azar places an in X one of the boxes in a
3-by-3 array of boxes, then Carl places an O in one of the remaining boxes. After that,
Azar places an X in one of the remaining boxes, and so on until all boxes are filled or one
of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever
comes first, in which case that player wins the game. Suppose the players make their
moves at random, rather than trying to follow a rational strategy, and that Carl wins
the game when he places his third O. How many ways can the board look after the
game is over?
Video Solution
Example 3.7 (AIME)
Four ambassadors and one advisor for each of them are to be seated at a round table with
12 chairs numbered in order 1 to 12. Each ambassador must sit in an even-numbered
chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are
N ways for the 8 people to be seated at the table under these conditions. Find the
remainder when N is divided by 1000.
Solution 3.3
We will break this into cases based on where the ambassadors are.
Case 1: They are in adjacent even numbered chairs
In this case, there are 6 configurations possible by rotating the table.
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Chapter 3. Casework
How many ways are there to place the advisors?
There are 5 ways as shown below. Blue represents the ambassadors and red represents
the advisors.
In addition, there are 4! ways to arrange the ambassadors amongst each other so we get a
total of 4! × 6 × 5 = 720 for this case.
Case 2: 3 are in adjacent even numbered chairs and the other is separated
In this case, there are also 6 possible configurations by rotating.
How many ways are there to place the advisors?
We can find the number of advisor placements for the 3 ambassadors sitting adjacent
and for the 1 ambassador sitting away separately and multiply.
How many ways are there to place the advisors around the 3 adjacent ambassadors?
By similar logic to earlier, there are 3 placements of the advisors. For the 1 ambassador
sitting alone, there is simply 2 choices for the advisor. In total, we have 6 possibilities for
where the advisor can sit.
Like earlier, we multiply by 4! for ambassador arrangements. This gives 4! × 6 × 6 = 864.
Case 3: 2 groups of 2 adjacent ambassadors
This time, there are only 3 rotational orientations due to the symmetry between the 2
groups of ambassadors.
How many ways are there to place the advisors around the ambassadors?
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Chapter 3. Casework
For each group of 2, there are 3 ways to place the advisors giving us a total of 3 × 3 = 9.
For this case, we have 4! × 3 × 9 = 648.
In total, we have 720 + 648 + 864 = 2 520 .
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3.2
Chapter 3. Casework
Conditional Probability
Concept 3.2.1
If a problem asks you the probability an event A happens given that another event B
has already happened, the probability formula can be adjusted to
Probability =
Number of Ways where A and B happen
Number of Ways where B happens
.
Sometimes, it may be necessary to fully deal with probabilities.
Probability =
Probability A and B happen
Probability B happens
Example 3.8
What is the probability that the sum of two dice is 7 given that at least one of the dice
rolled a 6?
Solution 3.4
This is a standard conditional probability problem.
How many ways are there for the sum of two dice to be 7 and at least one
of the dice rolled to be a 6?
There are 2 ways: 61, 16. Any higher dice rolls along with 6 will exceed 7 and thus
not work.
How many ways with 2 dice are there such that at least one of them is 6?
If there is only one 6, then there are 5 choices for the other dice roll and 2 ways to order
the 2 dice rolls giving 5 × 2 = 10. If there are 2 6’s, there is just 1 way. This gives a total of 11.
Therefore, the probability is
2
.
11
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Chapter 3. Casework
Example 3.9 (HMMT Modified)
Victor has a drawer with two red socks, two green socks, two blue socks, and 4 orange
socks. He repeatedly draws two socks at a time from the drawer at random, and stops if
the socks are of the same color. However, Victor is red-green colorblind, so he also stops
if he sees a red and green sock. What is the probability that Victor stops with two socks
of the same color? Assume Victor returns both socks to the drawer at each step.
Solution 3.5
Note that Victor keeps drawing socks until he picks up a matching pair or red-green pair.
However, Victor returns the socks back to the drawer after every step. Eventually, he will
draw a matching pair or red-green pair so the problem is equivalent to Given that Victor
draws a matching pair or red-green pair, find the probability the pair is matching.
How many ways are there to draw a matching pair?
There is 22 = 1 way for each of red, green, or blue socks but there are
for the orange socks. In total, 9 ways.
4
2
= 6 ways
How many ways are there to draw a matching or red-green pair?
We already found that there are 9 matching pairs. In addition, since there are 2 red
and 2 green socks, there are 2 × 2 = 4 red-green pairs for a total of 13.
Therefore, our answer is
9
13
Example 3.10 (AMC 10)
In a particular game, each of 4 players rolls a standard 6-sided die. The winner is the
player who rolls the highest number. If there is a tie for the highest roll, those involved
in the tie will roll again and this process will continue until one player wins. Hugo is
one of the players in this game. What is the probability that Hugo’s first roll was a 5,
given that he won the game?
Solution 3.6
We can try to find the number of ways Hugo rolls a 5 and wins and the number of ways Hugo
wins as usual.
What are the complications with this?
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Chapter 3. Casework
For example, one possible outcome is that the 4 players roll 5, 4, 2, 1. Another possible outcome is that the 4 players roll 4, 4, 3, 2 and the 2 players who rolled a 4 roll again
and get 6, 3. Note that these outcomes do not have the same probability of happening!
That’s why in this question it’s better to deal directly with probabilities and we can find
Probability Hugo Rolls a 5 and Wins
Probability Hugo Wins
. We’ll start by finding the probability Hugo rolls a 5 and wins.
What are the possible cases in which this can happen?
Hugo can roll strictly higher than everyone else or tie with 1 or more people. Case 1:
Hugo rolls strictly higher than everyone
Firstly, the probability Hugo rolls a 5 is 16 . Next, the probability everyone else rolls a
64
.
4 or less is 64 . In total, we have ( 16 ) × ( 46 )3 = 1296
Case 2: Hugo ties with 1 other person
Again, the probability Hugo rolls a 5 is 16 . There are 3 ways to select the person Hugo ties
with and there is a 16 probability they roll a 5. Also, there is a 46 probability the other 2
48
people roll a 4 or lower. In total, we have 3 × ( 16 )2 × ( 46 )2 = 1296
.
What is the probability Hugo will win the tiebreaks?
24
This is just 12 by symmetry. In total, the probability Hugo ties with someone and wins is 1296
.
Case 3: Hugo ties with 2 other people
The probability Hugo rolls a 5 is 16 . There are 32 ways to select the 2 people Hugo
ties with and there is a 16 probability they each roll
a 5. Also, there is a 46 probability the
12
other person rolls a 4 or lower. In total, we have 32 × ( 16 )3 × ( 64 ) = 1296
.
What is the probability Hugo will win the tiebreaks?
4
This is 13 by symmetry. In total, the probability Hugo ties with 2 other people and wins is 1296
.
Case 4: Hugo ties with 3 other people
1
The probability they all roll 5’s is ( 16 )4 = 1296
and there is a 14 chance Hugo wins the
1
tiebreaks giving us 4×1296 .
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Chapter 3. Casework
Now, what is the probability Hugo wins (ignoring any other constraints)?
This is 14 by symmetry! Therefore, our answer is
64
24
4
1
+ 1296
+ 1296
+ 4×1296
1296
1
4
3.3
=
256 + 96 + 16 + 1
369
41
=
=
1296
1296
144
Practice Problems
Problem 3.3.1 (AMC 10)
Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
Video Solution
Problem 3.3.2 (AMC 10)
Two tour guides are leading six tourists. The guides decide to split up. Each tourist
must choose one of the guides, but with the stipulation that each guide must take at
least one tourist. How many different groupings of guides and tourists are possible?
Video Solution
Problem 3.3.3 (AMC 10)
Three fair six-sided dice are rolled. What is the probability that the values shown
on two of the dice sum to the value shown on the remaining die?
Video Solution
Problem 3.3.4 (AMC 10)
How many rearrangements of abcd are there in which no two adjacent letters are also
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Chapter 3. Casework
adjacent letters in the alphabet? For example, no such rearrangements could include
either ab or ba.
Video Solution
Problem 3.3.5 (AMC 10)
Call a positive integer an uphill integer if every digit is strictly greater than the previous
digit. For example, 1357, 89, and 5 are all uphill integers, but 32, 1240, and 466 are not.
How many uphill integers are divisible by 15?
Video Solution
Problem 3.3.6 (AMC 10)
For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5, and
6, on each die are in the ratio 1 : 2 : 3 : 4 : 5 : 6. What is the probability of rolling a total
of 7 on the two dice?
Video Solution
Problem 3.3.7 (AMC 10)
Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric.
How many ways are there for the five of them to sit in a row of chairs under these
conditions?
Video Solution
Problem 3.3.8 (AMC 10)
Three young brother-sister pairs from different families need to take a trip in a van.
These six children will occupy the second and third rows in the van, each of which has
three seats. To avoid disruptions, siblings may not sit right next to each other in the
same row, and no child may sit directly in front of his or her sibling. How many seating
arrangements are possible for this trip?
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Chapter 3. Casework
Video Solution
Problem 3.3.9 (AMC 12)
How many three-digit numbers are composed of three distinct digits such that one
digit is the average of the other two?
Video Solution
Problem 3.3.10 (AMC 10)
There are 10 people standing equally spaced around a circle. Each person knows
exactly 3 of the other 9 people: the 2 people standing next to her or him, as well as the
person directly across the circle. How many ways are there for the 10 people to split up
into 5 pairs so that the members of each pair know each other?
Video Solution
Problem 3.3.11 (AIME)
The nine delegates to the Economic Cooperation Conference include 2 officials from
Mexico, 3 officials from Canada, and 4 officials from the United States. During the
opening session, three of the delegates fall asleep. Assuming that the three sleepers were
determined randomly, the probability that exactly two of the sleepers are from the same
country is m
, where m and n are relatively prime positive integers. Find m + n.
n
Video Solution
Problem 3.3.12 (AMC 10/12)
Arjun and Beth play a game in which they take turns removing one brick or two adjacent
bricks from one ”wall” among a set of several walls of bricks, with gaps possibly creating
new walls. The walls are one brick tall. For example, a set of walls of sizes 4 and 2 can be
changed into any of the following by one move: (3, 2), (2, 1, 2), (4), (4, 1), (2, 2), or(1, 1, 2).
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Chapter 3. Casework
Arjun plays first, and the player who removes the last brick wins. For which starting configuration is their a strategy that guarantees a win for Beth.
Video Solution
Problem 3.3.13 (AMC 10)
Let D(n) denote the number of ways of writing the positive integer n as a product
n = f1 · f2 · · · fk ,
where k ≥ 1, the fi are integers strictly greater than 1, and the order in which the factors
are listed matters (that is, two representations that differ only in the order of the factors
are counted as distinct). For example, the number 6 can be written as 6, 2 · 3, and 3 · 2,
so D(6) = 3. What is D(96)?
Video Solution
Problem 3.3.14 (AMC 10)
Consider systems of three linear equations with unknowns x, y, and z,
a1 x + b 1 y + c 1 z = 0
a2 x + b 2 y + c 2 z = 0
a3 x + b 3 y + c 3 z = 0
where each of the coefficients is either 0 or 1 and the system has a solution other than
x = y = z = 0. For example, one such system is {1x + 1y + 0z = 0, 0x + 1y + 1z =
0, 0x + 0y + 0z = 0} with a nonzero solution of {x, y, z} = {1, −1, 1}. How many such
systems are there? (The equations in a system need not be distinct, and two systems
containing the same equations in a different order are considered different.)
(A) 302
(B) 338
(C) 340
(D) 343
(E) 344
Video Solution
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Chapter 3. Casework
Problem 3.3.15 (AMC 10/12)
Each square in a 5 × 5 grid is either filled or empty, and has up to eight adjacent
neighboring squares, where neighboring squares share either a side or a corner. The grid
is transformed by the following rules: Any filled square with two or three filled neighbors
remains filled. Any empty square with exactly three filled neighbors becomes a filled
square. All other squares remain empty or become empty. A sample transformation is
shown in the figure below.
Initial
Transformed
Suppose the 5 × 5 grid has a border of empty squares surrounding a 3 × 3 subgrid.
How many initial configurations will lead to a transformed grid consisting of a single
filled square in the center after a single transformation? (Rotations and reflections of the
same configuration are considered different.)
?
?
?
?
?
?
?
?
?
Initial
Transformed
(A) 14 (B) 18 (C) 22 (D) 26 (E) 30
Video Solution
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Chapter 3. Casework
Additional Problems
Problem 3.3.16 (AIME)
Zou and Chou are practicing their 100-meter sprints by running 6 races against each
other. Zou wins the first race, and after that, the probability that one of them wins a
race is 23 if they won the previous race but only 13 if they lost the previous race. The
probability that Zou will win exactly 5 of the 6 races is m
, where m and n are relatively
n
prime positive integers. Find m + n.
Problem 3.3.17 (AIME)
How many different 4 × 4 arrays whose entries are all 1’s and -1’s have the property that
the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
Problem 3.3.18 (AMC 10)
In a particular game, each of 4 players rolls a standard 6-sided die. The winner is
the player who rolls the highest number. If there is a tie for the highest roll, those
involved in the tie will roll again and this process will continue until one player wins.
Hugo is one of the players in this game. What is the probability that Hugo’s first roll
was a 5, given that he won the game?
Problem 3.3.19 (AMC 10)
Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a
subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he
wins if and only if the sum of the numbers face up on the three dice is exactly 7. Jason
always plays to optimize his chances of winning. What is the probability that he chooses
to reroll exactly two of the dice?
Problem 3.3.20 (AMC 10)
Each of the 12 edges of a cube is labeled 0 or 1. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations
and/or reflections. For how many such labelings is the sum of the labels on the edges of
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Chapter 3. Casework
each of the 6 faces of the cube equal to 2?
Problem 3.3.21 (AMC 12)
Two different cubes of the same size are to be painted, with the color of each face
being chosen independently and at random to be either black or white. What is the
probability that after they are painted, the cubes can be rotated to be identical in
appearance?
Problem 3.3.22 (AMC 10/12)
A cube is constructed from 4 white unit cubes and 4 blue unit cubes. How many
different ways are there to construct the 2 × 2 × 2 cube using these smaller cubes? (Two
constructions are considered the same if one can be rotated to match the other.)
Answers
3.1 26
3.2 84
3.3 683
3.4 2220
3.5 84
3.6 148
3.7 520
2
3.8 11
9
3.9 13
41
3.10 144
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Chapter 3. Casework
35
3.3.1 128
3.3.2 62
5
3.3.3 24
3.3.4 2
3.3.5 6
8
3.3.6 63
3.3.7 28
3.3.8 96
3.3.9 112
3.3.10 13
3.3.11 139
3.3.12 (6, 2, 1)
3.3.13 112
3.3.14 338
3.3.15 22
3.3.17 90
41
3.3.18 144
7
3.3.19 36
3.3.20 20
147
3.3.21 1024
3.3.22 7
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Chapter 4
Complementary Counting
Video Lecture
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Chapter 4. Complementary Counting
Complementary counting is the problem solving technique of counting the opposite of
what we want and subtracting that from the total number of cases. The keyword “at least”
indicates that complementary counting may be helpful.
Remark 4.0.1
Often times, we have to use casework in conjuction with other techniques like complementary counting.
Example 4.1
How many subsets of the set {1, 2, 3, 4, 5, 6, 7, 8 , 9, 10} have at least 1 even number
and do not contain all of the elements?
Solution 4.1
How can we use complementary counting?
We can find the total number of subsets and subtract the subsets which have no even
numbers and contain all of the elements. Note that both having no even numbers and
containing all of the elements cannot happen, so we don’t have to worry about overlap
between the subsets that don’t work.
Case 1: Subset contains all of the elements
There is only 1 subset that contains all of the elements (the original set).
Case 2: Subset contains no even elements
If there are no even elements, then we simply have to pick some odd numbers from {1,
3, 5, 7, 9} for our subset.
How many ways are there to do this?
Remember that each element has 2 choices (to be in subset or not in subset), so the
number of ways to form our subset from just odd numbers is 25 = 32.
Therefore in total, 1 + 32 = 33 ways don’t work.
Next, how many total subsets are there?
Since there are 10 elements and each element has 2 choices, there are 210 = 1024 subsets.
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Therefore the number of valid subsets is 1024 − 33 = 991
Example 4.2 (Omega Learn)
There are 3 sections of a dart board. The outermost section is worth 2 point if hit. The
middle section is worth 5 points if hit. The innermost section is worth 10 points if hit.
The probability you hit inner section is 41 , the probability you hit the middle section is
3
, the probability you hit the outermost section is 28 , and the probability you miss the
8
dartboard entirely is 18 . You throw 3 darts at the dartboard. What is the probability
you get less than 20 points?
Video Solution
Example 4.3 (AIME)
A positive integer is called ascending if, in its decimal representation, there are at least
two digits and each digit is less than any digit to its right. How many ascending positive
integers are there?
Video Solution
Example 4.4 (AIME)
An integer is called snakelike if its decimal representation a1 a2 a3 · · · ak satisfies ai < ai+1
if i is odd and ai > ai+1 if i is even. How many snakelike integers between 1000 and
9999 have four distinct digits?
Solution 4.2
Let’s first begin by choosing the 4 distinct digits we will use.
How many ways are there to do this?
There are
10
4
ways to select 4 digits.
Given 4 digits, how many orderings of the digits are possible?
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Chapter 4. Complementary Counting
From the condition, the 2nd digit must larger than the 1st digit, the 3rd digit must be
smaller than the 2nd digit, and the 4th digit must be larger than the 3rd digit. Essentially,
the 2nd and 4th digits are larger than their neighboring digits.
Say we choose the 4 digits a, b, c, d where a < b < c < d.
How can we find the number of orderings?
There are many ways, but we will do casework on the location of d. d must be either
the 2nd or 4th digit. If d is the 4th digit, then the 2nd digit must be c as it has to be
larger than the first and third digits. Then, there are 2 ways to flip a and b amongst the
1st and 3rd digits giving a total of 2 ways for this case (acbd, bcad). If d is the 2nd digit,
then the 4th digit has to be b or c. If the 4th digit is c, then we have 2 ways to order a
and b as before ((adbc, bdac). If the 4th digit is b, then the 3rd digit must be a in order
for it to be smaller than the 4th digit making the 1st digit c. This
gives us 1 more way
(cdab). In total, we have 5 orderings. Multiplying the 5 orderings by 10
gives 1050 numbers.
4
Are we overcounting any 4 digit numbers that have 0 as a starting digit?
Yes! If 0 is chosen as one of our digits, there are 93 ways to choose the remaining 3 digits.
0 in this case is the smallest digit and therefore must be a. Therefore, the number of invalid orderings are the ones that have a as the first digit. Looking at the possible
orderings above, we
find 2 such orderings. Therefore, the number of overcounted numbers is 93 ×2 = 84×2 = 168.
Our answer is 1050 − 168 = 882
Example 4.5 (AMC 10)
Frieda the frog begins a sequence of hops on a 3 × 3 grid of squares, moving one square
on each hop and choosing at random the direction of each hop-up, down, left, or right.
She does not hop diagonally. When the direction of a hop would take Frieda off the grid,
she ”wraps around” and jumps to the opposite edge. For example if Frieda begins in the
center square and makes two hops ”up”, the first hop would place her in the top row
middle square, and the second hop would cause Frieda to jump to the opposite edge,
landing in the bottom row middle square. Suppose Frieda starts from the center square,
makes at most four hops at random, and stops hopping if she lands on a corner square.
What is the probability that she reaches a corner square on one of the four hops?
Video Solution
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4.1
Chapter 4. Complementary Counting
Complementary Counting with Symmetry
Example 4.6
You flip 10 fair coins. What is the probability of flipping more heads than tails?
Solution 4.3
One way to do this is with casework. However, this will require a lot of cases. Let’s try to
find a simpler solution with symmetry.
How can we use the symmetry between heads and tails?
Keep in mind that the probability of flipping heads is the same as the probability of flipping
tails. Therefore, heads and tails are symmetric and the probability of flipping more heads
than tails is the same as the probability of flipping more tails then heads.
Is the answer 12 then?
Unfortunately, no because it’s also possible that an equal number of heads and tails and
flipped.
How can we account for this in our symmetry argument?
Note that P (Total) = 1 = P (More Heads) + P (More Tails) + P (Equal Heads and Tails)
and since P (More Heads) = P (More Tails) we can rewrite our equation as
1 = 2P (More Heads) + P (Equal Heads and Tails)
. Once we find the value of P (Equal Heads and Tails), we can easily solve for P (More Heads)!
What is P (Equal Heads and Tails)?
There are 10
= 252 ways to flip 5 heads and 5 tails. The total number of possible
5
252
63
1
flips is 2 0 = 1024. Therefore, this probability is 1024
= 256
.
From this, what is P (More Heads)?
We can plug this into our equation
1 = 2P (More Heads) +
63
256
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Chapter 4. Complementary Counting
and solving we get P (More Heads) =
63
(1− 256
)
=
2
193
512
Example 4.7 (AMC 10/12)
Bernardo randomly picks 3 distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and
arranges them in descending order to form a 3-digit number. Silvia randomly picks 3
distinct numbers from the set {1, 2, 3, 4, 5, 6, 7, 8} and also arranges them in descending
order to form a 3-digit number. What is the probability that Bernardo’s number is
larger than Silvia’s number?
Solution 4.4
In this problem, casework would be extremely tedious. We will try to do a symmetry approach
as we did last time. However, Bernado can pick from the numbers 1 to 9 while Silvia can
only pick from the numbers 1 to 8.
How can we deal with the case one of Bernado’s numbers is a 9 separately?
Bernado selects 3 numbers from 1 to 9 so the probability that 9 is one of them is 39 = 31 .
What happens if Bernado selects a 9?
Bernado’s number must be larger as his number will have 9 at the hundreds digit. Regardless of Silvia’s largest number, her number can have a hundreds digit of at most 8 so will
always be smaller. Therefore, there is a 31 chance Bernado automatically wins.
In the 23 chance Bernado does not pick a 9, is there symmetry?
Yes! In the case where Bernado does not pick a 9, he is picking from the digits 1 to 8
so the situation is analagous to Silvia’s. Like in the previous problem, we write the equation
1 = P (Bernado number larger) + P (Silvia number larger) + P (Numbers Same
. Note that P (Bernado number larger) = P (Silvia number larger) so we get
1 = 2P (Bernado number larger) + P (Numbers Same)
so it suffices to find the probability the numbers are the same.
What is this probability?
They both have
8
3
= 56 options for their 3 digits. For every choice of 3 digits by Bernado,
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there is exactly 1 possible choice by Silvia that will result in the same number. Therefore,
1
.
this probability is 56
We can now solve to get P (Bernado number larger) =
1
1− 56
2
55
= 112
.
55
In total, amongst both cases, our probability is 31 + 23 × 112
= 111
=
168
37
.
56
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4.2
Chapter 4. Complementary Counting
Practice Problems
Problem 4.2.1 (AMC 10/12)
How many subsets of {2, 3, 4, 5, 6, 7, 8, 9} contain at least one prime number?
Video Solution
Problem 4.2.2 (AMC 10)
How many four-digit positive integers have at least one digit that is a 2 or a 3?
Video Solution
Problem 4.2.3 (AMC 10)
A radio program has a quiz consisting of 3 multiple-choice questions, each with 3
choices. A contestant wins if he or she gets 2 or more of the questions right. The
contestant answers randomly to each question. What is the probability of winning?
Video Solution
Problem 4.2.4 (AMC 10)
Let (a, b, c, d) be an ordered quadruple of not necessarily distinct integers, each one
of them in the set 0, 1, 2, 3. For how many such quadruples is it true that a · d − b · c is
odd? (For example, (0, 3, 1, 1) is one such quadruple, because 0 · 1 − 3 · 1 = −3 is odd.)
Video Solution
Problem 4.2.5 (AMC 10/12)
How many strings of length 5 formed from the digits 0,1,2,3,4 are there such that
for each j ∈ {1, 2, 3, 4}, at least j of the digits are less than j? (For example, 02214
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Chapter 4. Complementary Counting
satisfies the condition because it contains at least 1 digit less than 1, at least 2 digits less
than 2, at least 3 digits less than 3, and at least 4 digits less than 4. The string 23404
does not satisfy the condition because it does not contain at least 2 digits less than 2.)
(A) 500
(B) 625
(C) 1089
(D) 1199
(E) 1296
Video Solution
Additional Problems
Problem 4.2.6
A strange coin lands heads 13 of the time, tails 13 of the time, and on its side 13 of
the time. What’s the probability that after 3 flips, more heads than tails were flipped?
Problem 4.2.7 (AIME)
Let N be the number of ordered pairs of nonempty sets A and B that have the following
properties:
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, A ∩ B = ∅, The number of elements of A is
not an element of A, The number of elements of B is not an element of B. Find N .
Problem 4.2.8 (AIME)
An integer is called snakelike if its decimal representation a1 a2 a3 · · · ak satisfies ai < ai+1
if i is odd and ai > ai+1 if i is even. How many snakelike integers between 1000 and 9999
have four distinct digits?
Problem 4.2.9 (AIME)
Find the number of permutations of 1, 2, 3, 4, 5, 6 such that for each k with 1 ≤ k
≤ 5, at least one of the first k terms of the permutation is greater than k.
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Answers
4.1 991
4.2 189/256
4.3 902
4.4 882
4.5 25
32
4.6 193
512
4.7 37
56
4.2.1 240
4.2.2 5416
7
4.2.3 27
4.2.4 96
4.2.5 1296
4.2.8 882
4.2.9 461
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Chapter 5
Principle of Inclusion and Exclusion
Video Lecture
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Chapter 5. Principle of Inclusion and Exclusion
Definition 5.0.1. The principle of inclusion and exclusion (PIE) is a counting technique
that computes the number of elements that satisfy at least one of several properties while
guaranteeing that elements satisfying more than one property are not counted twice.
Definition 5.0.2 (Union Symbol). |A ∪ B| is the union of elements in both A and B
(duplicates are only written once)
Definition 5.0.3 (Intersection Symbol). |A ∩ B| is the intersection of elements in both A
and B (only those elements which are in both sets)
Theorem 5.0.4 (Principle of Inclusion Exclusion for 2 Sets)
Given two sets, |A1 | and |A2 |
|A1 ∪ A2 | = |A1 | + |A2 | − |A1 ∩ A2 |
Basically, we count the number of possibilities in 2 ”things” and subtract the duplicates.
Example 5.1
Find how many numbers from 1 to 100 (inclusive) that are divisible by 2 or 7.
Solution 5.1
To solve this problem, let’s first find how many numbers are divisible by 2 and how many
numbers are divisible by 7
How many numbers from 1 to 100 are divisible by 2?
Every 2nd number is divisible by 2, so all the numbers from 2 × 1 to 2 × 50 work. Therefore,
from 1 to 100 there are 50 numbers divisible by 2.
How many numbers from 1 to 100 are divisible by 7?
Every 7th number is divisible by 7, so all of the numbers from 7 × 1 to 7 × 14 work.
We calculate this by seeing that 100
is a little more than 14 so 7 × 14 is the largest multiple
7
of 7 less than 100. Therefore, there from 1 to 100 are 14 numbers divisible by 7.
Are there any numbers overcounted between the multiples of 2 and the multiples
of 7?
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Chapter 5. Principle of Inclusion and Exclusion
For a number to be both a multiple of 2 and a multiple of 7, it must be a multiple of
7 × 2 = 14. Therefore, we are overcounting these numbers since we are counting them twice
as part of the multiples of 2 and multiples of 7.
How many numbers from 1 to 100 are divisible by 14?
Every 14th number is divisible by 14, so all of the numbers from 14 × 1 to 14 × 7 work.
Therefore, from 1 to 100 there are 7 numbers divisible by 14.
Therefore, in total, the number of numbers that are multiple of 2 or 7 are the number
of multiples of 2 plus the number of multiples of 7 minus the multiples of 14.
This gives us an answer of 50 + 14 − 7 = 57 .
Example 5.2 (AIME)
Many states use a sequence of three letters followed by a sequence of three digits as their
standard license-plate pattern. Given that each three-letter three-digit arrangement is
equally likely, the probability that such a license plate will contain at least one palindrome
(a three-letter arrangement or a three-digit arrangement that reads the same left-to-right
m
as it does right-to-left) is , where m and n are relatively prime positive integers. Find
n
m + n.
Video Solution
Example 5.3 (Omega Learn)
How many subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} have at most 4 even numbers
and at most 2 elements that are multiples of 3?
This is another problem we will solve using the principle of inclusion and exclusion and
the subsets formula we learned earlier.
Solution 5.2
How we can apply the Principle of Inclusion and Exclusion to this problem?
Notice how there are 5 even numbers, and we have to find how many subsets have at
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Chapter 5. Principle of Inclusion and Exclusion
most 4 even numbers. Similarly, there are 3 multiples of 3, and we have to find how many
elements have at most 2 multiples of 3.
We can see that there are fewer cases that don’t satisfy the condition, so it is better to use
complementary counting. We will first count the number of subsets that have more than 4
even numbers and have more than 2 multiple of 3.
First, how many subsets have more than 4 even numbers?
Since there are 5 multiples of 2 from 1 to 10 (2, 4, 6, 8, 10), to have more than 4 multiples of
2 means having all of the multiples of 2 (only 1 way to choose this). From the remaining 5
numbers, there are 25 possible subsets. Therefore, in total, there are 1 × 25 = 25 subsets that
have more than 4 multiples of 2.
Next, how many subsets have more than 2 multiples of 3?
Since there are 3 multiples of 3 from 1 to 10 (3, 6, 9), to have more than 2 multiples
of 3 means having all of the multiples of 3 (only 1 ways to choose this). From the remaining
7 numbers, there are 27 possible subsets. By similar logic, in total, there will be 1 × 27 = 27
possible subsets.
How many subsets are we overcounting that have all of the even numbers
and multiples of 3 from 1 to 10?
For a subset to have all the multiples of 2 and 3, it must contain all the elements from 2, 3, 4,
6, 8, 9, 10 (only 1 way to choose this). From the remaining 3 numbers, there are 1 × 23 = 23
possible subsets.
Therefore, in total, the number of subsets that don’t satisfy our original condition is
27 + 25 − 23 = 128 + 32 − 8 = 152.
The total number of possible subsets of the original set of 10 numbers is 210 = 1024, so the
number of valid subsets is 1024 − 152 = 872
Remark 5.0.5
When the number of cases that don’t work is much less than the number of cases that
do work, it can often be helpful to use complementary counting along with PIE.
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Chapter 5. Principle of Inclusion and Exclusion
Example 5.4 (Omega Learn)
How many different ways are there to arrange the letters in the word LOLLIPOP if each
of the L’s must be next to an O?
Video Solution
Example 5.5 (AIME)
Let S be a set with six elements. Let P be the set of all subsets of S. Subsets A and
B of S, not necessarily distinct, are chosen independently and at random from P. The
probability that B is contained in one of A or S − A is nmr , where m, n, and r are positive
integers, n is prime, and m and n are relatively prime. Find m + n + r. (The set S − A
is the set of all elements of S which are not in A.)
Solution 5.3
Let’s first find the number of ways to choose subsets so that B is contained entirely in A.
How many ways are there to do this?
If we were just counting ways to choose a subset for A, there would be 26 ways as each
element could go in or not in A. Let’s try to do something similar in this case.
As seen in the picture, there are 3 regions where the elements can end up: in A and B, in A
but not B, or not in A or B (equivalent to in S − A). Therefore, for each element, we have 3
choices so 36 = 729 ways.
How many ways are there choose subsets so that B is contained in S − A?
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Chapter 5. Principle of Inclusion and Exclusion
Like before, there are 3 regions: in S − A and B, in S − A but not B, not in S − A or B
(equivalent to in A). Therefore, there are again 36 = 729 ways.
How many cases are we overcounting?
For a case to be overcounted, B must be in both A and S − A. However, A and S − A are
disjoint (they share no elements in common) so the only way for this to happen is if B is
the empty set. Therefore, the number of overcounted cases is simply the number of ways to
choose A which is 26 = 64 as B is fixed.
By PIE, we have 2 × 729 − 64 = 1394 valid cases. In total, there are 26 ways to pick
each of A and B so the probability is 1394
= 697
giving an answer of 697 + 2 + 11 = 710 .
21 2
21 1
Theorem 5.0.6 (Principle of Inclusion Exclusion for 3 Sets)
Given three sets, |A1 |, |A2 |, |A3 |,
|A1 ∪ A2 ∪ A3 | = |A1 | + |A2 | + |A3 | − |A1 ∩ A2 | − |A1 ∩ A3 | − |A2 ∩ A3 | + |A1 ∩ A2 ∩ A3 |
In this formula, we count the number of possibilities in 3 ”things”, subtract the possibilities
that are duplicates in all 3 pairs of sets, and add back the number of duplicates in all 3
sets.
Example 5.6
How many numbers less than or equal to 193 are multiples of 2, 3, or 5?
Solution 5.4
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Chapter 5. Principle of Inclusion and Exclusion
First, let’s find how many numbers are divisible by 2, how many numbers are divisible by 3,
and how many numbers are divisible by 5.
Every 2nd number is divisible by 2, so all the numbers from 2 × 1 to 2 × 96 work. Therefore,
there are 96 numbers divisible by 2 less than or equal to 193
Every 3rd number is divisible by 3, so all the numbers from 3 × 1 to 3 × 64 work. Therefore,
there are 64 numbers divisible by 3 less than or equal to 193
Every 5th number is divisible by 5, so all the numbers from 5 × 1 to 5 × 38 work. Therefore,
there are 38 numbers divisible by 5 less than or equal to 193.
Next, we must subtract numbers divisible by two of the numbers 2, 3, or 5.
For this to happen, what values must the numbers be divisible by?
Numbers divisible by 2 and 3 or numbers divisible by 2 × 3 = 6 are counted twice. Similarly,
numbers divisible by 2 and 5 or 2 × 5 = 10 are counted twice and numbers divisible by 3 and
5 or 3 × 5 = 15 are counted twice. Therefore, we are overcounting multiples of each of 6, 10,
and 15, so we must subtract them.
Every 6th number is divisible by 6, so all the numbers from 6 × 1 to 6 × 32 work. Therefore,
there are 32 numbers divisible by 6 less than or equal to 193
Every 10th number is divisible by 10, so all the numbers from 10 × 1 to 10 × 19 work.
Therefore, there are 19 numbers divisible by 10 less than or equal to 193
Every 15th number is divisible by 5, so all the numbers from 15 × 1 to 15 × 12 work. Therefore,
there are 12 numbers divisible by 15 less than or equal to 193.
If we subtract the overcounted numbers above, did we remove too much?
Next, we look at how many times multiples of 2, 3, and 5, or multiples of 2 × 3 × 5 = 60
were counted. Well, they were originally counted 3 times as part of the multiples of 2, 3, and
5. Then, they were subtracted 3 times as part of the multiples of 6, 10, and 15. Therefore,
they have been counted 0 times! But, we need to count them exactly once, so we must add
back the number of multiples of 2 × 3 × 5 = 60.
Every 60th number is divisible by 60, so all the numbers from 60 × 1 to 60 × 3 work.
Therefore, there are 3 numbers divisible by 60 less than or equal to 193
In total, to find the number of multiples of 2, 3, or 5 we have to add the number of
multiples of 2, the number of multiples of 3, the number of multiples of 5. Then, we have
to subtract the number of multiples of 6, 10, and 15. And finally, we have to add back the
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Chapter 5. Principle of Inclusion and Exclusion
number of multiples of 60.
This gives us an answer of 96 + 64 + 38 − 32 − 19 − 12 + 3 = 138 .
Example 5.7 (AMC 8)
Five different awards are to be given to three students. Each student will receive at least
one award. In how many different ways can the awards be distributed?
Video Solution
Example 5.8 (Modified AIME)
Call a set S product-free if there do not exist a, b, c ∈ S (not necessarily distinct) such
that ab = c. For example, the empty set and the set {16, 20} are product-free, whereas
the sets {4, 16} and {2, 8, 16} are not product-free. Find the number of product-free
subsets of the set {1, 2, 3, 4, ..., 7, 8, 9}.
Video Solution
Theorem 5.0.7 (Principle of Inclusion Exclusion Generalized)
Stated more formally, if (Ai )1≤i≤n are finite sets, then:
n
[
i=1
Ai =
n
X
i=1
|Ai | −
X
|Ai ∩ Aj | +
i<j
X
|Ai ∩ Aj ∩ Ak | − · · · + (−1)n−1 |A1 ∩ · · · ∩ An |
i<j<k
.
Example 5.9 (AIME)
Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For
each square, either color is equally likely to be used. The probability of obtaining a grid
that does not have a red 2 × 2 square is m
, where m and n are relatively prime positive
n
integers. Find m + n.
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Chapter 5. Principle of Inclusion and Exclusion
Solution 5.5
Since we are asked to find the probability the grid does not have red 2 × 2 square, it’s a good
idea to use complementary counting and find how many colorings have at least 1 2x2 red
2 × 2 square.
Amongst all choices of a red 2x2 square, how many colorings exist?
There are four 2 × 2 squares as shown above. Then, there are 25 ways to color the rest of the
squares red or blue giving a total of 4 × 32 = 128. However, we are overcounting possibilities
when there are 2 red 2x2 squares.
Amongst all choices of 2 red 2x2 squares, how many colorings exist?
There are 4 cases as shown above. In the 4 cases on the top, there are 3 remaining squares
giving 23 possibilities. In the 2 cases on the right, there are 2 remaining squares giving
22 possibilities. In total, we have 4 × 23 + 2 × 22 = 40. However, we are undercounting
possibilities when there are 3 red 2x2 squares.
Amongst all choices of 3 red 2 × 2 squares, how many colorings exist?
In each of the 4 cases shown above, there is 1 remaining square which has 2 choices so in
total, there are 4 × 21 = 8 ways in this case. We are overcounting possibilities when there are
4 red 2x2 squares.
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Chapter 5. Principle of Inclusion and Exclusion
There is simply 1 way to color the grid in this case.
Summing everything up, we get 128 − 40 + 8 − 1 = 95 configurations with a red 2 × 2
square. By complementary counting, there are 29 − 95 = 417 configurations without such a
square giving a probability of 417
so the answer is 417 + 512 = 929 .
512
5.1
Practice Problems
Problem 5.1.1 (AIME)
Many states use a sequence of three letters followed by a sequence of three digits
as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one
palindrome (a three-letter arrangement or a three-digit arrangement that reads the same
left to right as it does right to left) is m
, where m and n are relatively prime positive
n
integers. Find m + n.
Video Solution
Problem 5.1.2 (AMC 10)
Seven distinct pieces of candy are to be distributed among three bags. The red bag and
the blue bag must each receive at least one piece of candy; the white bag may remain
empty. How many arrangements are possible?
Video Solution
Problem 5.1.3 (AMC 10)
In how many ways can the sequence 1, 2, 3, 4, 5 be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
Video Solution
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Chapter 5. Principle of Inclusion and Exclusion
Problem 5.1.4 (AMC 10)
In how many ways can the sequence 1, 2, 3, 4, 5 be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
Video Solution
Additional Problems
Problem 5.1.5 (AIME)
Find the number of positive integers that are divisors of at least one of 1010 , 157 , 1811 .
Problem 5.1.6 (AIME)
In a five-team tournament, each team plays one game with every other team. Each team
m
has a 50% chance of winning any game it plays. (There are no ties.) Let
be the
n
probability that the tournament will produce neither an undefeated team nor a winless
team, where m and n are relatively prime integers. Find m + n.
Problem 5.1.7 (AIME)
Call a set S product-free if there do not exist a, b, c ∈ S (not necessarily distinct) such
that ab = c. For example, the empty set and the set {16, 20} are product-free, whereas
the sets {4, 16} and {2, 8, 16} are not product-free. Find the number of product-free
subsets of the set {1, 2, 3, 4, ..., 7, 8, 9, 10}.
Problem 5.1.8 (AIME)
Define a regular n-pointed star to be the union of n line segments P1 P2 , P2 P3 , . . . , Pn P1
such that
1. the points P1 , P2 , . . . , Pn are coplanar and no three of them are collinear,
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Chapter 5. Principle of Inclusion and Exclusion
2. each of the n line segments intersects at least one of the other line segments at a
point other than an endpoint, all of the angles at P1 , P2 , . . . , Pn are congruent,
3. all of the n line segments P1 P2 , P2 P3 , . . . , Pn P1 are congruent, and the path
P1 P2 , P2 P3 , . . . , Pn P1 turns counterclockwise at an angle of less than 180 degrees at
each vertex.
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars
are similar, but there are two non-similar regular 7-pointed stars. How many non-similar
regular 1000-pointed stars are there?
Problem 5.1.9 (AIME)
Find the number of permutations of 1, 2, 3, 4, 5, 6 such that for each k with 1 ≤ k
≤ 5, at least one of the first k terms of the permutation is greater than k.
Problem 5.1.10 (AIME)
While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order
in a row of six chairs. During the break, they went to the kitchen for a snack. When
they came back, they sat on those six chairs in such a way that if two of them sat next
to each other before the break, then they did not sit next to each other after the break.
Find the number of possible seating orders they could have chosen after the break.
Answers
5.1 57
5.2 059
5.3 872
5.4 108
5.5 710
5.6 138
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Chapter 5. Principle of Inclusion and Exclusion
5.7 150
5.8 252
5.9 929
5.1.1 059
5.1.2 1932
5.1.3 32
5.1.4 32
5.1.5 435
5.1.6 49
5.1.7 252
5.1.9 461
5.1.10 90
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Chapter 6
Stars & Bars
Video Lecture
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6.1
Chapter 6. Stars & Bars
Stars and Bars Fundamentals
Example 6.1
How many ways are there to distribute 8 identical computers amongst 3 students?
Solution 6.1
Make sure to keep in mind that the computers are identical. If the computers were not
identical, then we would just have 3 choices for each computer, so our answer would just be 38 .
However, when the computers are identical, only the number of computers each student has
matters. To start, let’s say we have 8 computers in a line: CCCCCCCC. Therefore, we must
divide these computers amongst 3 students.
How can we represent the 8 computers being divided amongst the 3 students?
We can think about having 3 groups. We can simply let the leftmost group be for student one, the middle group for student 2, and the rightmost group for student 3 because only
the number of computers each student has matters, so ordering is irrelevant.
How can we divide the 8 C’s amongst 3 groups?
We can do this by just placing 2 bars somewhere in between the 8 C’s! Everything left of the
first bar goes to student 1, everything in the middle goes to student 2, and everything to
the right of the 2nd bar goes to student 3. Therefore, the number of ways to distribute the
computers is simply the number of ways to insert 2 bars in between 8 C’s. Examples:
CC|CCCCC|C - 2 for student 1, 5 for student 2, 1 for student 3
|CCCC|CCCC - 0 for student 1, 4 for student 2, 4 for student 3
C|CCCCCCC| - 1 for student 1, 7 for student 2, 0 for student 3
CCCCCCCC|| - 8 for student 1, 0 for student 2, 0 for student 3
How do we find the number of ways to insert 2 bars in between 8 C’s?
We can think of having 10 slots (each slot for either a bar or a C). Then, we can choose 2 of
these slots for bars.
Therefore, the numbers of ways to select two slots out of the 10 slots (i.e. choose two
slots to put the two bars out of the 10 slots) is
10
10 × 9
=
= 45
2
2
!
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Chapter 6. Stars & Bars
Theorem 6.1.1 (Stars and Bars)
The number of ways to place n identical objects into k distinguishable bins is
n+k−1
n
!
Remark 6.1.2
Stars and Bars is very useful, and can often be adapted based on situations. For example,
if each bin has to have at least 1 object in it we assign 1 object to each bin to start off,
and apply our formula to the remaining n − k objects and k distinguishable bins.
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6.2
Chapter 6. Stars & Bars
Stars and Bars With Constraints
Concept 6.2.1
Stars and bars is extremely useful, and can often be adapted based on situations. For
example, if each bin has to have at least 1 object in it we assign each bin 1 object to
start off with and apply our formula with n − k objects and k distinguishable bins.
Example 6.2
7 astronauts are stranded in space with only 10 identical meals left. Every astronaut
must receive at least 1 meal or else, they will starve. How many ways are there to
distribute meals under this constraint?
Solution 6.2
This problem is very similar to the previous problem, except we now have an additional
constraint, so the formula won’t work. Let’s focus on the constraint, since it’s the most
complicated part of the problem.
How can we deal with the condition that every astronaut must get at least 1 meal?
If every astronaut must receive at least 1 meal and all the meals are identical, we can
simply pre-distribute 7 of the meals and give 1 to each astronaut. Notice that now, no matter
where the remaining meals are distributed, every astronaut will get 1 meal or more.
Now, how can we use stars on bars on the remaining problem?
After 7 meals are distributed, only 3 remain. There is no constraint on who these 3 meals
can go to, so it can go to any of the 7 astronauts. To divide these 3 meals into 7 groups, we
can place 6 bars in between the 3 meals. Then, out of the 3+6=9 slots for meals and bars,
we must choose 6 of them for bars.
We can do this in
9
6
=
9
3
= 9×8×7
= 9×8×7
= 84 .
3!
6
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Chapter 6. Stars & Bars
Example 6.3 (Modified AMC 10)
For some particular value of N , when (a + b + c + d + 1)N is expanded and like terms
are combined, the resulting expression contains exactly 1001 terms that include all four
variables a, b, c, and d, each to some positive power. What is N ?
Video Solution
Example 6.4
What’s the probability that the sum of the top faces on five regular 6-sided dice is 24?
Solution 6.3
Before we begin, note that each dice must be a value from 1 to 6. There is no way to use
stars and bars directly because it will not account for the constraint that the maximum of
each dice roll is a 6.
How can we use stars and bars for this problem?
Let the default for each dice be a 6. Then the maximum score is 30, we must essentially distribute 6 negatives to the 5 dice (-1, -1, -1, -1, -1, -1).
We can see that to distribute 6 negatives to 5 dice, we must have 4 bars to separate the
negatives into 5 groups for the dice. We then have 9 slots and must choose 4 of them for
bars. We can do this in 94 = 126 ways.
Are there any cases we are overcounting?
Because the minimum value of a dice roll is 1, each dice can only take at most 5 negatives. Therefore, we are overcounting the possibilities where 1 dice receives 6 negatives.
How many ways are there for 1 dice to have 6 negatives?
There are 6 dice, so there are 6 choice for which dice will have all 6 negatives. From
here, we can calculate our final
answer by subtracting the 6 ways that result in 1 dice
receiving all 6 negatives from 94 , the number of ways to distribute 6 negatives to 5 dice so
that no dice gets more than 5 negatives.
This gives us
9
5
− 6 = 120
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Chapter 6. Stars & Bars
Example 6.5
How many ways are there to choose positive integers a, b, c, and d such that a+b+c+d <
13
Video Solution
6.3
Stars and Bars but Stars are groups
Example 6.6 (Omega Learn)
You have 11 apples and 5 bananas. You place them into 3 baskets. How many ways are
there to do this if each basket must have more apples than bananas and at least 1 of
any fruit?
Video Solution
Example 6.7 (Omega Learn Math Contest)
Three squirrels have to distribute 19 identical acorns amongst themselves.
• The first squirrel demands to have a positive odd number of acorns.
• The second squirrel is fine with any non-negative even number of acorns.
• The third squirrel insists he must get an even number of acorns greater than or
equal to 4.
How many different ways can they distribute the acorns?
Video Solution
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6.4
Chapter 6. Stars & Bars
Stars and Bars to Count Orderings
Example 6.8
How many ways are there for 7 people to sit in a line if 3 of them are enemies and refuse
to sit next to each other?
We will worry about the ordering about the 3 enemies and remaining 4 people later. For
now, let’s find the number of possible ways to choose 3 locations for the 7 enemies.
Without the condition, how many ways are there to do this?
There are 7 possible locations and the enemies must sit at 3 of them so
10
3
.
What must happen so that the enemies don’t sit next to each other?
There must be at least 1 other person between each of the enemies.
Let w be the number of people left of the 1st enemy, let x be the number of people between
the 1st and 2nd enemies, let y be the number of people between the 2nd and 3rd enemies,
and let z be the number of people right of the 3rd enemy.
What conditions must w, x, y, z satisfy?
Firstly, since there are 4 people who are not enemies, we must have w + x + y + z = 4.
However, by our observation earlier, x and y must be at least 1.
How can we use stars and bars to find the number of solutions to this equation?
We start by assigning 1 star to x and y. Then, there are
5
3
solutions by stars and bars.
How many orderings of the enemies and non-enemies are possible?
There are 3! ways to order the enemies and 4! ways to order the other people so our
answer is 53 × 3! × 4! = 1440
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Chapter 6. Stars & Bars
Example 6.9 (AIME)
Ed has five identical green marbles, and a large supply of identical red marbles. He
arranges the green marbles and some of the red ones in a row and finds that the number
of marbles whose right hand neighbor is the same color as themselves is equal to the
number of marbles whose right hand neighbor is the other color. An example of such
an arrangement is GGRRRGGRG. Let m be the maximum number of red marbles for
which such an arrangement is possible, and let N be the number of ways he can arrange
the m + 5 marbles to satisfy the requirement. Find the remainder when N is divided by
1000.
Solution 6.4
At most how many marbles can have a different colored right neighbor?
From the 5 red marbles, for the right marble to be a different color, it must be green.
Therefore, at most 5 red marbles can have a different colored right neighbor. For the m green
marbles, for the right marble to be a different color it must be red and since there are 5 red marbles, at most 5 green marbles can have a different colored right neighbor. In total, we have 10.
From this, what is the maximum value of m, the total number of green marbles?
If there are an equal number of different colored and same colored right neighbors, there are
at most a total of 20 right neighbors. This means there can be at most 21 total marbles or
16 green marbles.
What must happen so the condition is satisfied with 5 red and 16 green marbles?
As discovered earlier, each red marble must be surrounded by a green marble on both
the right and left sides in order for each red marble to have a green marble to it’s right and
each green marble to have a red marble to it’s right.
How many ways are there to do this?
This is similar to the previous problem. We are dividing the 16 green marbles into 6
groups. However, in order for red marbles to have green marble neighbors on both sides,
there must be at least 1 green marble in each group. After assigning 1 green marble to each
group,
we have 10 green marbles left to distribute amongst 6 groups. We can do this in
15
=
2
003 ways.
5
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Chapter 6. Stars & Bars
Example 6.10 (AIME)
There are two distinguishable flagpoles, and there are 19 flags, of which 10 are identical
blue flags, and 9 are identical green flags. Let N be the number of distinguishable
arrangements using all of the flags in which each flagpole has at least one flag and no two
green flags on either pole are adjacent. Find the remainder when N is divided by 1000.
Solution 6.5
We will do complementary counting and first find the total number of flag configurations and
then subtract off the possibilities where all the flags are on 1 pole. We will first find how
many cases must be subtracted off.
How many ways are possible to put the 19 flags on 1 flagpole?
There are 9 green flags and we much distribute the 10 blue flags such that there is at
least 1 blue flag between each green flag. We assign 8 blue flags to place in between the green
flags and distribute
the remaining 2 blue flags amongst the 10 possible locations by stars and
11
bars giving 2 . We will multiply this by 2 as there are 2 possible flagpoles where all of the
flags can be on.
For the 2 flagpole case, how can we try and combine the flagpoles to make
it into a scenario similar to the previous one?
Imagine taking all the flags from 1st flagpole and all of the flags from the 2nd flagpole
and placing them on a third flagpole. Between the 2 flagpoles, place a red flag to distinguish
the flags between the 2 poles.
If there are no consecutive green flags on the original 2 flag poles, what must be
true about the green flags on the 3rd flag pole?
Like earlier, no 2 green flags can be consecutive. However, we can also separate the green flags
with a red flag! As seen in the diagram above, this corresponds to when there is a green flag
on bottom of the first flagpole and a green flag on the top of the 2nd flagpole, which is allowed!
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Chapter 6. Stars & Bars
Using the same stars and bars argument, how many ways are there to place the
blue and red flags in between the 9 green flags?
There are 10 regions around the green flags to place them in and 11 flags to place in
these groups. The middle 8 groups require a flag so we have
to distribute the remaining 3
flags amongst the 10 groups. By stars and bars, this is 12
. However, out of all the flags
3
that are either red or blue, we must choose 1 of them to be blue.
How many such choices are there?
In total, we have 11 ×
12
3
−2×
11
2
= 2420 − 110 = 2 310
newline
6.5
Practice Problems
Problem 6.5.1 (AMC)
Alice has 24 apples. In how many ways can she share them with Becky and Chris
so that each of the three people has at least two apples?
Video Solution
Problem 6.5.2 (AMC 10)
When 7 fair standard 6-sided dice are thrown, the probability that the sum of the
numbers on the top faces is 10 can be written as
n
,
67
where n is a positive integer. What is n?
Video Solution
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Chapter 6. Stars & Bars
Problem 6.5.3 (AMC 10)
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and
peanut butter cookies. There are at least six of each of these three kinds of cookies on
the tray. How many different assortments of six cookies can be selected?
Video Solution
Problem 6.5.4 (AMC 10)
Let D(n) denote the number of ways of writing the positive integer n as a product
n = f1 · f2 · · · fk ,
where k ≥ 1, the fi are integers strictly greater than 1, and the order in which the factors
are listed matters (that is, two representations that differ only in the order of the factors
are counted as distinct). For example, the number 6 can be written as 6, 2 · 3, and 3 · 2,
so D(6) = 3. What is D(96)?
Video Solution
Additional Problems
Problem 6.5.5 (Omega Learn)
Find the number of solutions to a + b + c + d < 19 where a, b, c, d are positive
integers, a and c must be odd, and b and d must be even.
Problem 6.5.6
How many terms will be in the expansion of (w + x + y + z)10 after simplifying?
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Chapter 6. Stars & Bars
Problem 6.5.7 (AMC 12)
The expression
(x + y + z)2006 + (x − y − z)2006
is simplified by expanding it and combining like terms. How many terms are in the
simplified expression?
Problem 6.5.8 (AIME)
A fair die is rolled four times. The probability that each of the final three rolls is
at least as large as the roll preceding it may be expressed in the form m
where m and n
n
are relatively prime positive integers. Find m + n.
Problem 6.5.9
In how many ways can 8 students and 3 teachers sit in the front row if between every
pair of teachers, there must be at least 2 students?
Problem 6.5.10 (AIME)
A fair die is rolled four times. The probability that each of the final three rolls is
at least as large as the roll preceding it may be expressed in the form m
where m and n
n
are relatively prime positive integers. Find m + n.
Problem 6.5.11
The wheel shown below consists of two circles and five spokes, with a label at each
point where a spoke meets a circle. A bug walks along the wheel, starting at point A. At
every step of the process, the bug walks from one labeled point to an adjacent labeled
point. Along the inner circle the bug only walks in a counterclockwise direction, and
along the outer circle the bug only walks in a clockwise direction. For example, the bug
could travel along the path AJABCHCHIJA, which has 10 steps. Let n be the number
of paths with 15 steps that begin and end at point A. Find the remainder when n is
divided by 1000.
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Chapter 6. Stars & Bars
J
A
I
F
B
E
C
H
D
G
Answers
6.1 45
6.2 84
6.3 14
6.4 120
6.5 495
6.6 210
6.7 36
6.9 003
6.10 310
6.5.1 190
6.5.2 84
6.5.3 28
6.5.4 112
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Chapter 6. Stars & Bars
6.5.5 210
6.5.6 286
6.5.7 1, 008, 016
6.5.8 79
6.5.9 35 × 11!
6.5.10 79
6.5.11 4
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Chapter 7
Combinatorial Identities
Video Lecture
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7.1
Chapter 7. Combinatorial Identities
Basic Identities
Theorem
7.1.1
n
k
=
n
n−k
n
Proof. nk represents the number of ways to choose k objects from n objects whereas n−k
represents the number of ways to not choose n − k objects from n objects. Therefore, the 2
quantities must be equal!
Theorem 7.1.2 (Binomial Identity)
The binomial identity states that
!
!
!
n
n
n
+
+ ··· +
= 2n
0
1
n
Proof. We will again use double counting.
How many ways are there to select some subset of n objects?
For each object, we have 2 choices: choose it, don’t choose it. Therefore, the total number of
ways is 2n .
How many ways are there to select 0 objects? 1 object? 2 objects? n objects?
There are
n
0
2 objects, . . . ,
ways to select 0 objects,
n
n
n
1
ways to select 1 object,
n
2
ways to select
to select n objects.
By summing this up, we get that the number of ways to choose any number of objects
is
!
!
!
n
n
n
+
+ ··· +
0
1
n
which must be equal to 2n !
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Chapter 7. Combinatorial Identities
Theorem 7.1.3 (Choosing Odd Even Identity)
This identity states
m
X
n
n−1
= (−1)m
k
m
!
(−1)k
k=0
!
Proof. Suppose we have n students of which we must select some of them to go on the field trip.
One of the students Bob is sick and can’t go. Out of all possible subsets of the remaining
n − 1 students, say a of the subsets have an even number of people and b of them have an
odd number of people.
Later, Bob recovers and can is now going on the field trip. In terms of a
and b, how many subsets of all n people have an odd number of people and an
even number of people?
All of the a subsets that originally had an even number of people now have an odd number
of people. All of the b subsets that originally had an odd number of people now have an even
number of people.
In total, amongst all subsets of n people in terms of a and b, how many have an
odd number of people and how many have an even number of people?
By adding up the quantities from the previous 2 scenarios, we get a + b subsets have
an odd number of people and b + a subsets have an even number of people. They are the
same!
What is this statement in terms of choose symbols?
The total number of subsets with an odd number of people is
n
+ n3 + . . . and the
1
total number of groups with an even number of people is n0 + n2 + . . . . These quantities
must be equal giving us our desired identity.
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7.2
Chapter 7. Combinatorial Identities
Vandermonde’s Identity
Theorem 7.2.1 (Vandermonde’s Identity)
Vandermonde’s Identity states that
n
0
!
!
!
m
n
+
r
1
!
m
n
+ ··· +
r−1
r
!
m
m+n
=
0
r
!
!
.
Proof. Imagine you have m oranges and n bananas.
How many ways are there to select any r fruits?
There are
m+n
r
ways.
How many ways are there to select r oranges?
From the m oranges, there are
m
r
to select r of them.
How many ways are there to select r − 1 oranges and 1 banana?
There are
m
r−1
tal, there are
ways to select the oranges and
m
r−1
n
1
n
1
ways to select the banana. In to-
ways.
Using this logic, can you find another formula for the number of ways to select r
fruits?
We can have r oranges and 0 bananas, r − 1 oranges and 1 banana, r − 2 oranges and
2 bananas, . . . , 0 orange and r bananas. Using the same logic as above, we get
n
0
. However,
be equal!
m+n
r
!
!
m
n
+
1
r
!
!
m
n
+ ··· +
r−1
r
!
!
m
0
also counts the number of ways to choose r fruits so these quantities must
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Remark 7.2.2
If you understand the logic behind this proof, applying Vandermonde’s will become
intuitive and you will not have to memorize this identity.
Theorem 7.2.3 (Special Case of Vandermonde’s Identity)
k
X
i=0
k
i
!2
2k
=
k
!
Example 7.1 (AIME)
A club consisting of 11 men and 12 women needs to choose a committee from among its
members so that the number of women on the committee is one more than the number
of men on the committee. The committee could have as few as 1 member or as many as
23 members. Let N be the number of such committees that can be formed. If N = ab ,
find a + b.
Video Solution
Example 7.2 (AIME)
Find the remainder when
!
3
2
2
!
+
4
2
2
!
40
+ ··· +
2
2
is divided by 1000.
Video Solution
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7.3
Chapter 7. Combinatorial Identities
Pascal’s and Hockey Stick Identity
Concept 7.3.1 (Pascal’s Triangle)
Pascal’s Triangle is a triangular array of binomial coefficients and contains numerous
patterns that can be used to make complex calculations much easier.
Pascal’s triangle is pictured below. It can be represented in terms of combinations,
which is depicted in the image below.
Theorem 7.3.2 (Pascal’s Identity)
Pascal’s identity states that
n
n
n+1
+
=
r
r+1
r+1
!
!
!
Proof. Imagine you have n + 1 people and you must choose k + 1 leaders.
How many ways are there to do this?
This is just
n+1
r+1
.
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If Bob, one of the n + 1 people, is a leader, how many ways are there to choose
the other r leaders from the remaining n people?
This is
n
r
.
If Bob is not a leader, how many ways are there to select the r + 1 leaders
from the remaining n people?
This is
n
r+1
.
Now, we have counted the number of ways to select r + 1 people in 2 different ways!
Equating, we get
!
!
!
n
n
n+1
+
=
r
r+1
r+1
.
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Chapter 7. Combinatorial Identities
Theorem 7.3.3 (Hockey Stick Identity)
Hockey Stick Identity states
k
k+1
n
n+1
+
+ ··· +
=
k
k
k
k+1
!
!
!
!
Proof.
Again, like earlier, we will count some quantity in 2 ways to prove this.
How many ways to select k + 1 leaders amongst n + 1 people?
This is
n+1
k+1
.
Now, assume the n + 1 people are ranked in order of skill level from 1 to n + 1.
If the lowest ranked leader has a rank of k + 1, how many ways to choose
the other leaders?
There are k people who have a higher ranking so
k
k
.
If the lowest ranked leader has a rank of k + 2, how many ways to choose
the other leaders?
There are k + 1 people who have a higher ranking so
k+1
k
.
We can continue with this all the way until the end.
If the lowest ranked leader has a rank of n + 1, how many ways to choose
the other leaders?
There are n + 1 people who have a higher ranking so
n
k
.
In total, we have that there are
k
k+1
n
+
+ ··· +
k
k
k
!
!
!
ways to select k + 1 people
by doing casework on the lowest ranked leader. We also found
out that there are n+1
ways
to select the leaders so they must be equal!
k+1
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Chapter 7. Combinatorial Identities
Theorem 7.3.4 (Hockey Stick Identity Generalization)
j
j+1
n
n+1
j
+
+ ··· +
=
−
k
k
k
k+1
k+1
!
!
!
!
!
Remark 7.3.5
This is derived by simply applying hockey stick theorem twice and subtracting. See if
you can prove it!
Example 7.3 (AIME Modified)
Melinda has ten empty boxes and 75 textbooks, three of which are mathematics textbooks.
The first box holds 3 textbooks, the 2nd box holds 4 textbooks, the 3rd box holds 5
textbooks, all the way until the 10th box with holds 12 textbooks. If Melinda packs her
textbooks into these boxes in random order, the probability that all three mathematics
textbooks end up in the same box can be written as m
, where m and n are relatively
n
prime positive integers. Find m + n.
Solution 7.1
Among the 75 book slots, we must choose 3 of them to be math textbooks.
How many ways are there to do this?
This is
75
3
.
Next, let’s find the number of ways each box can hold the 3 math textbooks.
How many ways can the first box hold the math books? 2nd box? 10th box?
The first box has only
3
3
ways to hold the textbooks. The 2nd box has 4 slots for textbooks
so there 43 ways it can hold the textbooks. This continues all the way until the 10th box
which has 10
ways to hold the 3 math textbooks.
3
How can we add these up in a smart way?
We can use the hockey stick identity!
Therefore, the probability is
3
3
+
4
3
+ ··· +
10
3
=
11
4
.
11
330
66
4
=
=
75
67525
11525
3
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Chapter 7. Combinatorial Identities
giving us an answer of 66 + 11525 = 11591 .
Remark 7.3.6
Imagine how tedious that would be to add if we had to do it manually!
Example 7.4 (AIME)
Consider all 1000-element subsets of the set {1, 2, 3, ..., 2015}. From each such subset
choose the least element. The arithmetic mean of all of these least elements is pq , where
p and q are relatively prime positive integers. Find p + q.
Solution 7.2
To find the average value of the least elements over all subsets, let’s first find the sum of the
least elements across all subsets. To do this, we will find the number of subsets with each
element as it’s least element.
How many subsets have 1 as a least element?
We
choose 999 more elements from the 2014 remaining elements, which can be done in
must
2014
ways.
999
How many subsets have 2 as a least element? 3? 4? 1016?
For 2 to bethe least element, we must choose
999
elements from the 2013 remaining
2013
2012
elements so 999 . By similar logic, there are 999 ways 3 can be the least element. This
continues all the way until 1016 which has
999
999
it can be the least element.
Based on this, what is the total sum of least elements across all the sets?
We get
2014
2013
999
1×
+2×
+ · · · + 1016 ×
999
999
999
!
!
!
. This reminds us of the hockey stick identity. However, the coefficients make it so that we
can’t apply it directly.
How can we divide the sum so that we can apply the hockey stick identity?
By the Hockey Stick Identity,
2014
2013
999
2015
+
+ ··· +
=
999
999
999
1000
!
!
!
!
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Chapter 7. Combinatorial Identities
.
After subtracting this from our original sum, what are we left with?
We are left with
2013
999
1×
+ · · · + 1015 ×
999
999
!
!
. By Hockey Stick Identity again, we find
2013
2013
999
2014
+1×
+ ··· + 1 ×
=
999
999
999
1000
!
!
!
!
. We can again subtract this from our original sum.
As shown above, we can continue this logic! We end up with
by Hockey Stick Identity again.
There are
2016
=
1001
288
143
2015
1000
2015
1000
+ 2014
+· · ·+ 1000
=
1000
1000
possible subsets so to find the average value we divide
2016
1001
2016!
(2016
1001)
= 1001!×1015!
=
2015!
2015
(1000)
1000!×1015!
Remark 7.3.7
These identities can be helpful in combinatorics problem, but their applications may not
always be straightforward, so a good approach to many combinatorics problem may be
to just manipulate expressions.
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7.4
Chapter 7. Combinatorial Identities
Practice Problems
Problem 7.4.1
A choir director must select a group of singers from among his 6 tenors and 8 basses. The
only requirements are that the difference between the number of tenors and basses must
be a multiple of 4, and the group must have at least one singer. Let N be the number of
different groups that could be selected. What is the remainder when N is divided by
100?
Video Solution
Problem 7.4.2 (AMC 12)
A choir director must select a group of singers from among his 6 tenors and 8 basses. The
only requirements are that the difference between the number of tenors and basses must
be a multiple of 4, and the group must have at least one singer. Let N be the number of
groups that could be selected. What is the remainder when N is divided by 100?
Video Solution
Problem 7.4.3 (AMC 10/12)
Suppose that 13 cards numbered 1, 2, 3, · · · , 13 are arranged in a row. The task is
to pick them up in numerically increasing order, working repeatedly from left to right.
In the example below, cards 1, 2, 3 are picked up on the first pass, 4 and 5 on the second
pass, 6 on the third pass, 7, 8, 9, 10 on the fourth pass, and 11, 12, 13 on the fifth pass.
For how many of the 13! possible orderings of the cards will the 13 cards be picked up in
exactly two passes?
7
(A) 4082
11
8
(B) 4095
6
4
5
(C) 4096
9
12
1
(D) 8178
13
10
2
3
(E) 8191
Video Solution
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Additional Problems
Problem 7.4.4 (AIME)
There are two distinguishable flagpoles, and there are 19 flags, of which 10 are identical
blue flags, and 9 are identical green flags. Let N be the number of distinguishable
arrangements using all of the flags in which each flagpole has at least one flag and no
two green flags on either pole are adjacent. Find the remainder when N is divided by
1000. (Solve this problem using Vandermonde’s Identity)
Answers
7.1 35(original81)
7.2 4
7.3 11591
7.4 431
288
7.4 143
7.4.1 95
7.4.2 95
7.4.3 8178
7.4.4 310
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Chapter 8
Geometric Counting
Video Lecture
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8.1
Chapter 8. Geometric Counting
Triangle Counting
Example 8.1
How many triangles can be formed by connecting vertices in a regular octagon?
Solution 8.1
You could solve this by casework on what type of triangle you will select. However, there is a
very smart and clever solution.
How many vertices defines a triangle?
This seems like a silly question, obviously 3.
How many ways are there to select 3 vertices from an octagon?
Since an octagon has 8 vertices, this will just be
8
3
.
Do all combinations of 3 vertices form a triangle?
This is a very important question to ask because it’s not always true. For example, if
3 points are collinear (you can draw a line through all 3 of them) then the answer would
actually be no, and we would have to subtract the extra cases. Luckily, in an octagon, no 3
points are collinear, so all combinations of 3 vertices do indeed form a triangle.
Therefore, our answer is just
8
8×7×6
8×7×6
=
=
= 56
3
3!
6
!
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Chapter 8. Geometric Counting
Example 8.2 (AMC 10)
Eight points are chosen on a circle, and chords are drawn connecting every pair of points.
No three chords intersect in a single point inside the circle. How many triangles with all
three vertices in the interior of the circle are created?
Example 8.3 (AMC 10)
How many triangles with positive area have all their vertices at points (i, j) in the
coordinate plane, where i and j are integers between 1 and 5, inclusive?
8.2
Rectangle Counting
Example 8.4
How many rectangles are in a 3x5 grid of rectangles?
Solution 8.2
This is another problem that can be solved by tedious casework but has a very slick solution.
What defines a rectangle in a grid?
4 points? Not quite because 4 points don’t necessarily make a rectangle. 2 vertical lines and
2 horizontal lines define a rectangle as shown in the picture below. In other words, any 2
vertical lines and horizontal lines in this grid will form a rectangle.
How can we count the number of rectangles in a grid based on this?
We simply have to choose 2 vertical lines and 2 horizontal lines!
There are 4 horizontal lines, so there are
4
2
ways to select 2 horizontal lines that de-
fine a rectangle. Similarly, there are 6 vertical lines, so there are
lines that define a rectangle.
6
2
ways to select 2 vertical
Therefore, the number of rectangles is just
4
6
4×3 6×5
×
=
×
= 6 × 15 = 90
2
2
2
2
!
!
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Chapter 8. Geometric Counting
Theorem 8.2.1
The general formula for the number of rectangles of all sizes in a rectangular grid of size
m × n is
!
!
m+1
n+1
×
2
2
.
Remark 8.2.2
Each combination of two horizontal lines and two vertical lines creates a unique rectangle.
We have
!
m+1
2
ways to choose two horizontal lines and
n+1
2
!
ways to choose two vertical lines.
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8.3
Chapter 8. Geometric Counting
Square Counting
Example 8.5
How many squares of all sizes can be formed from a 4x4 grid of unit squares?
Solution 8.3
Keep in mind that the answer is not 16 because we can’t just count unit squares!
What’s the easiest and most systematic way to count all possible squares?
We can break the problem down into 1x1, 2x2, 3x3, and 4x4 squares.
Case 1: 1x1 squares
For this case, there are just 4 × 4 = 16 squares.
Case 2: 2x2 squares
Notice how there are 3 possible choices of rows and 3 possible choices of columns to form a
2x2 square. In total, there are 3 × 3 = 9 possible 2x2 squares since each choice of two rows
and two columns produces a 2x2 square.
Case 3: 3x3 squares
Notice how there are 2 possible choices of rows and 2 possible choices of columns to form a
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3x3 square. In total, there are 2 × 2 = 4 possible 3x3 squares since each choice of two rows
and two columns produces a 3x3 square.
Case 4: 4x4 squares
There is just one 4x4 square (the entire square).
In total, the number of squares of all sizes is
16 + 9 + 4 + 1 = 30
Example 8.6 (Omega Learn)
How many rectangles, which are not squares, are in a 5 × 4 grid of squares?
Video Solution
Example 8.7 (Omega Learn)
How many squares with side length
√
5 can be found in a 5 by 5 grid of points?
Video Solution
Example 8.8 (AMC 10)
The 5 × 5 grid shown contains a collection of squares with sizes from 1 × 1 to 5 × 5. How
many of these squares contain the black center square?
Video Solution
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Chapter 8. Geometric Counting
Example 8.9 (Modified AMC 10)
How many rectangles formed by the gridlines in this 5 × 5 grid contain the black center
square?
Video Solution
Example 8.10 (AMC 10)
As shown in the figure below, a regular dodecahedron (the polyhedron consisting of 12
congruent regular pentagonal faces) floats in space with two horizontal faces. Note that
there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted
faces adjacent to the bottom face. How many ways are there to move from the top face
to the bottom face via a sequence of adjacent faces so that each face is visited at most
once and moves are not permitted from the bottom ring to the top ring?
Video Solution
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8.4
Chapter 8. Geometric Counting
Path Counting
Example 8.11
A rabbit is walking in a coordinate plane and has to get from (0,0) to the carrot at the
point (5,5). If the rabbit can only move 1 unit up or 1 unit right on any given move,
how many ways are there for the rabbit to get to the carrot?
Solution 8.4
This is yet another problem that is tedious to solve with casework. Instead, let’s find a
simpler way to solve this problem. Lets denote a right move by R and an up move by U.
How many right and up moves does the rabbit need to get to the carrot?
Because it’s going from (0,0) to (5,5), it needs 5 right moves and 5 up moves. Therefore,
we just need to count the number of ways to make 5 right moves and 5 up moves in any order.
How can we represent the number of ways to make 5 right moves and 5 up
moves (in any order)?
This can be thought of as rearrangements of 5 R and 5 U: RRRRRUUUUU. We just
need to find all possible rearrangements of this word. Given the total number of characters
is 10, and counting the duplicates (5 R’s and 5 U’s), we can use the word rearrangement
formula:
10!
= 252
5! × 5!
Theorem 8.4.1
The number of ways to get from (0,0) to a point (m,n) moving only up and right is
m+n
m
!
Remark 8.4.2
Imagine any arrangement string of m R’s and n U’s. Each string would correspond to a
unique path, so to count this, we can use the word rearrangement formula.
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Example 8.12
How many ways to go from (0, 0) to (5, 6) moving only right or up if you cannot visit
all of (1, 2), (3, 4), and (5, 5) because there is a monster that will eat you if you visit all
3 locations?
Video Solution
Example 8.13 (BMMT)
Sally is inside a pen consisting of points (a, b) such that 0 ≤ a ≤ b ≤ 4. If she is currently
on the point (x, y), she can move to either (x, y + 1), (x, y − 1), or (x + 1, y). Given that
she cannot revisit any point she has visited before, find the number of ways she can
reach (4, 4) from (0, 0).
Video Solution
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Chapter 8. Geometric Counting
Example 8.14 (AMC 8)
A game board consists of 64 squares that alternate in color between black and white.
The figure below shows square P in the bottom row and square Q in the top row. A
marker is placed at P. A step consists of moving the marker onto one of the adjoining
white squares in the row above. How many 7-step paths are there from P to Q? (The
figure shows a sample path.)
Q
P
Video Solution
Example 8.15 (HMMT)
The cells of a 5 × 5 grid are each colored red, white, or blue. Sam starts at the bottom-left
cell of the grid and walks to the top-right cell by taking steps one cell either up or to
the right. Thus, he passes through 9 cells on his path, including the start and end cells.
Compute the number of colorings for which Sam is guaranteed to pass through a total of
exactly 3 red cells, exactly 3 white cells, and exactly 3 blue cells no matter which route
he takes.
Solution 8.5
How must the grid be colored so that the condition is satisfied?
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A possible diagram is shown below.
How can use the fact that every path goes through the same colors?
We can construct 2 paths that are nearly identical and perhaps find some squares that
must be the same color. Consider the 2 paths shown below.
They are the same except for 2 steps (green and pink). Both of those paths must pass through
3 blue, 3 red, and 3 white squares. However, because they pass through the same colored
squares for the 8 squares their paths share, squares A and B must be the same color.
Why does this imply each diagonal must be colored the same color throughout?
We can apply the same logic: constructing 2 nearly identical paths that differ by an adjacent
diagonal square for all diagonally adjacent squares. This gives us that all the squares in a
diagonal must be the same color.
Now, the problem is equivalent to choosing 3 of the 9 diagonals to be each of red, white, and
blue.
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How can we do this?
9!
= 1680
This is just the word rearrangement formula! This gives an answer of 3!×3!×3!
8.5
Practice Problems
Problem 8.5.1 (AMC 10)
A set of 25 square blocks is arranged into a 5×5 square. How many different combinations
of 3 blocks can be selected from that set so that no two are in the same row or column?
Video Solution
Problem 8.5.2 (AMC)
From a regular octagon, a triangle is formed by connecting three randomly chosen
vertices of the octagon. What is the probability that at least one of the sides of the
triangle is also a side of the octagon?
Video Solution
Problem 8.5.3 (AMC 10)
How many ways are there to place 3 indistinguishable red chips, 3 indistinguishable
blue chips, and 3 indistinguishable green chips in the squares of a 3 × 3 grid so that
no two chips of the same color are directly adjacent to each other, either vertically or
horizontally.
Video Solution
Problem 8.5.4 (AMC 12)
Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can
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be rotated so that their appearances are identical. How many distinguishable colorings
are possible?
Video Solution
Additional Problems
Problem 8.5.5 (AMC)
A game board consists of 64 squares that alternate in color between black and white.
The figure below shows square P in the bottom row and square Q in the top row. A
marker is placed at P. A step consists of moving the marker onto one of the adjoining
white squares in the row above. How many 7-step paths are there from P to Q? (The
figure shows a sample path.)
Q
P
Problem 8.5.6 (AIME)
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A
large equilateral triangle is constructed from four of these paper triangles. Two large
triangles are considered distinguishable if it is not possible to place one on the other, using
translations, rotations, and/or reflections, so that their corresponding small triangles are
of the same color.
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Given that there are six different colors of triangles from which to choose, how many
distinguishable large equilateral triangles may be formed?
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Problem 8.5.7 (AIME)
Let m and n be odd integers greater than 1. An m × n rectangle is made up of unit
squares where the squares in the top row are numbered left to right with the integers 1
through n, those in the second row are numbered left to right with the integers n + 1
through 2n, and so on. Square 200 is in the top row, and square 2000 is in the bottom
row. Find the number of ordered pairs (m, n) of odd integers greater than 1 with the
property that, in the m × n rectangle, the line through the centers of squares 200 and
2000 intersects the interior of square 1099.
Answers
8.1 56
8.4 90
8.5 30
8.6 110
8.7 8
8.8 19
8.9 81
8.10 810
8.11 252
8.12 408
8.13 625
8.14 28
8.15 1680
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8.5.1 600
8.5.2 57
8.5.3 36
8.5.4 15
8.5.6 336
8.5.7 248
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Chapter 9
Geometric Probability
Video Lecture
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Chapter 9. Geometric Probability
Definition 9.0.1. Geometric probability is a way to calculate probability by measuring the
number of outcomes geometrically, in terms of length, area, or volume. The key to solving
geometric probability is
1. Try a few examples for the different cases, make sure to always mark the extreme cases
2. Try to figure out the region the shape maps out
3. Use geometry to find the area of this region
Remark 9.0.2
Geometric probability can be useful when the number of possible outcomes is infinite.
Example 9.1
Alex and Bob arrive at a random time from 3 PM to 4 PM. They will each wait for 15
minutes before leaving. What is the probability that they will meet?
Solution 9.1
What must happen for the 2 people to meet?
Alex and Bob’s arrival time must be within 15 minutes of each other. In other words,
Alex can not arrive more than 15 minutes after Bob and Bob can not arrive more than 15
minutes after Alex.
If Alex arrives at 3, when can Bob arrive?
Alex will wait 15 minutes so Bob must arrive by 3:15.
If Alex arrives at 3:15, when can Bob arrive?
In order to be within 15 minutes of Alex’s time, Bob must arrive between 3 and 3:30.
If Alex arrives at 3:30, when can Bob arrive? 3:45? 4?
If Alex arrives at 3:30, Bob must arrive from 3:15 to 3:45. If Alex arrives at 3:45, Bob
must arrive from 3:30 to 4. If Alex arrives at 4, Bob must arrive from 3:45 to 4.
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Using this, we draw a picture. The x-axis represents Alex’s arrival time and the y-axis
represents Bob’s arrival time. Black rectangles are shown for the 4 cases discussed and try to
find the area of the region where they meet.
How can we generalize this for all possible arrival times for Alex?
The green line denotes the points where Bob arrives at the same time as Alex. As discussed, Bob can arrive up to 15 minutes before or after.
We then construct the blue region as shown showing all possible valid regions.
If the side length of the square is 1, what is the area of the blue region?
To do this, we will subtract off the areas of the 2 red triangles. Each of them has an
7
9
9
area of 12 × 34 × 34 = 32
giving a total area of 16
. Therefore, the area of the blue region is
16
and since the area of the square is 1 (represents total probability), this is also the probability
they will meet.
Example 9.2 (AMC 10)
Chloé chooses a real number uniformly at random from the interval [0, 2017]. Independently, Laurent chooses a real number uniformly at random from the interval [0, 4034].
What is the probability that Laurent’s number is greater than Chloé’s number? (Assume
they cannot be equal)
Video Solution
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Chapter 9. Geometric Probability
Example 9.3 (AMC 10)
Real numbers x and y are chosen independently and uniformly at random from the
interval [0, 1]. Which of the following numbers is closest to the probability that x, y, and
1 are the side lengths of an obtuse triangle?
(A) 0.21
(B) 0.25
(C) 0.29
(D) 0.50
(E) 0.79
Video Solution
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Chapter 9. Geometric Probability
Example 9.4 (AMC 10)
A square with side length 8 is colored white except for 4 black isosceles right triangular
regions with
√ legs of length 2 in each corner of the square and a black diamond with side
length 2 2 in the center of the square, as shown in the diagram. A circular coin with
diameter 1 is dropped onto the square and lands in a random location where the coin is
completely contained within the square. The probability
will cover part of
that
√ the coin
1
the black region of the square can be written as 196 a + b 2 + π , where a and b are
positive integers. What is a + b?
Video Solution
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Chapter 9. Geometric Probability
Example 9.5 (AMC 10/12)
Let S be a square of side length 1. Two points are chosen independently at random on
the sides of S. The probability that the straight-line distance between the points is at
1
a − bπ
least is
, where a, b, and c are positive integers with gcd(a, b, c) = 1. What is
2
c
a + b + c?
Solution 9.2
We break this into 3 cases:
Case 1: 2 points on the same side
There is 14 chance that the 2nd point is on the same side as the 1st point, so this case
has a 14 chance of happening. Let x and y be the distances of those 2 points from a vertex.
The difference between x and y must be bigger than 12 .
How can we find the probability this happens?
We can use our standard geometric probability argument.
The area of the blue region shown represents the values of x and y that work. It has
an area of 2× 12 ×( 12 )2 = 14 which means the probability our condition is satisfied in this case is 14 .
Case 2: 2 points on adjacent sides
Out of 4 possible sides, 2 sides for the 2nd point result in it being on an adjacent side
to the 1st point giving us a probability of 12 that this case happens.
Let x and y be the distances between the 2 points to a common vertex.
What must be true about x and y?
By the pythagorean theorem, we must have
√ 2
x + y 2 ≥ 12 or x2 + y 2 ≥ 41 .
How can we find the area of points that satisfy this region?
This is the equation for a circle! Therefore, all points above the quarter circle shown
π
will work as shown. The quarter circle’s area is 14 × ( 12 )2 π = 16
. The area of the blue region
that works (and thus the probability the points are more than 12 away from each in other) is
π
1 − 16
in this case.
Case 3: 2 points on opposite sides
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Chapter 9. Geometric Probability
Out of 4 possible sides for the 2nd point, only 1 will result in it being on an opposite
side. Therefore, the probability this case happens is 14 .
Why are the points always more than 12 away in this case?
As shown, the distance between the 2 points will always be greater than 1 and thus 12
by the pythagorean theorem.
π
In total, we have that the probability is 14 × 14 + 12 × (1 − 16
+ 14 × 1 = 26−π
giving an
32
answer of 26 + 32 + 1 = 59 .
9.1
Geometric Probability in 3D
Example 9.6 (AMC 10/12)
Ant Amelia starts on the number line at 0 and crawls in the following manner. For
n = 1, 2, 3, Amelia chooses a time duration tn and an increment xn independently and
uniformly at random from the interval (0, 1). During the nth step of the process, Amelia
moves xn units in the positive direction, using up tn minutes. If the total elapsed time
has exceeded 1 minute during the nth step, she stops at the end of that step; otherwise,
she continues with the next step, taking at most 3 steps in all. What is the probability
that Amelia’s position when she stops will be greater than 1?
Video Solution
9.2
Practice Problems
Problem 9.2.1
Alex and Bob decide to meet at the park between 3 and 4 pm. They both will arrive at a random time during the hour but will only wait up to 15 minutes for the other
person and then leave the park. What is the probability that they will meet?
Video Solution
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Chapter 9. Geometric Probability
Problem 9.2.2 (AMC)
From a regular octagon, a triangle is formed by connecting three randomly chosen
vertices of the octagon. What is the probability that at least one of the sides of the
triangle is also a side of the octagon?
Video Solution
Problem 9.2.3 (AMC 10)
Real numbers between 0 and 1, inclusive, are chosen in the following manner. A
fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0
if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first
coin flip is tails, then the number is chosen uniformly at random from the closed interval
[0, 1]. Two random numbers x and y are chosen independently in this manner. What is
the probability that |x − y| > 12 ?
Video Solution
Additional Problems
Problem 9.2.4 (AIME)
Two mathematicians take a morning coffee break each day. They arrive at the cafeteria
independently, at random times between 9 a.m. and 10 a.m., and stay for exactly m
minutes. The probability
that either one arrives while the other is in the cafeteria is
√
40%, and m = a − b c, where a, b, and c are positive integers, and c is not divisible by
the square of any prime. Find a + b + c.
Problem 9.2.5 (AIME)
A circle of radius 1 is randomly placed in a 15-by-36 rectangle ABCD so that the
circle lies completely within the rectangle. Given that the probability that the circle will
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Chapter 9. Geometric Probability
not touch diagonal AC is m/n, where m and n are relatively prime positive integers.
Find m + n.
Answers
7
9.1 16
9.2 34
9.3 π4 − 12 = 0.29
9.4 68
9.5 59
7
9.2.1 16
9.2.2 57
7
9.2.3 12
9.2.5 817
131
Chapter 10
Expected Value
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Chapter 10. Expected Value
Definition 10.0.1. Expected value is the weighted average of outcomes.
Theorem 10.0.2 (Expected Value)
The expected value of some event X is
X
xi · P (xi )
where xi are the possible values of X and P (xi ) is the probability they occur.
Basically the expected value is just the weighted sum of probabilities of events
Remark 10.0.3
Often times, in finding the expected value, we can just look for symmetry instead
of summing each individual probability times number. For example, to calculate the
expected value of a dice roll rather than evaluating
1
1
1
1
1
1
· 1 + · 2 + · 3 + · 4 + · 5 + · 6 = 3.5
6
6
6
6
6
6
we can see that since all rolls from 1 to 6 are equally likely, the expected value is just
the average roll which is just the average of the 2 middle terms which is 3.5. (See the
arithmetic sequences section)
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Chapter 10. Expected Value
Theorem 10.0.4 (Linearity of Expectation)
For independent or dependent events,
E(x1 + x2 + · · · + xn ) = Ex1 + Ex2 + · · · + Exn
Basically, what this means is that the total expected value of n events is just the sum of
the expected values of each individual event.
Remark 10.0.5
This theorem is powerful as it allows us to find the expected value of the individual
events rather than of the whole thing at once.
Example 10.1 (AMC 10)
Five balls are arranged around a circle. Chris chooses two adjacent balls at random
and interchanges them. Then Silva does the same, with her choice of adjacent balls to
interchange being independent of Chris’s. What is the expected number of balls that
occupy their original positions after these two successive transpositions?
Video Solution
Example 10.2 (AMC 12)
What is the average number of pairs of consecutive integers in a randomly selected
subset of 5 distinct integers chosen from the set {1, 2, 3, · · · , 30}? (For example the set
{1, 17, 18, 19, 30} has 2 pairs of consecutive integers.)
Video Solution
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10.1
Chapter 10. Expected Value
Practice Problems
Problem 10.1.1
Mark is playing a game at the casino. The game costs $100. There is a 5% chance he
will win $500 and a 20% chance he will win $250 and 75% chance he will win $0. What
is the expected value for the money he will lose?
Video Solution
Problem 10.1.2 (AMC 10)
A player pays $5 to play a game. A die is rolled. If the number on the die is odd,
the game is lost. If the number on the die is even, the die is rolled again. In this case
the player wins if the second number matches the first and loses otherwise. How much
should the player win if the game is fair? (In a fair game the probability of winning times
the amount won is what the player should pay.)
Video Solution
Additional Problems
Problem 10.1.3 (AMC 12)
A school has 100 students and 5 teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are
50, 20, 20, 5, and 5. Let t be the average value obtained if a teacher is picked at random
and the number of students in their class is noted. Let s be the average value obtained if
a student was picked at random and the number of students in their class, including the
student, is noted. What is t − s?
Problem 10.1.4 (AIME)
For each permutation a1 , a2 , a3 , · · · , a10 of the integers 1, 2, 3, · · · , 10, form the sum
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Chapter 10. Expected Value
|a1 − a2 | + |a3 − a4 | + |a5 − a6 | + |a7 − a8 | + |a9 − a10 |.
p
The average value of all such sums can be written in the form , where p and q are
q
relatively prime positive integers. Find p + q.
Answers
10.1 2.2
10.2 23
10.1.1 25
10.1.2 60
10.1.3 −13.5
10.1.4 058
136
Chapter 11
Recursion
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Chapter 11. Recursion
Concept 11.0.1 (Recursion)
Recursion is the process solving the problem for small values and writing a recurrence
equation to iteratively calculate the values for larger values.
Steps to Solve Recursion Problems:
1. Base Cases: Manually find the values for small values of n
2. Recursion Equation: Look at the different cases for any general value of n (ex.
whether the last digit is 0 or 1)
• If you are stuck, you can try a few small cases and look for a pattern
3. Iteratively calculate higher values of n until you reach your answer
Remark 11.0.2
Note that you can get the answer to many recursion problems by using engineering
induction (see the meta-solving section).
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Chapter 11. Recursion
Example 11.1 (AMC 10/12)
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a
0, contain no two consecutive 0s, and contain no three consecutive 1s?
Video Solution
Example 11.2 (AIME)
A collection of 8 cubes consists of one cube with edge-length k for each integer k, 1 ≤
k ≤ 8. A tower is to be built using all 8 cubes according to the rules:
• Any cube may be the bottom cube in the tower.
• The cube immediately on top of a cube with edge-length k must have edge-length
at most k + 2.
Let T be the number of different towers than can be constructed. What is the
remainder when T is divided by 1000?
Video Solution
Example 11.3 (AMC 12)
Call a set of integers spacy if it contains no more than one out of any three consecutive
integers. How many subsets of {1, 2, 3, . . . , 12}, including the empty set, are spacy?
Video Solution
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11.1
Chapter 11. Recursion
Practice Problems
Problem 11.1.1 (AMC)
Everyday at school, Jo climbs a flight of 6 stairs. Jo can take the stairs 1, 2, or 3
at a time. For example, Jo could climb 3, then 1, then 2. In how many ways can Jo
climb the stairs?
Video Solution
Problem 11.1.2
Arjun and Beth play a game in which they take turns removing one brick or two
adjacent bricks from one ”wall” among a set of several walls of bricks, with gaps possibly
creating new walls. The walls are one brick tall. For example, a set of walls of sizes 4 and
2 can be changed into any of the following by one move: (3, 2), (2, 1, 2), (4), (4, 1), (2, 2),
or (1, 1, 2).
,
,
,...
Arjun plays first, and the player who removes the last brick wins. For which starting
configuration is there a strategy that guarantees a win for Beth?
(A) (6, 1, 1)
(B) (6, 2, 1)
(C) (6, 2, 2)
(D) (6, 3, 1)
(E) (6, 3, 2)
Video Solution
Additional Problems
Problem 11.1.3 (AMC 10)
Square ABCD in the coordinate plane has vertices at the points A(1, 1), B(−1, 1), C(−1, −1),
and D(1, −1). Consider the following four transformations:
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Chapter 11. Recursion
•
L, a rotation of 90◦ counterclockwise around the origin;
•
R, a rotation of 90◦ clockwise around the origin;
•
H, a reflection across the x-axis; and
•
V, a reflection across the y-axis.
Each of these transformations maps the squares onto itself, but the positions of the
labeled vertices will change. For example, applying R and then V would send the vertex
A at (1, 1) to (−1, −1) and would send the vertex B at (−1, 1) to itself. How many
sequences of 20 transformations chosen from {L, R, H, V } will send all of the labeled
vertices back to their original positions? (For example, R, R, V, H is one sequence of 4
transformations that will send the vertices back to their original positions.)
Problem 11.1.4 (AIME)
There are 210 = 1024 possible 10-letter strings in which each letter is either an A
or a B. Find the number of such strings that do not have more than 3 adjacent letters
that are identical.
Problem 11.1.5 (AIME)
The figure below shows a ring made of six small sections which you are to paint on a wall.
You have four paint colors available and you will paint each of the six sections a solid
color. Find the number of ways you can choose to paint the sections if no two adjacent
sections can be painted with the same color.
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Chapter 11. Recursion
Problem 11.1.6 (AIME)
Call a permutation a1 , a2 , . . . , an of the integers 1, 2, . . . , n quasi-increasing if ak ≤ ak+1 +2
for each 1 ≤ k ≤ n − 1. For example, 53421 and 14253 are quasi-increasing permutations of the integers 1, 2, 3, 4, 5, but 45123 is not. Find the number of quasi-increasing
permutations of the integers 1, 2, . . . , 7.
Problem 11.1.7 (AMC 12)
Let △A0 B0 C0 be a triangle whose angle measures are exactly 59.999◦ , 60◦ , and 60.001◦ .
For each positive integer n, define An to be the foot of the altitude from An−1 to line
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Chapter 11. Recursion
Bn−1 Cn−1 . Likewise, define Bn to be the foot of the altitude from Bn−1 to line An−1 Cn−1 ,
and Cn to be the foot of the altitude from Cn−1 to line An−1 Bn−1 . What is the least
positive integer n for which △An Bn Cn is obtuse?
Problem 11.1.8 (AIME)
A mail carrier delivers mail to the nineteen houses on the east side of Elm Street.
The carrier notices that no two adjacent houses ever get mail on the same day, but that
there are never more than two houses in a row that get no mail on the same day. How
many different patterns of mail delivery are possible?
Problem 11.1.9 (AIME)
Find the number of permutations of 1, 2, 3, 4, 5, 6 such that for each k with 1 ≤ k
≤ 5, at least one of the first k terms of the permutation is greater than k.
Answers
11.1 65
11.2 458
11.3 129
11.1.1 24
11.1.2 (6, 2, 1)
11.1.3 238
11.1.4 548
11.1.5 732
11.1.6 486
11.1.7 15
11.1.8 351
11.1.9 461
143
Chapter 12
Probability States
Video Lecture
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Chapter 12. Probability States
Concept 12.0.1 (States)
We use states in problems when we are trying to find the probability of ”a win” from
different positions or turns
When encountering states problems we use the following steps
1. Assign variables to the probabilities of winning from the different positions
2. Write your equations for the probability of winning from each of these positions in
terms of the other states
3. Solve your system of equations
Concept 12.0.2 (Symmetry in States)
Always be on the lookout for symmetry in states problems (positions where you have an
equal probability of winning from) to help simplify your equations.
Remark 12.0.3
Often times in state problems when you have a lot of states, you may have to write a
state recursion equation.
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Chapter 12. Probability States
Example 12.1 (AMC 12)
Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips
a fair coin repeatedly until he gets his first head, at which point he stops. What is the
probability that all three flip their coins the same number of times?
Video Solution
Example 12.2 (AMC 10)
Frieda the frog begins a sequence of hops on a 3 × 3 grid of squares, moving one square
on each hop and choosing at random the direction of each hop-up, down, left, or right.
She does not hop diagonally. When the direction of a hop would take Frieda off the grid,
she ”wraps around” and jumps to the opposite edge. For example if Frieda begins in the
center square and makes two hops ”up”, the first hop would place her in the top row
middle square, and the second hop would cause Frieda to jump to the opposite edge,
landing in the bottom row middle square. Suppose Frieda starts from the center square,
makes at most four hops at random, and stops hopping if she lands on a corner square.
What is the probability that she reaches a corner square on one of the four hops?
Video Solution
Example 12.3 (AMC 10)
Raashan, Sylvia, and Ted play the following game. Each starts with $1. A bell rings every
15 seconds, at which time each of the players who currently have money simultaneously
chooses one of the other two players independently and at random and gives $1 to that
player. What is the probability that after the bell has rung 2019 times, each player will
have $1? (For example, Raashan and Ted may each decide to give $1 to Sylvia, and
Sylvia may decide to give her dollar to Ted, at which point Raashan will have $0, Sylvia
will have $2, and Ted will have $1, and that is the end of the first round of play. In the
second round Rashaan has no money to give, but Sylvia and Ted might choose each
other to give their $1 to, and the holdings will be the same at the end of the second
round.)
Video Solution
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12.1
Chapter 12. Probability States
Practice Problems
Problem 12.1.1 (AMC 10)
Amelia has a coin that lands heads with probability 13 , and Blaine has a coin that
lands on heads with probability 25 . Amelia and Blaine alternately toss their coins until
someone gets a head; the first one to get a head wins. All coin tosses are independent.
Amelia goes first. The probability that Amelia wins is pq , where p and q are relatively
prime positive integers. What is q − p?
Video Solution
Problem 12.1.2 (AMC 10)
Debra flips a fair coin repeatedly, keeping track of how many heads and how many
tails she has seen in total, until she gets either two heads in a row or two tails in a row,
at which point she stops flipping. What is the probability that she gets two heads in a
row but she sees a second tail before she sees a second head?
Video Solution
Problem 12.1.3 (AMC 10)
A frog sitting at the point (1, 2) begins a sequence of jumps, where each jump is
parallel to one of the coordinate axes and has length 1, and the direction of each jump
(up, down, right, or left) is chosen independently at random. The sequence ends when
the frog reaches a side of the square with vertices (0, 0), (0, 4), (4, 4), and (4, 0). What is
the probability that the sequence of jumps ends on a vertical side of the square?
Video Solution
Problem 12.1.4 (AIME)
Lily pads 1, 2, 3, . . . lie in a row on a pond. A frog makes a sequence of jumps starting on
pad 1. From any pad k the frog jumps to either pad k + 1 or pad k + 2 chosen randomly
with probability 12 and independently of other jumps. The probability that the frog visits
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Chapter 12. Probability States
pad 7 is pq , where p and q are relatively prime positive integers. Find p + q.
Video Solution
Problem 12.1.5 (AMC 12)
In a small pond there are eleven lily pads in a row labeled 0 through 10. A frog
is sitting on pad 1. When the frog is on pad N , 0 < N < 10, it will jump to pad N − 1
N
N
with probability 10
and to pad N + 1 with probability 1 − 10
. Each jump is independent
of the previous jumps. If the frog reaches pad 0 it will be eaten by a patiently waiting
snake. If the frog reaches pad 10 it will exit the pond, never to return. What is the
probability that the frog will escape without being eaten by the snake?
Video Solution
Additional Problems
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Chapter 12. Probability States
Problem 12.1.6 (AMC 12)
A square is drawn in the Cartesian coordinate plane with vertices at (2, 2), (−2, 2),
(−2, −2), (2, −2). A particle starts at (0, 0). Every second it moves with equal probability
to one of the eight lattice points (points with integer coordinates) closest to its current
position, independently of its previous moves. In other words, the probability is 1/8
that the particle will move from (x, y) to each of (x, y + 1), (x + 1, y + 1), (x + 1, y),
(x + 1, y − 1), (x, y − 1), (x − 1, y − 1), (x − 1, y), or (x − 1, y + 1). The particle will
eventually hit the square for the first time, either at one of the 4 corners of the square
or at one of the 12 lattice points in the interior of one of the sides of the square. The
probability that it will hit at a corner rather than at an interior point of a side is m/n,
where m and n are relatively prime positive integers. What is m + n?
Answers
12.1 17
12.2 25
32
12.3 14
12.1.1 4
1
12.1.2 24
12.1.3 58
12.1.4 107
63
12.1.5 146
12.1.6 39
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Algebra
150
Chapter 13
Algebraic Manipulations
Video Lecture
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Chapter 13. Algebraic Manipulations
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Chapter 13. Algebraic Manipulations
Theorem 13.0.1 (Exponent Rules)
x−a =
1
xa
xa × xb = xa+b
xa ÷ xb = xa−b
(xa )b = xab
13.1
Binomial Expansions
Theorem 13.1.1 (Binomial Square Expansions)
(x + y)2 = x2 + 2xy + y 2 = (x − y)2 + 4xy
(x − y)2 = x2 − 2xy + y 2 = (x + y)2 − 4xy
(x + y)2 + (x − y)2 = 2(x2 + y 2 )
(x + y)2 − (x − y)2 = 4xy
(x + y + z)2 = x2 + y 2 + z 2 + 2(xy + yz + xz)
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Chapter 13. Algebraic Manipulations
Theorem 13.1.2 (Binomial Cube Expansions)
(x + y)3 = x3 + 3xy(x + y) + y 3
(x + y)3 = x3 + 3x2 y + 3xy 2 + y 3
(x − y)3 = x3 − 3xy(x − y) − y 3
(x − y)3 = x3 − 3x2 y + 3xy 2 − y 3
x3 + y 3 + z 3 − 3xyz = (x + y + z)(x2 + y 2 + z 2 − xy − xz − yz)
Theorem 13.1.3
If
x+
then
1
=a
x
1
= a2 − 2
x2
1
x3 + 3 = a3 − 3a
x
1
x4 + 4 = (a2 − 2)2 − 2
x
x2 +
Example 13.1 (AMC 10)
Suppose that the number a satisfies the equation 4 = a + a−1 . What is the value of
a4 + a−4
Video Solution
Example 13.2
Given that x − x1 = 3, find x3 − x13
Video Solution
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13.2
Chapter 13. Algebraic Manipulations
Quadratic Factorizations
Theorem 13.2.1 (Difference of Squares)
x2 − y 2 = (x − y)(x + y)
13.3
Cubic Factorizations
Theorem 13.3.1 (Difference of Cubes)
x3 − y 3 = (x − y)(x2 + xy + y 2 )
Theorem 13.3.2 (Sum of Cubes)
x3 + y 3 = (x + y)(x2 − xy + y 2 )
Example 13.3
Given that x − x1 = 3, find x3 − x13
Video Solution
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13.4
Chapter 13. Algebraic Manipulations
Higher Power Factorizations
Theorem 13.4.1 (nth power Factorizations)
Sum of odd powers
x2n+1 + y 2n+1 = (x + y)(x2n − x2n−1 y + x2n−2 y 2 − · · · − xy 2n−1 + y 2n )
Note: The signs in the second term alternate between positive and negative
xn − y n = (x − y)(xn−1 + xn−2 y + xn−3 y 2 + · · · + xy n−2 + y n−1 )
Note: The signs in second term are all positive
Theorem 13.4.2 (Simon’s Favorite Factoring Trick)
xy + kx + jy + jk = (x + j)(y + k)
Remark 13.4.3
You can generally apply this factorization when you have xy, x, and y terms. After
applying the factorization, you can then find all possible values for each of your terms in
your factorization (remember negatives!).
Example 13.4 (Omega Learn)
√
If x > y, x + y = 8, and xy = 4, the value of x4 − y 4 is a b. Find a + b.
Video Solution
13.5
Algebraic Equations
Concept 13.5.1 (Algebraic Equations Techniques)
1. Substitution
2. Elimination
3. Adding/Subtracting/Multiplying Equations
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Chapter 13. Algebraic Manipulations
4. Group terms together
5. Find common denominator
6. Don’t always have to find every variable
7. Use symmetry between variables
8. Make smart substitutions that can simplify your expression
9. Look for common factorizations
Example 13.5 (Omega Learn)
Solve for a in
a2 b = 4
b2 c = 3
c2 a = 18
Video Solution
Example 13.6 (AMC 12)
If x, y, and z are positive numbers satisfying
x+
1
= 4,
y
y+
1
= 1,
z
and
z+
1
7
=
x
3
Then what is the value of xyz ?
Video Solution
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Chapter 13. Algebraic Manipulations
Example 13.7 (AIME)
Find the positive solution to
1
x2 − 10x − 29
+
1
x2 − 10x − 45
−
2
x2 − 10x − 69
=0
Video Solution
Example 13.8 (AIME)
Let (a, b, c) be the real solution of the system of equations x3 − xyz = 2, y 3 − xyz = 6,
z 3 − xyz = 20. The greatest possible value of a3 + b3 + c3 can be written in the form m
,
n
where m and n are relatively prime positive integers. Find m + n.
Video Solution
13.6
Practice Problems
Problem 13.6.1 (AMC 10)
Assuming a ̸= 3, b ̸= 4, and c ̸= 5, what is the value in simplest form of the following expression?
a−3 b−4 c−5
·
·
5−c 3−a 4−b
Video Solution
Problem 13.6.2 (AMC 10)
Supposed that x and y are nonzero real numbers such that 3x+y
= −2. What is
x−3y
x+3y
the value of 3x−y ?
Video Solution
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Chapter 13. Algebraic Manipulations
Problem 13.6.3 (AMC 10)
Real numbers a and b satisfy the equations 3a = 81b+2 and 125b = 5a−3 . What is
ab?
Problem 13.6.4 (AMC 10)
Let a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5. What is a + b + c + d?
Video Solution
Problem 13.6.5 (AMC 10)
If y + 4 = (x − 2)2 , x + 4 = (y − 2)2 , and x ̸= y, what is the value of x2 + y 2 ?
Video Solution
Problem 13.6.6 (AMC 10)
Real numbers x and y satisfy x + y = 4 and x · y = −2. What is the value of
x+
x3 y 3
+
+ y?
y 2 x2
Video Solution
Problem 13.6.7 (AMC 10)
For what value x does 10x · 1002x = 10005 ?
Video Solution
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Chapter 13. Algebraic Manipulations
Problem 13.6.8 (AMC 12)
Given that x and y are distinct nonzero real numbers such that x + x2 = y + y2 , what is
xy?
Video Solution
Problem 13.6.9 (AMC 10)
Suppose that real number x satisfies
√
√
49 − x2 − 25 − x2 = 3
√
√
What is the value of 49 − x2 + 25 − x2 ?
Video Solution
Problem 13.6.10 (AMC 10)
Two non-zero real numbers, a and b, satisfy ab = a − b. Which of the following is
a possible value of ab + ab − ab?
Video Solution
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Chapter 13. Algebraic Manipulations
Problem 13.6.11 (AMC 10)
Real numbers x and y satisfy the equation x2 + y 2 = 10x − 6y − 34. What is x + y?
Video Solution
Additional Problems
Problem 13.6.12
If x + x1 = 4, then what is the value of x8 + x18 ?
Problem 13.6.13 (MATHCOUNTS)
If (a − a1 )2 = 4, what is the absolute value of (a3 − a13 )?
Problem 13.6.14 (AMC 10)
Which of the following is equivalent to
(A) − x
(B) x
(C) 1
q
(D)
x
1− x−1
x
q
x
2
when x < 0?
√
(E) x −1
Problem 13.6.15 (AIME)
Find x2 + y 2 if x and y are positive integers such that
xy + x + y = 71,
x2 y + xy 2 = 880.
Problem 13.6.16 (AIME)
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Chapter 13. Algebraic Manipulations
Let x1 ≤ x2 ≤ · · · ≤ x100 be real numbers such that |x1 | + |x2 | + · · · + |x100 | = 1
and x1 + x2 + · · · + x100 = 0. Among all such 100-tuples of numbers, the greatest value
that x76 − x16 can achieve is m
, where m and n are relatively prime positive integers.
n
Find m + n.
Problem 13.6.17 (AIME)
Suppose that x, y, and z are three positive numbers that satisfy the equations xyz = 1,
, where m and n are relatively prime positive
x + z1 = 5, and y + x1 = 29. Then z + y1 = m
n
integers. Find m + n.
Problem 13.6.18 (AIME)
Find ax5 + by 5 if the real numbers a, b, x, and y satisfy the equations
ax + by = 3,
ax2 + by 2 = 7,
ax3 + by 3 = 16,
ax4 + by 4 = 42.
Problem 13.6.19 (AIME)
The equation 2333x−2 + 2111x+2 = 2222x+1 + 1 has three real roots. Given that their
sum is m/n where m and n are relatively prime positive integers, find m + n.
Problem 13.6.20 (AIME)
√
What is the product of the real roots of the equation x2 + 18x + 30 = 2 x2 + 18x + 45?
Problem 13.6.21 (AIME)
Find A2 , where A is the sum of the absolute values of all roots of the following equation:
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√
x = 19 + √19+ √
Chapter 13. Algebraic Manipulations
91
91
91
19+ √
19+ √ 91 91
19+ x
Problem 13.6.22 (AIME)
Let a, b, c, and d be real numbers that satisfy the system of equations
a + b = −3,
ab + bc + ca = −4,
abc + bcd + cda + dab = 14,
abcd = 30.
There exist relatively prime positive integers m and n such that
a2 + b 2 + c 2 + d 2 =
m
.
n
Find m + n.
Problem 13.6.23 (AIME)
Let m be the largest real solution to the equation
3
5
17
19
+ x−5
+ x−17
+ x−19
= x2 − 11x − 4
x−3
There are positive integers a, b, and c such that m = a +
q
b+
√
c. Find a + b + c.
Problem 13.6.24 (AIME)
Let x and y be real numbers satisfying x4 y 5 + y 4 x5 = 810 and x3 y 6 + y 3 x6 = 945.
Evaluate 2x3 + (xy)3 + 2y 3 .
Problem 13.6.25 (AIME)
The equation √2000x6 + 100x5 + 10x3 + x − 2 = 0 has exactly two real roots, one
of which is m+r n , where m, n and r are integers, m and r are relatively prime, and r > 0.
Find m + n + r.
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Chapter 13. Algebraic Manipulations
Answers
13.1 194
13.2 36
13.3 36
13.4 1795
13.5 2
13.6 1
13.7 013
13.8 158
13.6.1 −1
13.6.2 2
13.6.3 60
13.6.4 −10
3
13.6.5 15
13.6.6 440
13.6.7 3
13.6.8 2
13.6.9 8
13.6.10 2
13.6.11 2
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Chapter 13. Algebraic Manipulations
13.6.13 14
13.6.14 −x
13.6.15 146
13.6.16 841
13.6.17 5
13.6.18 20
13.6.19 113
13.6.20 20
13.6.21 383
13.6.22 145
13.6.23 263
13.6.24 89
13.6.25 200
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Chapter 14
Vieta’s Formulas for Polynomials
Video Lecture
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14.1
Chapter 14. Vieta’s Formulas for Polynomials
Polynomials
Theorem 14.1.1 (Quadratic Formula)
The solutions to the quadratic equation
ax2 + bx + c = 0
are
x=
Concept 14.1.2 (Discriminant)
In the quadratic formula,
−b ±
√
b2 − 4ac
2a
b2 − 4ac
(the part inside the square root) is the discriminant of the quadratic.
1. If the discriminant b2 − 4ac is 0, then the quadratic has a double or repeated root
2. If the discriminant b2 − 4ac is positive, the quadratic has 2 different real roots
3. If the discriminant b2 − 4ac is negative, the quadratic has no real roots
Theorem 14.1.3 (Vieta’s Formula For Quadratics)
In a quadratic equation
the sum of its roots is
and the product of its roots is
ax2 + bx + c = 0
−b
a
c
a
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Chapter 14. Vieta’s Formulas for Polynomials
Example 14.1
If the roots of the quadratic 3x2 + 6x − 5 are r and s, find r3 + s3 .
Video Solution
Theorem 14.1.4 (Vieta’s Formula For Higher Degree Polynomials)
In a polynomial
an xn + an−1 xn−1 + ... + a1 x + a0 = 0
with roots
r1 , r2 , r3 , ...rn
the following holds:
r1 + r2 + r3 + ... + rn (the sum of all terms) = −
an−1
an
r1 r2 + r1 r3 + .. + rn−1 rn (the sum of all products of 2 terms) =
an−2
an
r1 r2 r3 + r1 r2 r4 + ... + rn−2 rn−1 rn (the sum of all products of 3 terms) = −
an−3
an
..
.
r1 r2 r3 . . . rn (the sum of all products of n terms) = (−1)n
a0
an
Note that the negative and positive signs alternate. When summing the products for
odd number of terms, we will have a negative sign otherwise we will have a positive sign.
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Chapter 14. Vieta’s Formulas for Polynomials
Example 14.2
If r, s, and t are roots of the cubic x3 − 6x − 5 = 0, find r3 + s3 + t3 .
Video Solution
Example 14.3 (AMC 12)
The polynomial p(x) = x3 + ax2 + bx + c has the property that the average of its zeros,
the product of its zeros, and the sum of its coefficients are all equal. The y-intercept of
the graph of y = p(x) is 2. What is b?
Video Solution
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14.2
Chapter 14. Vieta’s Formulas for Polynomials
Practice Problems
Problem 14.2.1
Find the sum and product of the roots of the quadratic 2x2 − 24x + 17?
Video Solution
Problem 14.2.2 (AMC 12)
There are two values of a for which the equation 4x2 + ax + 8x + 9 = 0 has only
one solution for x. What is the sum of these values of a?
Video Solution
Problem 14.2.3 (AMC 10)
Let a and b be the roots of the equation x2 − mx + 2 = 0. Suppose that a + 1b
and b + a1 are the roots of the equation x2 − px + q = 0. What is q?
Video Solution
Problem 14.2.4 (AMC 10)
What is the sum of the reciprocals of the roots of the equation 2003
x + 1 + x1 = 0?
2004
Video Solution
Problem 14.2.5 (AMC 10)
The quadratic equation x2 + mx + n has roots twice those of x2 + px + m, and none of
m, n, and p is zero. What is the value of np ?
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Chapter 14. Vieta’s Formulas for Polynomials
Video Solution
Problem 14.2.6 (AMC 10)
Let f be a function for which f
z for which f (3z) = 7.
x
3
= x2 + x + 1. Find the sum of all values of
Video Solution
Problem 14.2.7 (AMC 10)
What is the sum of all real numbers x for which |x2 − 12x + 34| = 2?
Video Solution
Problem 14.2.8 (AMC 10)
The polynomial x3 − ax2 + bx − 2010 has three positive integer roots. What is the
smallest possible value of a?
Video Solution
Problem 14.2.9 (AMC 10)
Suppose that a and b are nonzero real numbers, and that the equation x2 + ax + b = 0
has solutions a and b. Then the pair (a, b) is
Video Solution
Problem 14.2.10 (AIME)
Suppose that the roots of x3 + 3x2 + 4x − 11 = 0 are a, b, and c, and that the roots of
x3 + rx2 + sx + t = 0 are a + b, b + c, and c + a. Find t.
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Chapter 14. Vieta’s Formulas for Polynomials
Video Solution
Problem 14.2.11 (AMC 10)
For certain real numbers a, b, and c, the polynomial
g(x) = x3 + ax2 + x + 10
has three distinct roots, and each root of g(x) is also a root of the polynomial
f (x) = x4 + x3 + bx2 + 100x + c.
What is f (1)?
Video Solution
Problem 14.2.12 (AMC 10)
Let a, b, and c be three distinct one-digit numbers. What is the maximum value
of the sum of the roots of the equation (x − a)(x − b) + (x − b)(x − c) = 0?
Video Solution
Problem 14.2.13
[AMC 10] All the roots of the polynomial z 6 − 10z 5 + Az 4 + Bz 3 + Cz 2 + Dz + 16
are positive integers, possibly repeated. What is the value of B?
Video Solution
Problem 14.2.14 (USAMO)
In the polynomial x4 − 18x3 + kx2 + 200x − 1984 = 0, the product of 2 of its roots is
−32. Find k.
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Chapter 14. Vieta’s Formulas for Polynomials
Video Solution
Problem 14.2.15 (AMC 10/12)
All the roots of polynomial z 6 − 10z 5 + Az 4 + Bz 3 + Cz 2 + Dz + 16 are positive integers,
possibly repeated. What is the value of B?
Video Solution
14.3
Additional Problems
Problem 14.3.1 (AMC 10)
The zeros of the function f (x) = x2 − ax + 2a are integers. What is the sum of
the possible values of a?
Problem 14.3.2 (AIME)
Let r, s, and t be the three roots of the equation
8x3 + 1001x + 2008 = 0.
Find (r + s)3 + (s + t)3 + (t + r)3 .
Problem 14.3.3 (AIME)
Find the sum of the roots, real and non-real, of the equation x2001 +
given that there are no multiple roots.
1
−x
2
2001
= 0,
Problem 14.3.4 (AHSME)
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Chapter 14. Vieta’s Formulas for Polynomials
How many distinct ordered triples (x, y, z) satisfy the following equations?
x + 2y + 4z = 12
xy + 4yz + 2xz = 22
xyz = 6
Problem 14.3.5 (AIME)
Consider the polynomials P (x) = x6 − x5 − x3 − x2 − x and Q(x) = x4 − x3 − x2 − 1.
Given that z1 , z2 , z3 , and z4 are the roots of Q(x) = 0, find P (z1 ) + P (z2 ) + P (z3 ) + P (z4 ).
Problem 14.3.6 (AIME)
For integers a, b, c and d, let f (x) = x2 + ax + b and g(x) = x2 + cx + d. Find the
number of ordered triples (a, b, c) of integers with absolute values not exceeding 10 for
which there is an integer d such that g(f (2)) = g(f (4)) = 0.
Problem 14.3.7 (AIME)
Let P (x) = x2 − 3x − 7, and let Q(x) and R(x) be two quadratic polynomials also
with the coefficient of x2 equal to 1. David computes each of the three sums P + Q,
P + R, and Q + R and is surprised to find that each pair of these sums has a common
root, and these three common roots are distinct. If Q(0) = 2, then R(0) = m
, where m
n
and n are relatively prime positive integers. Find m + n.
Answers
14.1 −18
14.2 15
14.3 −11
14.2.1 12, 17
2
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Chapter 14. Vieta’s Formulas for Polynomials
14.2.2 −16
14.2.3 92
14.2.4 −1
14.2.5 8
14.2.6 − 91
14.2.7 18
14.2.8 78
14.2.9 (1, −2)
14.2.10 023
14.2.11 −7007
14.2.12 16.5
14.2.13 −88
14.2.14 86
14.2.15 −88
14.3.1 16
14.3.2 753
14.3.3 500
14.3.4 6
14.3.5 6
14.3.6 510
14.3.7 71
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Chapter 15
Polynomial Roots
Video Lecture
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Chapter 15. Polynomial Roots
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15.1
Chapter 15. Polynomial Roots
Polynomial Root Representation
Theorem 15.1.1 (Representation of Polynomial in terms of roots)
In a polynomial
P (x) = an xn + an−1 xn−1 + · · · + a1 x1 + a0
it can be expressed in the form
an (x − r1 )(x − r2 )(x − r3 ) . . . (x − rn )
where r1 , r2 , r3 , . . . , rn are the n roots of the polynomial.
Corollary 15.1.2 (Representation of Monic Polynomial in terms of roots)
In a polynomial
P (x) = xn + an−1 xn−1 + . . . a1 x1 + a0
(where the leading coeffecient is 1), it can be expressed in the form
(x − r1 )(x − r2 )(x − r3 ) . . . (x − rn )
where r1 , r2 , r3 , . . . , rn are the n roots of the polynomial.
Concept 15.1.3
When a problem asks you to find an expression like (k − r)(k − s)(k − t) where r, s, and
t are roots of the polynomial for a monic polynomial, it would just be equal to P (k) by
the above definition. The same will work for non monic polynomials except it would be
P (k)
where an is the coefficient of the xn term in the polynomial.
an
Theorem 15.1.4 (Fundamental theorem of Algebra)
A polynomial of degree n (the largest term is to the power of n) has n complex roots
including multiplicity (for example, a double root would be counted as 2 roots when
including multiplicity)
Theorem 15.1.5 (Remainder Theorem)
The remainder when when a polynomial P (x) is divided by x − r is P (r)
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Chapter 15. Polynomial Roots
Corollary 15.1.6 (Factor Theorem)
x − r will divide a polynomial P (x) if P (r) = 0
This is a direct consequence of the remainder theorem.
Example 15.1 (AHSME)
Find the remainder when x51 + 51 is divided by x + 1.
Video Solution
Example 15.2 (USAMO)
In the polynomial x4 − 18x3 + kx2 + 200x − 1984 = 0, the product of 2 of its roots is
−32. Find k.
Video Solution
Example 15.3 (AIME)
For distinct complex numbers z1 , z2 , . . . , z673 , the polynomial
(x − z1 )3 (x − z2 )3 · · · (x − z673 )3
can be expressed as x2019 + 20x2018 + 19x2017 + g(x), where g(x) is a polynomial with
complex coefficients and with degree at most 2016. The sum
X
zj zk
1≤j<k≤673
can be expressed in the form m
, where m and n are relatively prime positive integers.
n
Find m + n.
Video Solution
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15.2
Chapter 15. Polynomial Roots
Symmetric Polynomials
Definition 15.2.1 (Symmetric Polynomials). A polynomial
P (x) = an xn + an−1 xn−1 + ... + a1 x + a0
is symmetric if
an = a0
an−1 = a1
an−2 = a2
an−3 = a3
etc.
Basically, opposite coefficients are equal.
Concept 15.2.2 (Solving Symmetric Polynomials of Even Degree)
To solve a symmetric polynomial P (x) = a0 xn + a1 xn−1 + ... + a1 x + a0 of even degree,
n
• Divide by x 2
• Group the xk and x1k terms together
• Make the substitution
1
x
and write all the terms in your expression that way
y =x+
• Solve the reduced polynomial
Example 15.4 (AMC 10)
The real number x satisfies the equation x + x1 =
√
5. What is the value of x11 − 7x7 + x3 ?
Video Solution
Video Solution
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Chapter 15. Polynomial Roots
Example 15.5 (Omega Learn)
Solve for both roots of 10000x4 − 5000x3 + 825x2 − 50x + 1 = 0.
Video Solution
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15.3
Chapter 15. Polynomial Roots
Polynomial Manipulations
Concept 15.3.1 (Reciprocal Roots)
In a polynomial
P (x) = an xn + an−1 xn−1 + ... + a1 x + a0
with roots r1 , r2 , r3 , . . . , rn ,
Q(x) = a0 xn + a1 xn−1 + ... + an−1 x + an
will have roots
1 1
1
, ,...,
r1 r2
rn
Essentially, when flipping the coefficients of a polynomial, it will have roots that are
reciprocals of the original roots.
Concept 15.3.2 (Roots That Are More or Less)
In a polynomial
P (x) = an xn + an−1 xn−1 + ... + a1 x + a0
with roots r1 , r2 , r3 , . . . , rn ,
Q(x) = an (x − k)n + an−1 (x − k)n−1 + ... + a1 (x − k) + a0
will have roots
r1 + k, r2 + k, r3 + k, . . . , rn + k
Remark 15.3.3
Remember, if the roots are k more, than we subtract k from each of the x terms in our
polynomial.
Remark 15.3.4
Polynomial manipulations are useful when evaluating complex expressions in terms of
roots. For example, in order to evaluate
1
1
1
+
+
3
3
(r − 3)
(s − 3)
(t − 3)3
of a polynomial with roots r, s, t, rather than expanding it out and bashing with Vieta’s
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Chapter 15. Polynomial Roots
Formulas, we can simplify construct a new polynomial with roots
1
1
1
,
,
r−3 s−3 t−3
. Then, we can use standard Vieta tricks to find r3 + s3 + t3 . One way is to use the
factorization a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ac) in the new
polynomial.
Example 15.6 (AMC 12)
Let g(x) be a polynomial with leading coefficient 1, whose three roots are the reciprocals
of the three roots of f (x) = x3 + ax2 + bx + c, where 1 < a < b < c. What is g(1) in
terms of a, b, and c?
Video Solution
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Chapter 15. Polynomial Roots
Theorem 15.3.5 (Newton Sums)
In a polynomial
P (x) = an xn + an−1 xn−1 + ... + a1 x + a0
with roots
r1 , r2 , r3 , . . . , rn
let S1 = r1 + r2 + · · · + rn
S2 = r12 + r22 + · · · + rn2
..
.
Sk = r1k + r2k + · · · + rnk
..
.
then the following holds true
an S1 + an−1 = 0
an S2 + an−1 S1 + 2an−2 = 0
an S3 + an−1 S2 + an−2 S1 + 3an−3 = 0
..
.
Note that anegative number = 0 as that might show up in your expansion.
Essentially, what this is saying is
1. Start off with a Sk value and multiply by it by the leftmost polynomial coefficient.
2. Then, multiply Sn−1 by the polynomial’s coefficient right after it.
3. Continue doing so and summing the products until either
• k = 0 in which case instead of multiplying S0 by last the last a term we
multiply k
• an−i becomes 0 in which case we simply add the last term and stop
4. Set your final sum of terms to be equal to 0
Remark 15.3.6
Note that by each Newton Sum Equation, we can iteratively calculate each Pk rather
than having to bash with Vieta’s Formulas.
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15.4
Chapter 15. Polynomial Roots
Practice Problems
Problem 15.4.1 (AMC 10)
Let f be a function for which f
z for which f (3z) = 7.
x
3
= x2 + x + 1. Find the sum of all values of
Video Solution
Problem 15.4.2 (AIME)
Suppose that the roots of x3 + 3x2 + 4x − 11 = 0 are a, b, and c, and that the roots of
x3 + rx2 + sx + t = 0 are a + b, b + c, and c + a. Find t.
Video Solution
Problem 15.4.3 (AMC 10)
For certain real numbers a, b, and c, the polynomial
g(x) = x3 + ax2 + x + 10
has three distinct roots, and each root of g(x) is also a root of the polynomial
f (x) = x4 + x3 + bx2 + 100x + c.
What is f (1)?
Video Solution
Problem 15.4.4
[AMC 10/12B] Let P (x) be a polynomial with rational coefficients such that when
P (x) is divided by the polynomial x2 + x + 1, the remainder is x + 2, and when P (x) is
divided by the polynomial x2 + 1, the remainder is 2x + 1. There is a unique polynomial of
least degree with these two properties. What is the sum of the squares of the coefficients
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Chapter 15. Polynomial Roots
of that polynomial?
(A) 10
(B) 13
(C) 19
(D) 20
(E) 23
Video Solution
15.5
Additional Problems
Problem 15.5.1 (AHSME)
Given the equations x2 + kx + 6 = 0 and x2 − kx + 6 = 0. If, when the roots of
the equation are suitably listed, each root of the second equation is 5 more than the
corresponding root of the first equation, then k equals:
Problem 15.5.2 (AMC 10)
What is the product of all the roots of the equation
q
5|x| + 8 =
√
x2 − 16.
Problem 15.5.3 (AHSME)
Let P (x) be a polynomial such that when P (x) is divided by x − 19, the remainder is 99, and when P (x) is divided by x − 99, the remainder is 19. What is the remainder
when P (x) is divided by (x − 19)(x − 99)?
Problem 15.5.4 (AIME)
The
real root of the equation 8x3 − 3x2 − 3x − 1 = 0 can be written in the form
√
√
3 a+ 3 b+1
, where a, b, and c are positive integers. Find a + b + c.
c
Problem 15.5.5 (AIME)
Let x1 < x2 < x3 be the three real roots of the equation
√
2014x3 − 4029x2 + 2 = 0. Find
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Chapter 15. Polynomial Roots
x2 (x1 + x3 ).
Problem 15.5.6 (AMC 12)
The graph of y = x6 − 10x5 + 29x4 − 4x3 + ax2 lies above the line y = bx + c except at three values of x, where the graph and the line intersect. What is the largest of
these values?
Problem 15.5.7 (AIME)
The equation √2000x6 + 100x5 + 10x3 + x − 2 = 0 has exactly two real roots, one
of which is m+r n , where m, n and r are integers, m and r are relatively prime, and r > 0.
Find m + n + r.
Problem 15.5.8 (AIME)
Let P (x) be a nonzero polynomial such that (x − 1)P (x + 1) = (x + 2)P (x) for every real
x, and (P (2))2 = P (3). Then P ( 72 ) = m
, where m and n are relatively prime positive
n
integers. Find m + n.
Problem 15.5.9 (AIME)
Let P (x) be a quadratic polynomial with complex coefficients whose x2 coefficient
is 1. Suppose the equation P (P (x)) = 0 has four distinct solutions, x = 3, 4, a, b. Find
the sum of all possible values of (a + b)2 .
Problem 15.5.10 (AMC 10)
A quadratic polynomial with real coefficients and leading coefficient 1 is called disrespectful
if the equation p(p(x)) = 0 is satisfied by exactly three real numbers. Among all the
disrespectful quadratic polynomials, there is a unique such polynomial p̃(x) for which
the sum of the roots is maximized. What is p̃(1)?
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Chapter 15. Polynomial Roots
Answers
15.1 50
15.2 86
15.3 352
15.4 0
1 1
15.5 20
,5
15.6 1+a+b+c
c
15.4.1 − 91
15.4.2 023
15.4.3 −7007
15.4.4 23
15.5.1 5
15.5.2 −64
15.5.3 −x + 118
15.5.4 98
15.5.5 2
15.5.6 4
15.5.7 200
15.5.8 109
15.5.9 85
5
15.5.10 16
188
Chapter 16
Arithmetic Sequences
Video Lecture
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Chapter 16. Arithmetic Sequences
Definition 16.0.1 (Arithmetic Sequence). An arithmetic sequence is a sequence of numbers
with the same difference between consecutive terms.
1, 4, 7, 10, 13, . . . , 40
is an arithmetic sequence because there is always a difference of 3 between consecutive terms.
Remark 16.0.2
Note that an arithmetic sequence can also have a negative common difference. For
example, in the arithmetic sequence
40, 37, 34, . . . , 4, 1
the common difference is −3.
Definition 16.0.3 (Arithmetic Sequence Notation). In general, the terms of an arithmetic
sequence can be represented as:
a1 , a2 , a3 , a4 , . . . , an
where
• d is the common difference between consecutive terms
• n is the number of terms
Theorem 16.0.4 (nth term in an Arithmetic Sequence)
an = a1 + (n − 1)d
an = am + (n − m)d
Theorem 16.0.5 (Number of terms in an Arithmetic Sequence)
n=
an − a1
+1
d
Essentially,
Number of Terms =
Last Term − First Term
+1
Common Difference
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Chapter 16. Arithmetic Sequences
Theorem 16.0.6 (Average of terms in an Arithmetic Sequence)
Average of Terms =
a1 + an
2
Theorem 16.0.7 (Sum of all terms in an Arithmetic Sequence)
Sn =
Substitute
a1 + an
×n
2
an = a1 + (n − 1)d
to get
Sn =
2a1 + (n − 1)d
×n
2
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Chapter 16. Arithmetic Sequences
Example 16.1 (AIME)
For each positive integer k, let Sk denote the increasing arithmetic sequence of integers
whose first term is 1 and whose common difference is k. For example, S3 is the sequence
1, 4, 7, 10, . . . . For how many values of k does Sk contain the term 2005?
Video Solution
Example 16.2 (AIME)
If the integer k is added to each of the numbers 36, 300, and 596, one obtains the squares
of three consecutive terms of an arithmetic series. Find k.
Video Solution
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16.1
Chapter 16. Arithmetic Sequences
Practice Problems
Problem 16.1.1 (AMC 10)
What is the greatest number of consecutive integers whose sum is 45?
Video Solution
Problem 16.1.2 (AMC 10)
In how many ways can 345 be written as the sum of an increasing sequence of two
or more consecutive positive integers?
Video Solution
Problem 16.1.3 (AMC 10/12)
Suppose that S is a finite set of positive integers. If the greatest integer in S is removed from S, then the average value (arithmetic mean) of the integers remaining is 32.
If the least integer in S is also removed, then the average value of the integers remaining
is 35. If the greatest integer is then returned to the set, the average value of the integers
rises to 40. The greatest integer in the original set S is 72 greater than the least integer
in S. What is the average value of all the integers in the set S?
Video Solution
Problem 16.1.4 (AMC 10)
In the five-sided star shown, the letters A, B, C, D, and E are replaced by the numbers 3, 5, 6, 7, and 9, although not necessarily in this order. The sums of the numbers
at the ends of the line segments AB,BC,CD,DE, and EA form an arithmetic sequence,
although not necessarily in this order. What is the middle term of the arithmetic
sequence?
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Chapter 16. Arithmetic Sequences
A
C
D
E
B
Video Solution
Problem 16.1.5 (AMC 10)
A grocer makes a display of cans in which the top row has one can and each lower
row has two more cans than the row above it. If the display contains 100 cans, how many
rows does it contain?
Video Solution
Problem 16.1.6 (AIME)
The terms of an arithmetic sequence add to 715. The first term of the sequence is
increased by 1, the second term is increased by 3, the third term is increased by 5, and in
general, the kth term is increased by the kth odd positive integer. The terms of the new
sequence add to 836. Find the sum of the first, last, and middle terms of the original
sequence.
Video Solution
Problem 16.1.7 (AMC 10)
Suppose that {an } is an arithmetic sequence with
a1 + a2 + · · · + a100 = 100 and a101 + a102 + · · · + a200 = 200.
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Chapter 16. Arithmetic Sequences
What is the value of a2 − a1 ?
Video Solution
Additional Problems
Problem 16.1.8 (AMC 10)
Mary divides a circle into 12 sectors. The central angles of these sectors, measured
in degrees, are all integers and they form an arithmetic sequence. What is the degree
measure of the smallest possible sector angle?
Problem 16.1.9 (AMC 10)
How many non-similar triangles have angles whose degree measures are distinct positive
integers in arithmetic progression?
Problem 16.1.10 (AIME)
The degree measures of the angles in a convex 18-sided polygon form an increasing
arithmetic sequence with integer values. Find the degree measure of the smallest angle.
Problem 16.1.11
Find the roots of the polynomial x5 − 5x4 − 35x3 + ax2 + bx + c, given that the roots
form an arithmetic progression.
Problem 16.1.12 (AIME)
Find the eighth term of the sequence 1440, 1716, 1848, . . . , whose terms are formed by
multiplying the corresponding terms of two arithmetic sequences.
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Chapter 16. Arithmetic Sequences
Answers
16.1 12
16.2 925
16.1.1 90
16.1.2 7
16.1.3 36.8
16.1.4 12
16.1.5 10
16.1.6 195
16.1.7 0.01
16.1.8 8
16.1.9 59
16.1.10 143
16.1.11 −−
16.1.12 348
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Chapter 17
Geometric Sequences
Video Lecture
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Chapter 17. Geometric Sequences
Definition 17.0.1 (Geometric Sequence). A geometric sequence is a sequence of numbers
with the same ratio between consecutive terms.
1, 2, 4, 8, 16, 32 . . . , 1024
is a geometric sequence because there is always a ratio of 2 between consecutive terms.
Definition 17.0.2 (Geometric Sequence Notation). In general, the terms of a geometric
sequence can be represented as:
g1 , g2 , g3 , g4 , . . . , gn
where
• r is the common ratio between consecutive terms
• n is the number of terms
Remark 17.0.3
Note that a geometric sequence can also have a negative common ratio. For example the
sequence 1, −2, 4, −8, . . . , 256, −512, 1024 has a common ratio of −2.
Theorem 17.0.4 (nth term in a Geometric Sequence)
gn = g1 · rn−1
gn = gm · r(n−m)
Example 17.1 (AIME)
Two geometric sequences a1 , a2 , a3 , . . . and b1 , b2 , b3 , . . . have the same common ratio,
with a1 = 27, b1 = 99, and a15 = b11 . Find a9 .
Video Solution
Theorem 17.0.5 (Sum of all terms in a finite Geometric Sequence)
Sn = g1
(rn − 1)
r−1
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Chapter 17. Geometric Sequences
Theorem 17.0.6 (Sum of all Terms in an Infinite Geometric Sequence)
For −1 < r < 1,
S∞ =
g1
1−r
Remark 17.0.7
The reason the formula only works for |r| < 1 is because if |r| ≥ 1 the sum will diverge
or essentially be infinite. We can only find the sum of an infinite geometric sequence
which is converging as its sum approaches a constant value. Examples:
1+
1+
1 1 1
1
+ + + ··· =
=2
2 4 8
1 − 21
1
1
3
1 1
+ +
+ ··· =
1 =
3 9 27
2
1− 3
Example 17.2 (AIME)
For −1 < r < 1, let S(r) denote the sum of the geometric series
12 + 12r + 12r2 + 12r3 + · · · .
Let a between −1 and 1 satisfy S(a)S(−a) = 2016. Find S(a) + S(−a).
Video Solution
Example 17.3 (AIME)
An infinite geometric series has sum 2005. A new series, obtained by squaring each term
of the original series, has 10 times the sum of the original series. The common ratio of
the original series is m
where m and n are relatively prime integers. Find m + n.
n
Video Solution
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17.1
Chapter 17. Geometric Sequences
Practice Problems
Problem 17.1.1 (MATHCOUNTS)
The fifth term of a geometric sequence of positive numbers is 11 and the eleventh
term is 5. What is the eighth term of the sequence?
Video Solution
Problem 17.1.2 (MATHCOUNTS)
The third and fourth terms of an arithmetic sequence are the first and second terms of a
geometric sequence. If the first two terms of the arithmetic sequence are 5, 2, then what
is the fourth term of the geometric sequence?
Video Solution
Problem 17.1.3 (MATHCOUNTS)
The numbers a, b, c, and d form a geometric sequence, in that order. If b is three
more than a, and c is nine more than b, what is the value of d?
Video Solution
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Chapter 17. Geometric Sequences
Problem 17.1.4 (MATHCOUNTS)
The positive integers A, B, and C form an arithmetic sequence while the integers
C
B, C, and D form a geometric sequence. If B
= 53 , what is the smallest possible value of
A + B + C + D?
Video Solution
17.2
Additional Problems
Problem 17.2.1 (AIME)
Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to
right, form a geometric sequence. Find the difference between the largest and smallest
geometric numbers.
Problem 17.2.2 (AIME)
The sum of the first 2011 terms of a geometric sequence is 200. The sum of the
first 4022 terms is 380. Find the sum of the first 6033 terms.
Problem 17.2.3 (AIME)
Suppose n is a positive integer and d is a single digit in base 10. Find n if
n
= 0.d25d25d25 . . .
810
Problem 17.2.4 (AMC 10)
The sum of an infinite geometric series is a positive number S, and the second term in
the series is 1. What is the smallest possible value of S?
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Problem 17.2.5 (AMC 10)
Positive integers a, b, and 2009, with a < b < 2009, form a geometric sequence with an
integer ratio. What is a?
Problem 17.2.6 (AIME)
In an increasing sequence of four positive integers, the first three terms form an arithmetic
progression, the last three terms form a geometric progression, and the first and fourth
terms differ by 30. Find the sum of the four terms.
Problem 17.2.7 (AMC 12)
A high school basketball game between the Raiders and Wildcats was tied at the
end of the first quarter. The number of points scored by the Raiders in each of the four
quarters formed an increasing geometric sequence, and the number of points scored by
the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At
the end of the fourth quarter, the Raiders had won by one point. Neither team scored
more than 100 points. What was the total number of points scored by the two teams in
the first half?
Problem 17.2.8
Let p and q be real numbers with |p| < 1 and |q| < 1 such that
p + pq + pq 2 + pq 3 + · · · = 2andq + qp + qp2 + qp3 + · · · = 3
Find 100pq.
Problem 17.2.9 (AIME)
A sequence of positive integers with a1 = 1 and a9 + a10 = 646 is formed so that
the first three terms are in geometric progression, the second, third, and fourth terms are
in arithmetic progression, and, in general, for all n ≥ 1, the terms a2n−1 , a2n , a2n+1 are in
geometric progression, and the terms a2n , a2n+1 , and a2n+2 are in arithmetic progression.
Let an be the greatest term in this sequence that is less than 1000. Find n + an .
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Chapter 17. Geometric Sequences
Problem 17.2.10 (AIME)
Two distinct, real, infinite geometric series each have a sum of 1 and have the same
second term. The third term of √one of the series is 1/8, and the second term of both
series can be written in the form m−n
, where m, n, and p are positive integers and m is
p
not divisible by the square of any prime. Find 100m + 10n + p.
Answers
17.1 363
17.2 336
17.3 802
√
17.1.1 55
17.1.2 −64
17.1.3 81
2
17.1.4 52
17.2.1 840
17.2.2 542
17.2.3 750
17.2.4 4
17.2.5 41
17.2.6 129
17.2.7 34
17.2.8 48
17.2.9 973
17.2.10 518
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Chapter 18
Special Sequences
Video Lecture
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Chapter 18. Special Sequences
Theorem 18.0.1 (Sum of numbers formula)
1 + 2 + 3 + ··· + n =
(n)(n + 1)
2
Theorem 18.0.2 (Sum of odd numbers formula)
1 + 3 + 5 + · · · + (2n − 1) = n2
In simple terms, the
Sum of first n odd numbers = n2
Theorem 18.0.3 (Sum of even numbers formula)
2 + 4 + 6 + · · · + 2n = n(n + 1)
To intuitively think about it, just take 2 common from each term
2(1 + 2 + 3 + · · · + n) = 2
(n)(n + 1)
= n(n + 1)
2
Theorem 18.0.4 (Sum of squares formula)
12 + 22 + · · · + n2 =
(n)(n + 1)(2n + 1)
6
Theorem 18.0.5 (Sum of Cubes Formula)
1 + 2 + ··· + n =
3
3
3
(n)(n + 1)
2
!2
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18.1
Chapter 18. Special Sequences
Telescoping
Concept 18.1.1 (Telescoping)
Expand the first few and last few terms, and cancel out any terms you see.
Remark 18.1.2
Generally, whenever you have long expressions that seem to be hard or impossible to
compute manually, telescoping is probably a good option.
Example 18.1 (AIME)
1
Consider the sequence defined by ak = 2
for k ≥ 1. Given that am + am+1 + · · · +
k +k
1
an−1 = , for positive integers m and n with m < n, find m + n.
29
Video Solution
Example 18.2 (Omega Learn)
Let the sequence an be defined as an = 31+2+3+···+n for all positive integers n. Sohil is
trying to evaluate the following sequence: 12 × 32 × 34 ×. . . . Because of his bad handwriting,
he accidentally writes a plus sign instead of the multiplication sign before all fractions of
1
3(1+2)
n
the form ana+1
(for example, 313+1 , 3(1+2)
, . . . ). What is the value of the new expression?
+1
Video Solution
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Chapter 18. Special Sequences
Example 18.3 (AIME)
Given that
x1 = 211,
x2 = 375,
x3 = 420,
x4 = 523, and
xn = xn−1 − xn−2 + xn−3 − xn−4 when n ≥ 5,
find the value of x531 + x753 + x975 .
Video Solution
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18.2
Chapter 18. Special Sequences
Practice Problems
Problem 18.2.1 (AMC 8)
What is the value of the product
1·3
2·2
2·4
3·3
3·5
97 · 99
···
4·4
98 · 98
98 · 100
?
99 · 99
Video Solution
Problem 18.2.2 (AMC 10)
Let f (x) = x2 (1 − x)2 . What is the value of the sum
f
1
2
3
4
2017
2018
−f
+f
−f
+ ··· + f
−f
?
2019
2019
2019
2019
2019
2019
Video Solution
Problem 18.2.3 (AMC 10)
Seven cubes, whose volumes are 1, 8, 27, 64, 125, 216, and 343 cubic units, are stacked
vertically to form a tower in which the volumes of the cubes decrease from bottom to
top. Except for the bottom cube, the bottom face of each cube lies completely on top of
the cube below it. What is the total surface area of the tower (including the bottom) in
square units?
Video Solution
Problem 18.2.4 (AMC 10)
Hiram’s algebra notes are 50 pages long and are printed on 25 sheets of paper; the
first sheet contains pages 1 and 2, the second sheet contains pages 3 and 4, and so on.
One day he leaves his notes on the table before leaving for lunch, and his roommate
decides to borrow some pages from the middle of the notes. When Hiram comes back, he
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Chapter 18. Special Sequences
discovers that his roommate has taken a consecutive set of sheets from the notes and
that the average (mean) of the page numbers on all remaining sheets is exactly 19. How
many sheets were borrowed?
Video Solution
Additional Problems
Problem 18.2.5 (AHSME)
If Tn = 1 + 2 + 3 + · · · + n and
Pn =
T2
T3
T4
Tn
·
·
· ··· ·
T2 − 1 T3 − 1 T4 − 1
Tn − 1
for n = 2, 3, 4, · · · , then P1991 is closest to which of the following numbers?
(A) 2.0 (B) 2.3 (C) 2.6 (D) 2.9 (E) 3.2
Problem 18.2.6 (AMC 12)
Let A be the set of positive integers that have no prime factors other than 2, 3, or
5. The infinite sum
1 1 1 1 1 1 1 1
1
1
1
1
1
1
+ + + + + + + +
+
+
+
+
+
+ ···
1 2 3 4 5 6 8 9 10 12 15 16 18 20
of the reciprocals of the elements of A can be expressed as m
, where m and n are relatively
n
prime positive integers. What is m + n?
Problem 18.2.7 (AHSME)
1
1
1
1
Find the sum 1(3)
+ 3(5)
+ · · · + (2n−1)(2n+1)
+ · · · + 255(257)
.
Problem 18.2.8 (AMC 10)
The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, . . . starts with two 1s, and each term after-
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Chapter 18. Special Sequences
wards is the sum of its two predecessors. Which one of the ten digits is the last to appear
in the units position of a number in the Fibonacci sequence?
Problem 18.2.9 (AIME)
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of
the squares of the three terms equals the product of the middle term and the square of
the common difference. Find the sum of the third terms of all special sequences.
Problem 18.2.10
Evaluate
1
1−
3
2 !
1−
2 !
1
4
1−
2 !
1
5
...
Problem 18.2.11 (AIME)
A sequence of integers a1 , a2 , a3 , . . . is chosen so that an = an−1 − an−2 for each n ≥ 3.
What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms
is 1985, and the sum of the first 1985 terms is 1492?
Problem 18.2.12 (AMC 12)
Set u0 = 14 , and for k ≥ 0 let uk+1 be determined by the recurrence
uk+1 = 2uk − 2u2k .
This sequence tends to a limit; call it L. What is the least value of k such that
|uk − L| ≤
1
21000
?
Problem 18.2.13 (AMC 10)
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Chapter 18. Special Sequences
The sum of the first m positive odd integers is 212 more than the sum of the first
n positive even integers. What is the sum of all possible values of n?
Problem 18.2.14 (AIME)
The sequence {an } is defined by
a0 = 1, a1 = 1, and an = an−1 +
a2n−1
for n ≥ 2.
an−2
The sequence {bn } is defined by
b0 = 1, b1 = 3, and bn = bn−1 +
b2n−1
for n ≥ 2.
bn−2
Find ab32
.
32
Problem 18.2.15 (AIME)
n +2009
The terms of the sequence (ai ) defined by an+2 = a1+a
for n ≥ 1 are positive inn+1
tegers. Find the minimum possible value of a1 + a2 .
Answers
18.1 840
18.2 12
18.3 898
18.2.1 50
99
18.2.2 0
18.2.3 658
18.2.4 13
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Chapter 18. Special Sequences
18.2.5 2.9
18.2.6 19
18.2.7 128
257
18.2.8 6
18.2.9 31
18.2.10 23
18.2.11 986
18.2.12 10
18.2.13 255
18.2.14 561
18.2.15 90
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Chapter 19
Mean, Median, Mode
Video Lecture
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Chapter 19. Mean, Median, Mode
Definition 19.0.1 (Mean/Average).
Mean = average of all terms =
sum of all terms
number of terms
Example 19.1 (AIME)
A finite set S of distinct real numbers has the following properties: the mean of S ∪ {1}
is 13 less than the mean of S, and the mean of S ∪ {2001} is 27 more than the mean of
S. Find the mean of S.
Video Solution
Definition 19.0.2 (Median). After arranging the numbers in increasing or decreasing order:
If number of terms is odd,
Median = middle number
If number of terms is even,
Median = average of middle two numbers
Definition 19.0.3 (Mode).
Mode = Most common term(s)
Remark 19.0.4
There could be multiple modes. If the problem says “unique mode”, it means that there
is only one mode.
Definition 19.0.5. Geometric Mean of numbers a1 , a2 , a3 , . . . , an
√
n
a1 × a2 × a3 · · · × an
Definition 19.0.6. Harmonic Mean of numbers a1 , a2 , a3 , . . . , an
=
1
1
+ a1 +···+ a1
a1
n
2
n
n
= 1
1
+ a2 + · · · + a1n
a1
Definition 19.0.7. Quadratic Mean of numbers a1 , a2 , a3 , . . . , an
q
=
a21 + a22 + a23 + · · · + a2n
n
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Chapter 19. Mean, Median, Mode
Example 19.2 (AMC 10)
What is the sum of all real numbers x for which the median of the numbers 4, 6, 8, 17,
and x is equal to the mean of those five numbers?
Video Solution
Example 19.3 (AMC 10)
When the mean, median, and mode of the list
10, 2, 5, 2, 4, 2, x
are arranged in increasing order, they form a non-constant arithmetic progression. What
is the sum of all possible real values of x?
Video Solution
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19.1
Chapter 19. Mean, Median, Mode
Practice Problems
Problem 19.1.1 (AMC 10)
The quiz scores of a class with k > 12 students have a mean of 8. The mean of a
collection of 12 of these quiz scores is 14. What is the mean of the remaining quiz scores
in terms of k?
Video Solution
Problem 19.1.2 (AMC 10)
Suppose that S is a finite set of positive integers. If the greatest integer in S is removed from S, then the average value (arithmetic mean) of the integers remaining is 32.
If the least integer in S is also removed, then the average value of the integers remaining
is 35. If the greatest integer is then returned to the set, the average value of the integers
rises to 40. The greatest integer in the original set S is 72 greater than the least integer
in S. What is the average value of all the integers in the set S?
Video Solution
Problem 19.1.3 (AMC 10)
The quiz scores of a class with k > 12 students have a mean of 8. The mean of a
collection of 12 of these quiz scores is 14. What is the mean of the remaining quiz scores
of terms of k?
Video Solution
Problem 19.1.4 (AMC 10/12)
Ms. Blackwell gives an exam to two classes. The mean of the scores of the students
in the morning class is 84, and the afternoon class’s mean score is 70. The raio of the
number of students in the morning class to the number of students in the afternoon class
is 34 . What is the mean of the scores of all the students?
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Video Solution
Problem 19.1.5 (AMC 10/12)
In the following list of numbers, the integer n appears n times in the list for 1 ≤ n ≤ 200.
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ·, 200, 200, ·, 200
What is the median of the numbers in this list?
Video Solution
Additional Problems
Problem 19.1.6 (AMC 10)
The numbers 3, 5, 7, a, and b have an average (arithmetic mean) of 15. What is the
average of a and b?
Problem 19.1.7 (AMC 10)
The mean, median, and mode of the 7 data values 60, 100, x, 40, 50, 200, 90 are all
equal to x. What is the value of x?
Problem 19.1.8 (AMC 12)
Every high school in the city of Euclid sent a team of 3 students to a math contest.
Each participant in the contest received a different score. Andrea’s score was the median
among all students, and hers was the highest score on her team. Andrea’s teammates
Beth and Carla placed 37th and 64th, respectively. How many schools are in the city?
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Chapter 19. Mean, Median, Mode
Problem 19.1.9 (AMC 10)
What is the median of the following list of 4040 numbers?
1, 2, 3, . . . , 2020, 12 , 22 , 32 , . . . , 20202
Problem 19.1.10 (AMC 10)
A list of 2018 positive integers has a unique mode, which occurs exactly 10 times.
What is the least number of distinct values that can occur in the list?
Problem 19.1.11 (AMC 10)
Melanie computes the mean µ, the median M , and the modes of the 365 values that are
the dates in the months of 2019. Thus her data consist of 12 1s, 12 2s, . . . , 12 28s,
11 29s, 11 30s, and 7 31s. Let d be the median of the modes. Which of the following
statements is true?
(A) µ < d < M
(B) M < d < µ
(C) d = M = µ
(D) d < M < µ
(E) d < µ < M
Problem 19.1.12 (AIME)
Let S be a list of positive integers–not necessarily distinct–in which the number 68
appears. The average (arithmetic mean) of the numbers in S is 56. However, if 68 is
removed, the average of the remaining numbers drops to 55. What is the largest number
that can appear in S?
Problem 19.1.13 (AMC 10)
The arithmetic mean of two distinct positive integers x and y is a two-digit integer.
The geometric mean of x and y is obtained by reversing the digits of the arithmetic mean.
What is |x − y|?
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Problem 19.1.14 (AIME)
A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let D be the
difference between the mode and the arithmetic mean of the sample. What is the largest
possible value of ⌊D⌋? (For real x, ⌊x⌋ is the greatest integer less than or equal to x.)
Answers
19.1 651
19.2 −5
19.3 20
19.1.1 8k−168
k−12
19.1.2 36.8
19.1.3 8k−168
k−12
19.1.4 76
19.1.5 142
19.1.6 30
19.1.7 90
19.1.8 23
19.1.9 1976.5
19.1.10 225
19.1.11 extbf (E)d < µ < M
19.1.12 649
19.1.13 66
19.1.14 947
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System of Equations
Video Lecture
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Example 20.1 (AMC 10)
Teams A and B are playing in a basketball league where each game results in a win for
one team and a loss for the other team. Team A has won 23 of its games and team B
has won 58 of its games. Also, team B has won 7 more games and lost 7 more games
than team A. How many games has team A played?
Video Solution
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20.1
Chapter 20. System of Equations
Work and Rate
Theorem 20.1.1
Work = Rate × Time
Equivalently,
Rate =
Work
Time
Time =
Work
Rate
Theorem 20.1.2
The if one person can do something is a amount of time, and someone else can do it in b
amount of time, together they can do it in
ab
a+b
time.
Remark 20.1.3
This not only applies to work. For example, if a problem says 2 faucets take a and b
hours to fill a tub, together they can fill a tub in
ab
a+b
hours.
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Example 20.2 (AIME)
Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe
can paint twice as fast as Abe. Abe begins to paint the room and works alone for the
first hour and a half. Then Bea joins Abe, and they work together until half the room is
painted. Then Coe joins Abe and Bea, and they work together until the entire room is
painted. Find the number of minutes after Abe begins for the three of them to finish
painting the room.
Video Solution
Example 20.3 (AIME)
The workers in a factory produce widgets and whoosits. For each product, production
time is constant and identical for all workers, but not necessarily equal for the two
products. In one hour, 100 workers can produce 300 widgets and 200 whoosits. In two
hours, 60 workers can produce 240 widgets and 300 whoosits. In three hours, 50 workers
can produce 150 widgets and m whoosits. Find m.
Video Solution
20.2
Practice Problems
Problem 20.2.1 (AMC 10)
In an after-school program for juniors and seniors there is a debate team with an
equal number of students from each class on the team. Among the 28 students on the
program, 25% of the juniors and 10% of the seniors are on the debate team. How many
juniors are in the program?
Video Solution
Problem 20.2.2 (AMC 10)
Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent
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coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection
is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?
Video Solution
Problem 20.2.3 (AMC 10)
The sum of two natural numbers is 17,402. One of the two numbers is divisible by
10. If the units digit of that number is erased, the other number is obtained. What is
the difference of these two numbers?
Video Solution
Problem 20.2.4 (AMC 10)
Pablo, Sofia, and Mia got some candy eggs at a party. Pablo had three times as
many eggs as Sofia, and Sofia had twice as many eggs as Mia. Pablo decides to give some
of his eggs to Sofia and Mia so that all three will have the same number of eggs. What
fraction of his eggs should Pablo give to Sofia?
Video Solution
Problem 20.2.5 (AMC 10/12)
The sum of two natural numbers is 17,402. One of the two numbers is divisible by
10. If the units digit of that number is erased, the other number is obtained. What is
the difference of these two numbers?
Video Solution
Problem 20.2.6 (AMC 10/12)
At a math contest, 57 students are wearing blue shirts, and another 75 students are
wearing yellow shirts. The 132 students are assigned into 66 pairs. In exactly 23 of
these pairs, both students are wearing blue shirts. In how many pairs are both students
wearing yellow shirts?
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Video Solution
Problem 20.2.7 (AMC 10/12)
When a student multiplied the number 66 by the repeating decimal
1.abab · · · = 1.ab
Where a and b are digits. He did not notice the notation and just multiplied 66 times
1.ab. Later he found that his answer is 0.5 less than the correct answer. What is the
2-digit integer ab?
Video Solution
Problem 20.2.8 (AMC 10/12B)
Suppose that S is a finite set of positive integers. If the greatest integer in S is removed from S, then the average value (arithmetic mean) of the integers remaining is 32.
IF the least integer in S is also removed, then the average value of the integers remaining
is 35. IF the greatest integer is then returned to the set, the average value of the integers
rises to 40. The greatest integer in the original set S is 72 greater than the least integer
in S. Wha tis the average value of all the integers in the set S?
Video Solution
Additional Problems
Problem 20.2.9 (AMC 10)
Ana and Bonita were born on the same date in different years, n years apart. Last year
Ana was 5 times as old as Bonita. This year Ana’s age is the square of Bonita’s age.
What is n?
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Problem 20.2.10 (AMC 10)
At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins,
triplets, and quadruplets accounted for 1000 of the babies born. There were four times
as many sets of triplets as sets of quadruplets, and there was three times as many sets of
twins as sets of triplets. How many of these 1000 babies were in sets of quadruplets?
Problem 20.2.11 (AMC 10)
Two jars each contain the same number of marbles, and every marble is either blue or
green. In Jar 1 the ratio of blue to green marbles is 9 : 1, and the ratio of blue to green
marbles in Jar 2 is 8 : 1. There are 95 green marbles in all. How many more blue marbles
are in Jar 1 than in Jar 2?
Problem 20.2.12 (AMC 10)
In the United States, coins have the following thicknesses: penny, 1.55 mm; nickel,
1.95 mm; dime, 1.35 mm; quarter, 1.75 mm. If a stack of these coins is exactly 14 mm
high, how many coins are in the stack?
Problem 20.2.13 (AMC 10)
Boris has an incredible coin-changing machine. When he puts in a quarter, it returns five
nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it
returns five quarters. Boris starts with just one penny. Which of the following amounts
could Boris have after using the machine repeatedly?
(A)3.63
(B)5.13
(C)6.30
(D)7.45
(E)9.07
Problem 20.2.14 (AMC 10)
One morning each member of Angela’s family drank an 8-ounce mixture of coffee with
milk. The amounts of coffee and milk varied from cup to cup, but were never zero.
Angela drank a quarter of the total amount of milk and a sixth of the total amount of
coffee. How many people are in the family?
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Problem 20.2.15 (AMC 12)
The state income tax where Kristin lives is levied at the rate of p% of the first $28000
of annual income plus (p + 2)% of any amount above $28000. Kristin noticed that the
state income tax she paid amounted to (p + 0.25)% of her annual income. What was her
annual income?
Problem 20.2.16 (AMC 10/12)
Paula the painter and her two helpers each paint at constant, but different, rates.
They always start at 8:00 AM, and all three always take the same amount of time to
eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM.
On Tuesday, when Paula wasn’t there, the two helpers painted only 24% of the house
and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by
working until 7:12 P.M. How long, in minutes, was each day’s lunch break?
Problem 20.2.17 (AMC 12)
A group of 12 pirates agree to divide a treasure chest of gold coins among themselves as
k
follows. The k th pirate to take a share takes 12
of the coins that remain in the chest. The
number of coins initially in the chest is the smallest number for which this arrangement
will allow each pirate to receive a positive whole number of coins. How many coins does
the 12th pirate receive?
Problem 20.2.18 (AIME)
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters
of a solution that is 48% acid. Jar C contains one liter of a solution that is k% acid.
From jar C, m
liters of the solution is added to jar A, and the remainder of the solution
n
in jar C is added to jar B. At the end both jar A and jar B contain solutions that are
50% acid. Given that m and n are relatively prime positive integers, find k + m + n.
Problem 20.2.19 (AIME)
In order to complete a large job, 1000 workers were hired, just enough to complete
the job on schedule. All the workers stayed on the job while the first quarter of the work
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was done, so the first quarter of the work was completed on schedule. Then 100 workers
were laid off, so the second quarter of the work was completed behind schedule. Then
an additional 100 workers were laid off, so the third quarter of the work was completed
still further behind schedule. Given that all workers work at the same rate, what is the
minimum number of additional workers, beyond the 800 workers still on the job at the
end of the third quarter, that must be hired after three-quarters of the work has been
completed so that the entire project can be completed on schedule or before?
Problem 20.2.20 (AIME)
A group of clerks is assigned the task of sorting 1775 files. Each clerk sorts at a
constant rate of 30 files per hour. At the end of the first hour, some of the clerks are
reassigned to another task; at the end of the second hour, the same number of the
remaining clerks are also reassigned to another task, and a similar assignment occurs
at the end of the third hour. The group finishes the sorting in 3 hours and 10 minutes.
Find the number of files sorted during the first one and a half hours of sorting.
Answers
20.1 42
20.2 334
20.3 450
20.2.1 8
20.2.2 2
20.2.3 14, 238
20.2.4 1/6
20.2.5 14, 238
20.2.6 32
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20.2.7 75
20.2.8 36.8
20.2.9 12
20.2.10 100
20.2.11 5
20.2.12 8
20.2.13 $7.45
20.2.14 5
20.2.15 32000
20.2.16 48
20.2.17 1925
20.2.18 85
20.2.19 766
20.2.20 945
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Chapter 21
Speed, Distance, and Time
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Chapter 21. Speed, Distance, and Time
Theorem 21.0.1
Distance = Speed × Time
Equivalently,
Speed =
Distance
Time
Time =
Distance
Speed
Theorem 21.0.2
Average Speed =
Total Distance
Total Time
Remark 21.0.3
A common mistake is to assume that average speed is the averages of all speeds (especially
when the distance you are traveling at each of those speeds are the same). Remember,
that’s not true unless you are traveling at those speeds for the same amount of time!
Example 21.1 (AMC 10)
Henry decides one morning to do a workout, and he walks 34 of the way from his home to
his gym. The gym is 2 kilometers away from Henry’s home. At that point, he changes his
mind and walks 34 of the way from where he is back toward home. When he reaches that
point, he changes his mind again and walks 34 of the distance from there back toward the
gym. If Henry keeps changing his mind when he has walked 34 of the distance toward
either the gym or home from the point where he last changed his mind, he will get very
close to walking back and forth between a point A kilometers from home and a point B
kilometers from home. What is |A − B|?
Video Solution
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Example 21.2 (AIME)
Al walks down to the bottom of an escalator that is moving up and he counts 150 steps.
His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al’s speed
of walking (in steps per unit time) is three times Bob’s walking speed, how many steps
are visible on the escalator at a given time? (Assume that this value is constant.)
Video Solution
Example 21.3 (AIME)
A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps
onto the start of the walkway and stands. Bob steps onto the start of the walkway two
seconds later and strolls forward along the walkway at a constant rate of 4 feet per
second. Two seconds after that, Cy reaches the start of the walkway and walks briskly
forward beside the walkway at a constant rate of 8 feet per second. At a certain time,
one of these three persons is exactly halfway between the other two. At that time, find
the distance in feet between the start of the walkway and the middle person.
Video Solution
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21.1
Chapter 21. Speed, Distance, and Time
Practice Problems
Problem 21.1.1 (AMC 10)
A cart rolls down a hill, travelling 5 inches the first second and accelerating so that
during each successive 1-second time interval, it travels 7 inches more than during the
previous 1-second interval. The cart takes 30 seconds to reach the bottom of the hill.
How far, in inches, does it travel?
Video Solution
Problem 21.1.2 (AMC 10)
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at 4 miles per hour.
Halfway to the tower, the trail becomes really steep, and Chantal slows down to 2 miles
per hour. After reaching the tower, she immediately turns around and descends the steep
part of the trail at 3 miles per hour. She meets Jean at the halfway point. What was
Jean’s average speed, in miles per hour, until they meet?
Video Solution
Problem 21.1.3 (MATHCOUNTS)
A car passes point A driving at a constant rate of 60 km per hour. A second car,
traveling at a constant rate of 75 km per hour, passes the same point A a while later and
then follows the first car. It catches the first car after traveling a distance of 75 km past
point A. How many minutes after the first car passed point A did the second car pass
point A?
Video Solution
Problem 21.1.4 (AMC 10)
A cart rolls down a hill, travelling 5 inches the first second and accelerating so that
during each successive 1-second time interval, it travels 7 inches more than during the
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previous 1-second interval. The cart takes 30 seconds to reach the bottom of the hill.
How far, in inches, does it travel?
Video Solution
Problem 21.1.5 (AMC 10/12)
Chantal and Jean start hiking from a trailhead toward a fire tower. Jean is wearing a heavy backpack and walks slower. Chantal starts walking at 4 miles per hour.
Halfway to the tower, the trail becomes really steep, and Chantal slows down to 2 miles
per hour. After reaching the tower, she immediately turns around and descends the steep
part of the trail at 3 miles per hour. She meets Jean at the halfway point. What was
Jean’s average speed, in miles per hour, until they meet?
Video Solution
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Additional Problems
Problem 21.1.6 (AMC 10)
Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes
was 60 mph (miles per hour), and his average speed during the second 30 minutes was 65
mph. What was his average speed, in mph, during the last 30 minutes?
Problem 21.1.7 (AMC 10)
Samia set off on her bicycle to visit her friend, traveling at an average speed of 17
kilometers per hour. When she had gone half the distance to her friend’s house, a tire
went flat, and she walked the rest of the way at 5 kilometers per hour. In all it took her
44 minutes to reach her friend’s house. In kilometers rounded to the nearest tenth, how
far did Samia walk?
Problem 21.1.8 (AMC 10)
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its
battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip,
using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles
per gallon. How long was the trip in miles?
Problem 21.1.9 (AIME)
Rudolph bikes at a constant rate and stops for a five-minute break at the end of every
mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph
bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and
Rudolph begin biking at the same time and arrive at the 50-mile mark at exactly the
same time. How many minutes has it taken them?
Problem 21.1.10 (AMC 12)
Andrea and Lauren are 20 kilometers apart. They bike toward one another with Andrea
traveling three times as fast as Lauren, and the distance between them decreasing at a
rate of 1 kilometer per minute. After 5 minutes, Andrea stops biking because of a flat
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tire and waits for Lauren. After how many minutes from the time they started to bike
does Lauren reach Andrea?
Problem 21.1.11 (AMC 12)
While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water
comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in
more than 30 gallons of water. Steve starts rowing towards the shore at a constant rate
of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate,
in gallons per minute, at which LeRoy can bail if they are to reach the shore without
sinking?
Problem 21.1.12 (AIME)
Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers 74 kilometers after
biking for 2 hours, jogging for 3 hours, and swimming for 4 hours, while Sue covers 91
kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours. Their
biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find
the sum of the squares of Ed’s biking, jogging, and swimming rates.
Problem 21.1.13 (AIME)
Ana, Bob, and Cao bike at constant rates of 8.6 meters per second, 6.2 meters per
second, and 5 meters per second, respectively. They all begin biking at the same time
from the northeast corner of a rectangular field whose longer side runs due west. Ana
starts biking along the edge of the field, initially heading west, Bob starts biking along
the edge of the field, initially heading south, and Cao bikes in a straight line across the
field to a point D on the south edge of the field. Cao arrives at point D at the same
time that Ana and Bob arrive at D for the first time. The ratio of the field’s length to
the field’s width to the distance from point D to the southeast corner of the field can be
represented as p : q : r, where p, q, and r are positive integers with p and q relatively
prime. Find p + q + r.
Problem 21.1.14 (AIME)
Jon and Steve ride their bicycles along a path that parallels two side-by-side train
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tracks running the east/west direction. Jon rides east at 20 miles per hour, and Steve
rides west at 20 miles per hour. Two trains of equal length, traveling in opposite directions
at constant but different speeds each pass the two riders. Each train takes exactly 1
minute to go past Jon. The westbound train takes 10 times as long as the eastbound
train to go past Steve. The length of each train is m
miles, where m and n are relatively
n
prime positive integers. Find m + n.
Answers
21.1 1 15
21.2 120
21.3 52
21.1.1 3195
12
21.1.2 13
21.1.3 15min
21.1.4 3195
12
21.1.5 13
21.1.6 67
21.1.7 2.8
21.1.8 440
21.1.9 620
21.1.10 65
21.1.11 8
21.1.12 314
21.1.13 061
21.1.14 49
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Number Theory
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Chapter 22
Primes and Factors
Video Lecture
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Chapter 22. Primes and Factors
Definition 22.0.1 (Primes). Primes are numbers that have exactly two factors: 1 and the
number itself. Ex. 2, 3, 5, 7, 11, 13, 17, 19, 23, etc. are all primes
Remark 22.0.2
In order to check whether a number n is prime, we need to check all the primes that are
less than or equal to
√
n
Concept 22.0.3 (Prime Factorization)
Prime factorization is a way to express each number as a product of primes.
Example: prime factorization of 60 is 22 × 3 × 5
Example 22.1 (AMC 10)
Kiana has two older twin brothers. The product of their three ages is 128. What is the
sum of their three ages?
Video Solution
Theorem 22.0.4 (Number of Factors of a Number)
If the prime factorization of the number is expressed as:
pe11 × pe22 × · · · × pekk
then the number of factors of this number is
(e1 + 1)(e2 + 1) . . . (ek + 1)
Remark 22.0.5
Basically, in order to find the number of factors of a number:
1. Find the prime factorization of the number
2. Add 1 to all of the exponents
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3. Multiply them together
Example 22.2 (AIME)
Let n = 231 319 . How many positive integer divisors of n2 are less than n but do not
divide n?
Video Solution
Example 22.3 (AIME)
Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive
n
integral divisors, including 1 and itself. Find 75
.
Video Solution
22.1
Sum of Factors
Theorem 22.1.1 (Sum of Factors of a Number)
If the prime factorization of the number is expressed as:
pe11 × pe22 × · · · × pekk
then the number of factors of this number is
(1+p11 +p21 +· · ·+pe11 −1 +pe11 )(1+p12 +p22 +· · ·+pe22 −1 +pe22 ) . . . (1+p1k +p2k +· · ·+pkek −1 +pekk )
Concept 22.1.2 (Sum of Factors of a Number)
Essentially, for a prime p in the prime factorization, first find the sum of pk for all
possible exponents k in the prime factorization. Then, we will multiply all such sums for
all of the primes to get the sum of all the factors of the number.
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Chapter 22. Primes and Factors
Product of Factors
Theorem 22.2.1 (Product of Factors of a Number)
The product of factors of a number n where it has f factors (this can be calculated using
f
the number of factors formula) is n 2
Example 22.4 (AMC 12)
For n a positive integer, let f (n) be the quotient obtained when the sum of all positive
divisors of n is divided by n. For example,
f (14) = (1 + 2 + 7 + 14) ÷ 14 =
12
7
What is f (768) − f (384)?
Video Solution
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Chapter 22. Primes and Factors
Practice Problems
Problem 22.3.1 (AMC 10)
A positive integer divisor of 12! is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as m
, where m and n are relatively
n
prime positive integers. What is m + n?
Video Solution
Problem 22.3.2 (AMC 10)
A single bench section at a school event can hold either 7 adults or 11 children. When N
bench sections are connected end to end, an equal number of adults and children seated
together will occupy all the bench space. What is the least possible positive integer value
of N ?
(A) 9
(B) 18
(C) 27
(D) 36
(E) 77
Video Solution
Problem 22.3.3 (AMC 10)
For how many (not necessarily positive) integer values of n is the value of 4000 ·
integer?
n
2
5
an
Video Solution
Problem 22.3.4 (AMC 10)
The numbers from 1 to 8 are placed at the vertices of a cube in such a manner that the
sum of the four numbers on each face is the same. What is this common sum?
Video Solution
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Problem 22.3.5 (AMC 10)
How many positive even multiples of 3 less than 2020 are perfect squares?
Video Solution
Problem 22.3.6 (AMC 10)
How many positive integer divisors of 2019 are perfect squares or perfect cubes (or
both)?
Video Solution
Problem 22.3.7 (AMC 12)
What is the sum of the exponents of the prime factors of the square root of the largest
perfect square that divides 12! ?
Video Solution
Problem 22.3.8 (AMC 10)
How many ways are there to write 2016 as the sum of twos and threes, ignoring order?
(For example, 1008 · 2 + 0 · 3 and 402 · 2 + 404 · 3 are two such ways.)
Video Solution
Problem 22.3.9 (AIME)
Find the number of positive integers with three not necessarily distinct digits, abc,
with a ̸= 0 and c ̸= 0 such that both abc and cba are multiples of 4.
Video Solution
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Problem 22.3.10 (AMC 10)
For some positive integer n, the number 110n3 has 110 positive integer divisors, including 1 and the number 110n3 . How many positive integer divisors does the number
81n4 have?
Video Solution
Problem 22.3.11 (AMC 10)
Let S be the set of all positive integer divisors of 100, 000. How many numbers are
the product of two distinct elements of S?
Video Solution
Problem 22.3.12 (AMC 10)
The 25 integers from −10 to 14, inclusive, can be arranged to form a 5-by-5 square in
which the sum of the numbers in each row, the sum of the numbers in each column, and
the sum of the numbers along each of the main diagonals are all the same. What is the
value of this common sum?
Video Solution
Problem 22.3.13 (AMC 10)
All the numbers 2, 3, 4, 5, 6, 7 are assigned to the six faces of a cube, one number to each
face. For each of the eight vertices of the cube, a product of three numbers is computed,
where the three numbers are the numbers assigned to the three faces that include that
vertex. What is the greatest possible value of the sum of these eight products?
Video Solution
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Problem 22.3.14 (AMC 12)
How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?
Video Solution
Problem 22.3.15 (AMC 10)
How many positive two-digit integers are factors of 224 − 1?
Video Solution
Problem 22.3.16 (AIME)
There is a prime number p such that 16p + 1 is the cube of a positive integer. Find p.
Video Solution
Problem 22.3.17 (AMC 10)
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is 1 year
older than Chloe, and Zoe is exactly 1 year old today. Today is the first of the 9 birthdays
on which Chloe’s age will be an integral multiple of Zoe’s age. What will be the sum of
the two digits of Joey’s age the next time his age is a multiple of Zoe’s age?
Video Solution
Problem 22.3.18 (AMC 10)
What is the tens digit of 20152016 − 2017?
Video Solution
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Problem 22.3.19 (AMC 10)
Let N = 123456789101112 . . . 4344 be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when
N is divided by 45?
Video Solution
Problem 22.3.20 (AMC 10)
What is the hundreds digit of (20! − 15!)?
Video Solution
Problem 22.3.21 (AIME)
Find the number of ordered pairs of positive integers (m, n) such that m2 n = 2020 .
Video Solution
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Chapter 22. Primes and Factors
Problem 22.3.22 (AMC 10/12)
Let N = 34 · 34 · 63 · 270. What is the ratio of the sum of the odd divisors of N
to the sum of the even divisors of N ?
Video Solution
Additional Problems
Problem 22.3.23 (AMC 12)
In multiplying two positive integers a and b, Ron reversed the digits of the two-digit
number a. His erroneous product was 161. What is the correct value of the product of a
and b?
Problem 22.3.24 (AMC 10)
Positive integers a, b, and 2009, with a < b < 2009, form a geometric sequence with an
integer ratio. What is a?
Problem 22.3.25 (AMC 10)
How many positive cubes divide 3! · 5! · 7! ?
Problem 22.3.26 (AMC 12)
Let N = 34 · 34 · 63 · 270. What is the ratio of the sum of the odd divisors of N
to the sum of the even divisors of N ?
Problem 22.3.27 (AMC 10)
Suppose that m and n are positive integers such that 75m = n3 . What is the min-
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Chapter 22. Primes and Factors
imum possible value of m + n?
Problem 22.3.28 (AMC 10)
Elmo makes N sandwiches for a fundraiser. For each sandwich he uses B globs of
peanut butter at 4¢per glob and J blobs of jam at 5¢per blob. The cost of the peanut
butter and jam to make all the sandwiches is $2.53. Assume that B, J, and N are positive
integers with N > 1. What is the cost of the jam Elmo uses to make the sandwiches?
Problem 22.3.29 (AMC 12)
The largest prime factor of 16384 is 2 because 16384 = 214 . What is the sum of
the digits of the greatest prime number that is a divisor of 16383?
Problem 22.3.30 (AMC 12)
How many ways are there to paint each of the integers 2, 3, . . . , 9 either red, green,
or blue so that each number has a different color from each of its proper divisors?
Problem 22.3.31 (AMC 10)
The positive integers A, B, A − B, and A + B are all prime numbers. The sum of
these four primes is
(A) even
(B) divisible by 3
(C) divisible by 5
(D) divisible by 7
(E) prime
Problem 22.3.32 (AMC 10)
Given that 38 · 52 = ab , where both a and b are positive integers, find the smallest
possible value for a + b.
Problem 22.3.33 (AIME)
How many positive integer divisors of 20042004 are divisible by exactly 2004 positive
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Chapter 22. Primes and Factors
integers?
Problem 22.3.34 (AIME)
Find the number of positive integers that are divisors of at least one of 1010 , 157 , 1811 .
Problem 22.3.35 (AIME)
Let S be the sum of all numbers of the form a/b, where a and b are relatively prime
positive divisors of 1000. What is the greatest integer that does not exceed S/10?
Problem 22.3.36 (AMC 12)
For each positive integer n, let f1 (n) be twice the number of positive integer divisors of n, and for j ≥ 2, let fj (n) = f1 (fj−1 (n)). For how many values of n ≤ 50 is
f50 (n) = 12?
Problem 22.3.37 (AMC 10)
For each positive integer n > 1, let P (n) denote the greatest
prime factor of√n. For how
√
many positive integers n is it true that both P (n) = n and P (n + 48) = n + 48?
Answers
22.1 18
22.2 589
22.3 432
1
22.4 192
22.3.1 23
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22.3.2 18
22.3.3 9
22.3.4 18
22.3.5 7
22.3.6 37
22.3.7 8
22.3.8 337
22.3.9 040
22.3.10 325
22.3.11 117
22.3.12 10
22.3.13 729
22.3.14 96
22.3.15 12
22.3.16 307
22.3.17 11
22.3.18 0
22.3.19 9
22.3.20 0
22.3.21 231
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22.3.22 1 : 14
22.3.23 224
22.3.24 41
22.3.25 6
22.3.26 1 : 14
22.3.27 60
22.3.28 $1.65
22.3.29 10
22.3.30 432
22.3.31 prime
22.3.32 407
22.3.33 54
22.3.34 435
22.3.35 248
22.3.36 10
22.3.37 1
252
Chapter 23
Divisibility & Legendre’s Formula
Video Lecture
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Chapter 23. Divisibility & Legendre’s Formula
Concept 23.0.1 (Divisibility Rules)
2
Last digit is even
3
Sum of digits is divisible by 3
4
Last 2 digits divisible by 4
5
Last digit is 0 or 5
6
Divisible by 2 and 3
7
Take out factors of 7 until you reach a small number that is either
divisible or not divisible by 7
8
Last 3 digits are divisible by 8
9
Sum of digits is divisible by 9
10
Last digit is 0
11
Calculate the sum of odd digits (O) and even digits (E). If |O − E|
is divisible by 11, then the number is also divisible by 11
12
Divisible by 3 and 4
15
Divisible by 3 and 5
Example 23.1 (AMC 10)
Let n denote the smallest positive integer that is divisible by both 4 and 9, and whose
base-10 representation consists of only 4’s and 9’s, with at least one of each. What are
the last four digits of n?
Video Solution
Theorem 23.0.2 (Legendre’s Theorem)
$ %
$
%
n
n
vp (n!) = Number of powers of p in n! =
+ 2 + ...
p
p
Example 23.2 (AIME)
Let N be the number of consecutive 0’s at the right end of the decimal representation of
the product 1!2!3!4! · · · 99!100!. Find the remainder when N is divided by 1000.
Video Solution
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Chapter 23. Divisibility & Legendre’s Formula
Example 23.3 (AMC 10)
The base-ten representation for 19! is 121, 6T 5, 100, 40M, 832, H00, where T , M , and H
denote digits that are not given. What is T + M + H?
Video Solution
23.1
Practice Problems
Problem 23.1.1 (AMC 10)
A rectangular floor that is 10 feet wide and 17 feet long is tiled with 170 one-foot
square tiles. A bug walks from one corner to the opposite corner in a straight line.
Including the first and the last tile, how many tiles does the bug visit?
Video Solution
Problem 23.1.2 (AMC 12)
How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?
Video Solution
Problem 23.1.3 (AMC 10)
For all integers n ≥ 9, the value of
(n + 2)! − (n + 1)!
n!
is always which of the following?
(A) a multiple of 4
(B) a multiple of 10
(D) a perfect square
(E) a perfect cube
(C) a prime number
Video Solution
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Chapter 23. Divisibility & Legendre’s Formula
Problem 23.1.4 (IMO)
Prove that the fraction 21n+4
is irreducible for every natural number n.
14n+3
Video Solution
Problem 23.1.5 (AMC 10)
What is that largest positive integer n for which n3 + 100 is divisible by n + 10?
Video Solution
Problem 23.1.6 (AMC 10)
Call a positive integer an uphill integer if every digit is strictly greater than the previous
digit. For example, 1357, 89, 5 are all uphill integers but 32, 1240, 466 are not. How
many uphill integers are divisible by 15?
Video Solution
Additional Problems
Problem 23.1.7 (AMC 10)
What is the largest integer that is a divisor of
(n + 1)(n + 3)(n + 5)(n + 7)(n + 9)
for all positive even integers n?
Problem 23.1.8 (AMC 12)
The number 21! = 51, 090, 942, 171, 709, 440, 000 has over 60, 000 positive integer di-
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Chapter 23. Divisibility & Legendre’s Formula
visors. One of them is chosen at random. What is the probability that it is odd?
Problem 23.1.9 (AMC 10)
Call a positive integer an uphill integer if every digit is strictly greater than the previous
digit. For example, 1357, 89, and 5 are all uphill integers, but 32, 1240, and 466 are not.
How many uphill integers are divisible by 15?
Problem 23.1.10 (AMC 10)
Let N = 123456789101112 . . . 4344 be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when
N is divided by 45?
Problem 23.1.11 (AMC 10)
Let n denote the smallest positive integer that is divisible by both 4 and 9, and whose
base-10 representation consists of only 4’s and 9’s, with at least one of each. What are
the last four digits of n?
Problem 23.1.12 (AMC 10)
A finite sequence of three-digit integers has the property that the tens and units digits of
each term are, respectively, the hundreds and tens digits of the next term, and the tens
and units digits of the last term are, respectively, the hundreds and tens digits of the
first term. For example, such a sequence might begin with the terms 247, 475, and 756
and end with the term 824. Let S be the sum of all the terms in the sequence. What is
the largest prime factor that always divides S?
Problem 23.1.13 (AMC 10)
For how many positive integers n ≤ 1000 is
998
999
1000
+
+
n
n
n
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not divisible by 3? (Recall that ⌊x⌋ is the greatest integer less than or equal to x.)
Problem 23.1.14 (AIME)
Let a10 = 10, and for each positive integer n > 10 let an = 100an−1 + n. Find the
least positive n > 10 such that an is a multiple of 99.
Problem 23.1.15 (AIME)
Define n!! to be n(n − 2)(n − 4) · · · 3 · 1 for n odd and n(n − 2)(n − 4) · · · 4 · 2 for
n even. When
2009
X (2i − 1)!!
(2i)!!
i=1
is expressed as a fraction in lowest terms, its denominator is 2a b with b odd. Find
ab
.
10
Answers
23.1 4944
23.2 124
23.3 12
23.1.1 26
23.1.2 96
23.1.3 (D)aperf ectsquare
23.1.4 N A
23.1.5 890
23.1.6 6
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23.1.7 15
1
23.1.8 19
23.1.9 6
23.1.10 9
23.1.11 4944
23.1.12 37
23.1.13 22
23.1.14 45
23.1.15 401
259
Chapter 24
GCD & LCM
Video Lecture
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Chapter 24. GCD & LCM
Definition 24.0.1 (Greatest Common Divisor). The Greatest Common Divisor (GCD) of
two or more non-0 integers is the largest positive integer that divides each of the integers.
Note: This is also known as GCF (Greatest Common Factor), and the terms GCF and GCD
are often used interchangeably.
Definition 24.0.2 (Least Common Multiple). The Least Common Multiple (LCM) of two
or more non-0 integers is the smallest positive integer that is divisible by both the numbers.
Concept 24.0.3
GCD/LCM Greatest common divisor of m and n = GCD(m, n) can be found by taking
the lowest prime exponents from the prime factorizations of m and n.
Least common multiple of m and n = LCM (m, n) can be found by taking the highest prime exponents from the prime factorizations of m and n.
Theorem 24.0.4
The product of GCD and LCM of two numbers is equal to the product of the two
numbers:
GCD(m, n) · LCM (m, n) = m · n
Theorem 24.0.5
If two numbers have a common factor c, then
gcd(ac, bc) = c · gcd(a, b)
Example 24.1 (Omega Learn)
Two numbers a and b satisfy the condition
lcm(gcd(24, a), gcd(a, b)) = lcm(gcd(24, b), gcd(a, b))
If a = 2b, what is the smallest possible value of a + b?
Video Solution
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Chapter 24. GCD & LCM
Theorem 24.0.6 (Euclidean Algorithm)
The Euclidean algorithm states that
gcd(x, y) = gcd(x − ky, y)
where x > y and k is a positive integer.
Remark 24.0.7
We can apply the Euclidean Algorithm multiple times to easily find the GCD of large
numbers since after applying the Euclidean algorithm, we know have 2 smaller numbers
which we can apply the Euclidean Algorithm again until we get 2 very small numbers.
For example,
gcd(186, 92) = gcd((186 − (2 · 92)), 92)
= gcd(2, 92)
= gcd(2, (92 − (2 · 46)))
= gcd(2, 0)
=2
Example 24.2 (AIME)
The numbers in the sequence 101, 104, 109, 116,. . . are of the form an = 100 + n2 , where
n = 1, 2, 3, . . . For each n, let dn be the greatest common divisor of an and an+1 . Find
the maximum value of dn as n ranges through the positive integers.
Video Solution
Theorem 24.0.8 (Bezout’s Identity)
Integer solutions to the equation
ax + by = c
will only exist if and only if gcd(a, b) divides c
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24.1
Chapter 24. GCD & LCM
Practice Problems
Problem 24.1.1 (AMC 10)
Let a, b, c, and d be positive integers such that gcd(a, b) = 24, gcd(b, c) = 36, gcd(c, d) =
54, and 70 < gcd(d, a) < 100. Which of the following must be a divisor of a?
(A) 5
(B) 7
(C) 11
(D) 13
(E) 17
Video Solution
Problem 24.1.2 (IMO)
is irreducible for every natural number n.
Prove that the fraction 21n+4
14n+3
Video Solution
Problem 24.1.3 (AIME)
What is that largest positive integer n for which n3 + 100 is divisible by n + 10?
Video Solution
Problem 24.1.4 (AMC 10)
How many ordered pairs (a, b) of positive integers satisfy the equation
a · b + 63 = 20 · lcm(a, b) + 12 · gcd(a, b),
where gcd(a, b) denotes the greatest common divisor of a and b, and lcm(a, b) denotes
their least common multiple?
Video Solution
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Additional Problems
Problem 24.1.5 (AMC 12)
Let N be the second smallest positive integer that is divisible by every positive integer less than 7. What is the sum of the digits of N ?
Problem 24.1.6 (AMC 10)
There are 10 horses, named Horse 1, Horse 2, . . . , Horse 10. They get their
names from how many minutes it takes them to run one lap around a circular race track:
Horse k runs one lap in exactly k minutes. At time 0 all the horses are together at the
starting point on the track. The horses start running in the same direction, and they
keep running around the circular track at their constant speeds. The least time S > 0,
in minutes, at which all 10 horses will again simultaneously be at the starting point is
S = 2520. Let T > 0 be the least time, in minutes, such that at least 5 of the horses are
again at the starting point. What is the sum of the digits of T ?
Problem 24.1.7 (AMC 12)
Suppose a, b, c are positive integers such that
a + b + c = 23
and
gcd(a, b) + gcd(b, c) + gcd(c, a) = 9.
What is the sum of all possible distinct values of a2 + b2 + c2 ?
Problem 24.1.8 (AMC 12)
Let N be the second smallest positive integer that is divisible by every positive integer less than 7. What is the sum of the digits of N ?
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Problem 24.1.9 (AMC 12)
Let n be the smallest positive integer such that n is divisible by 20, n2 is a perfect
cube, and n3 is a perfect square. What is the number of digits of n?
Problem 24.1.10 (AMC 12)
Mary chose an even 4-digit number n. She wrote down all the divisors of n in inn
creasing order from left to right: 1, 2, . . . , , n. At some moment Mary wrote 323 as a
2
divisor of n. What is the smallest possible value of the next divisor written to the right
of 323?
Problem 24.1.11 (AMC 12)
How many positive integers n are there such that n is a multiple of 5, and the least
common multiple of 5! and n equals 5 times the greatest common divisor of 10! and n?
Problem 24.1.12 (AIME)
For how many ordered pairs of positive integers (x, y), with y < x ≤ 100, are both
x
and x+1
integers?
y
y+1
Problem 24.1.13 (AMC 10)
Let n be the least positive integer greater than 1000 for which
gcd(63, n + 120) = 21 and
gcd(n + 63, 120) = 60.
What is the sum of the digits of n?
Problem 24.1.14 (AIME)
Consider the sequence (ak )k≥1 of positive rational numbers defined by a1 = 2020
and for
2021
k ≥ 1, if ak = m
for
relatively
prime
positive
integers
m
and
n,
then
n
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Chapter 24. GCD & LCM
m + 18
.
n + 19
Determine the sum of all positive integers j such that the rational number aj can be
t
written in the form t+1
for some positive integer t.
ak+1 =
Problem 24.1.15 (AIME)
Find the number of ordered pairs (m, n) such that m and n are positive integers in
the set {1, 2, ..., 30} and the greatest common divisor of 2m + 1 and 2n − 1 is not 1.
Answers
24.1 24
24.2 401
24.1.1 13
24.1.2 −N A−
24.1.3 890
24.1.4 2
24.1.5 3
24.1.6 3
24.1.7 438
24.1.8 13
24.1.9 7
24.1.10 340
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24.1.11 48
24.1.12 85
24.1.13 18
24.1.14 059
24.1.15 295
267
Chapter 25
Modular Arithmetic
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Chapter 25. Modular Arithmetic
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Chapter 25. Modular Arithmetic
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Chapter 25. Modular Arithmetic
Video Lectures
Modular Arithmetic Euler’s & Binomial Theorems Chinese Remainder Theorem & Line
Definition 25.0.1.
n ≡ a (mod b)
means the number ’n’ leaves the same remainder as ’a’ when divided by b
Theorem 25.0.2
If a = x (mod n) and b ≡ y (mod n), then
ab ≡ xy
(mod n)
am ≡ x m
(mod n)
Theorem 25.0.3
If a ≡ x (mod n), then
Example 25.1 (AIME)
Find the remainder when 9 × 99 × 999 × · · · × 99
· · · 9} is divided by 1000.
| {z
999 9’s
Video Solution
Example 25.2 (AMC 12)
Let S be a subset of {1, 2, 3, . . . , 30} with the property that no pair of distinct elements
in S has a sum divisible by 5. What is the largest possible size of S?
Video Solution
Concept 25.0.4 (Digit Cycles)
To calculate large digit(s) of a number ab , a strategy that may work is to just look for a
pattern by computing the first few values of ab and then seeing that the pattern will
repeat for large values of b.
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Example 25.3 (AMC 12)
Let k = 20082 + 22008 . What is the units digit of k 2 + 2k ?
Video Solution
Theorem 25.0.5 (Euler’s Totient Function)
If number n has the prime factorization
pe11 · pe22 · pe33 . . . penn
then
1
ϕ(n) = n · 1 −
p1
1
1
1−
... 1 −
p2
pn
!
!
!
where ϕ(n) denotes the number of positive integers less than or equal to n that are
relatively prime to n.
Steps to find totient of a number
1. Find prime factorization
2. For all primes, calculate and multiply
1−
1
pi
3. Multiply this product to the number n to get the totient
Theorem 25.0.6 (Euler’s Totient Theorem)
aϕ(n) ≡ 1
if and only if
(mod n)
gcd(a, n) = 1
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Chapter 25. Modular Arithmetic
Corollary 25.0.7 (Fermat’s Little Theorem)
ap−1 ≡ 1
if an only if p is a prime and
(mod p)
gcd(a, n) = 1
Concept 25.0.8 (Modular Inverses)
If a is denoted as the modular inverse of b (mod n), then
ab ≡ 1
We also write that
(mod n)
a−1 ≡ b (mod n)
since a and b are inverses (mod n).
Theorem 25.0.9 (Wilson’s Theorem)
(p − 1)! ≡ p − 1 ≡ −1
(mod p)
Example 25.4 (AIME)
Let an = 6n + 8n . Determine the remainder upon dividing a83 by 49.
Video Solution
Theorem 25.0.10 (Binomial Theorem)
For non-negative n,
!
!
!
!
n n 0
n n−1
n n−2 2
n 0 n
(x + y) =
x y +
x y+
x y + ··· +
xy
0
1
2
n
n
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Chapter 25. Modular Arithmetic
Example 25.5 (AMC 10)
What is the hundreds digit of 20112011 ?
Video Solution
Theorem 25.0.11 (Chinese Remainder Theorem)
If a positive number x satisfies
x ≡ a1
(mod n1 )
x ≡ a2
(mod n2 )
..
.
x ≡ ak
(mod nk )
where all ni are relatively prime, then x has a unique solution (mod n1 · n2 · n3 . . . nk )
Remark 25.0.12
Be careful! This may not necessarily be true if any ni share common factors as then
congruences might contradict each other.
Concept 25.0.13 (Solving Linear Congruences)
To solve a linear congruence with 2 congruences you can either
• Guess and Check until you reach a value that works and satisfies both mods
• Algebraic Method
1. Find 2 congruences
n ≡ r1
(mod m1 )
n ≡ r2
(mod m2 )
such that m1 and m2 are relatively prime
2. Rewrite them algebraically
n = k(m1 ) + r1
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n = j(m2 ) + r2
3. Set them equal mod the smaller of m1 and m2 (in this case, say m1 < m2 )
k(m1 ) + r1 ≡ r2
(mod m2 ) =⇒ m1 ≡ (r2 − r1 ) · k −1
(mod m2 )
4. Guess and check to find the value of
k −1
(mod m2 )
5. Using the value of what b is (mod d), rewrite it algebraically.
6. Substitute it back into the expression
n = k(m1 ) + r1
7. Convert it back to mods to get the final congruence
Concept 25.0.14
The solution to
is
n ≡ r1
(mod m1 )
n ≡ r2
(mod m2 )
n ≡ r1 + m1 (r2 − r1 ) · i
where i ≡ m−1
(mod m2 )
1
Remark 25.0.15
To solve a general congruence of more than 2 congruences, just solve them 2 at a time
until you are left with just 1 congruence.
Example 25.6
Find the smallest positive integer greater than 100 that leaves a remainder of 5 when
divided by 9 and leaves a remainder of 4 when divided by 17.
Video Solution
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Chapter 25. Modular Arithmetic
Example 25.7 (AIME)
The positive integers N and N 2 both end in the same sequence of four digits abcd when
written in base 10, where digit a is not zero. Find the three-digit number abc.
Video Solution
25.1
Practice Problems
Problem 25.1.1
Find the remainder 369 when it’s divided by 5.
Video Solution
Problem 25.1.2
Find the remainder 349 when it’s divided by 5.
Video Solution
Problem 25.1.3
Find the remainder when 7803 is divided by 400.
Video Solution
Problem 25.1.4 (AMC 10)
An integer N is selected at random in the range 1 ≤ N ≤ 2020 . What is the probability
that the remainder when N 16 is divided by 5 is 1?
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Video Solution
Problem 25.1.5 (PUMAC)
If p, q and r are primes with pqr = 7(p + q + r), find p + q + r
Video Solution
Problem 25.1.6
Find the smallest positive number that leaves a remainder of 11 when divided by
13 and a remainder of 2 when divided by 5.
Problem 25.1.7 (AMC 10)
What is the hundreds digit of (20! − 15!)?
Video Solution
Problem 25.1.8 (AMC 10)
What is the tens digit of 20152016 − 2017?
Video Solution
Problem 25.1.9 (AMC 10)
Which of the following expressions is never a prime number when p is a prime number?
(A) p2 + 16
(B) p2 + 24
(C) p2 + 26
(D) p2 + 46
(E) p2 + 96
Video Solution
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Chapter 25. Modular Arithmetic
Problem 25.1.10 (AMC 12)
Let S(n) equal the sum of the digits of positive integer n. For example, S(1507) = 13.
For a particular positive integer n, S(n) = 1274. Which of the following could be the
value of S(n + 1)?
Video Solution
Problem 25.1.11 (AMC 10)
Let a1 , a2 , . . . , a2018 be a strictly increasing sequence of positive integers such that
a1 + a2 + · · · + a2018 = 20182018 .
What is the remainder when a31 + a32 + · · · + a32018 is divided by 6?
Video Solution
Problem 25.1.12 (AMC 10)
How many of the first 2018 numbers in the sequence 101, 1001, 10001, 100001, . . . are
divisible by 101?
Problem 25.1.13 (AMC 10)
Let a1 , a2 , . . . , a2018 be a strictly increasing sequence of positive integers such that
a1 + a2 + · · · + a2018 = 20182018 .
What is the remainder when a31 + a32 + · · · + a32018 is divided by 6?
Problem 25.1.14 (AMC 12)
Let N = 123456789101112 . . . 4344 be the 79-digit number that is formed by writing the integers from 1 to 44 in order, one after the other. What is the remainder when
N is divided by 45?
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Video Solution
Problem 25.1.15 (AMC 10/12)
One of the following numbers is not divisible by any prime number less than 10. Which
is it? (A) 2606 − 1 (B) 2606 + 1 (C) 2607 − 1 (D) 2607 + 1 (E) 2607 + 3607
Video Solution
Problem 25.1.16 (AMC 10/12)
Let x0 , x1 , x2 , · · · be a sequence of numbers, where each xk is either 0 or 1. For
each positive integer n, define
Sn =
n−1
X
xk 2k
k=0
Suppose 7Sn ≡ 1 (mod 2 ) for all n ≥ 1. What is the value of the sum
n
x2019 + 2x2020 + 4x2021 + 8x2022 ?
(A) 6
(B) 7
(C) 12
(D) 14
(E) 15
Video Solution
Additional Problems
Problem 25.1.17 (AMC 10)
What is the remainder when 30 + 31 + 32 + · · · + 32009 is divided by 8?
Problem 25.1.18 (AMC 10)
What is the greatest power of 2 that is a factor of 101002 − 4501 ?
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(A) 21002
Chapter 25. Modular Arithmetic
(B) 21003
(C) 21004
(D) 21005
(E) 21006
Problem 25.1.19 (AMC 10)
Let n be a 5-digit number, and let q and r be the quotient and the remainder, respectively, when n is divided by 100. For how many values of n is q + r divisible by
11?
Problem 25.1.20 (AMC 10)
How many distinct four-digit numbers are divisible by 3 and have 23 as their last
two digits?
Problem 25.1.21 (AIME)
For positive integers N and k, define N to be k-nice if there exists a positive integer a such that ak has exactly N positive divisors. Find the number of positive integers
less than 1000 that are neither 7-nice nor 8-nice.
Problem 25.1.22 (AIME)
Let S be the set of integers between 1 and 240 whose binary expansions have exactly two
1’s. If a number is chosen at random from S, the probability that it is divisible by 9 is
p/q, where p and q are relatively prime positive integers. Find p + q.
Problem 25.1.23 (AIME)
It is known that, for all positive integers k,
12 + 22 + 32 + . . . + k 2 = k(k+1)(2k+1)
. Find the smallest positive integer k such that
6
2
2
2
2
1 + 2 + 3 + . . . + k is a multiple of 200.
Problem 25.1.24 (AIME)
Find the sum of all positive integers n such that when 13 + 23 + 33 + · · · + n3 is
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divided by n + 5, the remainder is 17.
Problem 25.1.25 (AMC 10/12)
The number obtained from the last two nonzero digits of 90! is equal to n. What
is n?
Answers
25.1 109
25.2 13
25.3 6
25.4 035
25.5 6
25.6 140
25.7 937
25.1.1 1
25.1.2 −1
25.1.3 343
25.1.4 54
25.1.5 15
25.1.6 37
25.1.7 0
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25.1.8 0
25.1.9 extbf (C)p2 + 26
25.1.10 1239
25.1.11 4
25.1.12 505
25.1.13 4
25.1.14 9
25.1.15 2607 − 1
25.1.16 6
25.1.17 4
25.1.18 21005
25.1.19 8181
25.1.20 30
25.1.21 749
25.1.22 913
25.1.23 112
25.1.24 239
25.1.25 12
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Algebraic Number Theory
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Chapter 26. Algebraic Number Theory
Concept 26.0.1 (Algebraic Number Theory Techniques)
1. Make substitutions to convert numbers to variables
2. Group terms
3. Look for factorizations
4. Sometimes use modular arithmetic, divisibility, or other number theory techniques
Example 26.1 (AIME)
Find the number of ordered triples (a, b, c) where a, b, and c are positive integers, a is a
factor of b, a is a factor of c, and a + b + c = 100.
Video Solution
26.1
Quadratic Factorizations
Theorem 26.1.1 (Exponent Rules)
x−a =
1
xa
xa × xb = xa+b
xa ÷ xb = xa−b
(xa )b = xab
Theorem 26.1.2 (Difference of Squares)
x2 − y 2 = (x − y)(x + y)
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Example 26.2 (Omega Learn)
How many values of n from 1 to 100 inclusive make 16n − 2 × 4n + 1 a multiple of 49?
Video Solution
26.2
Cubic Factorizations
Theorem 26.2.1 (Difference of Cubes)
x3 − y 3 = (x − y)(x2 + xy + y 2 )
Theorem 26.2.2 (Sum of Cubes)
x3 + y 3 = (x + y)(x2 − xy + y 2 )
Example 26.3 (AIME)
Find the sum of all positive integers n such that when 13 + 23 + 33 + · · · + n3 is divided
by n + 5, the remainder is 17
Video Solution
26.3
Simon’s Favorite Factoring Trick
Theorem 26.3.1 (Simon’s Favorite Factoring Trick)
xy + kx + jy + jk = (x + j)(y + k)
Remark 26.3.2
You can generally apply this factorization when you have xy, x, and y terms. After
applying the factorization, you can then find all possible values for each of your terms in
your factorization (remember negatives!).
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Example 26.4 (AMC 12)
A rectangular floor measures a by b feet, where a and b are positive integers with b > a.
An artist paints a rectangle on the floor with the sides of the rectangle parallel to
the sides of the floor. The unpainted part of the floor forms a border of width 1 foot
around the painted rectangle and occupies half of the area of the entire floor. How many
possibilities are there for the ordered pair (a, b)?
26.4
Higher Power Factorizations
Theorem 26.4.1 (nth power Factorizations)
Sum of odd powers
x2n+1 + y 2n+1 = (x + y)(x2n − x2n−1 y + x2n−2 y 2 − · · · − xy 2n−1 + y 2n )
Note: The signs in the second term alternate between positive and negative
xn − y n = (x − y)(xn−1 + xn−2 y + xn−3 y 2 + · · · + xy n−2 + y n−1 )
Note: The signs in second term are all positive
26.5
Sophie Germain’s Identity
Theorem 26.5.1 (Sophie Germain’s Identity)
x4 + 4y 4 = (x2 − 2xy + 2y 2 )(x2 + 2xy + 2y 2 )
Remark 26.5.2
Be on the lookout for 4th powers to apply Sophie Germain’s Identity!
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26.6
Chapter 26. Algebraic Number Theory
Practice Problems
Problem 26.6.1 (AMC 10)
There is a positive integer n such that (n + 1)! + (n + 2)! = n! · 440. What is the
sum of the digits of n?
Video Solution
Problem 26.6.2 (AMC 10)
For all integers n ≥ 9, the value of
(n + 2)! − (n + 1)!
n!
is always which of the following?
(A) a multiple of 4
(B) a multiple of 10
(D) a perfect square
(E) a perfect cube
(C) a prime number
Video Solution
Problem 26.6.3 (AMC 10)
Let S(n) equal the sum of the digits of positive integer n. For example, S(1507) = 13.
For a particular positive integer n, S(n) = 1274. Which of the following could be the
value of S(n + 1)?
Video Solution
Problem 26.6.4 (AMC 10)
What is the greatest three-digit positive integer n for which the sum of the first n
positive integers is not a divisor of the product of the first n positive integers?
Video Solution
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Problem 26.6.5 (AMC 10)
When a student multiplied the number 66 by the repeating decimal,
1.a b a b . . . = 1.a b,
where a and b are digits, he did not notice the notation and just multiplied 66 times
1.a b. Later he found that his answer is 0.5 less than the correct answer. What is the
2-digit number a b?
Video Solution
Problem 26.6.6 (AMC 10)
What is the remainder when 2202 + 202 is divided by 2101 + 251 + 1?
Video Solution
Problem 26.6.7 (AIME)
Compute
(104 + 324)(224 + 324)(344 + 324)(464 + 324)(584 + 324)
.
(44 + 324)(164 + 324)(284 + 324)(404 + 324)(524 + 324)
Video Solution
Problem 26.6.8 (AMC 12)
A sequence of numbers is defined by D0 = 0, D1 = 0, D2 = 1 and Dn = Dn−1 + Dn−3
for n ≥ 3. What are the parities (evenness or oddness) of the triple of numbers
(D2021 , D2022 , D2023 ), where E denotes even and O denotes odd?
Video Solution
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Problem 26.6.9 (AMC 10/12)
Which of the following is equivalent to
(2 + 3)(22 + 32 )(24 + 34 )(28 + 38 )(216 + 316 )(232 + 332 )(264 + 364 )?
Video Solution
Problem 26.6.10 (AMC 10)
The ages of Jonie’s four cousins are distinct single-digit positive integers. Two of
the cousins ages multiplied together give 24, while the other two multiply to 30. What is
the sum of the ages of Jonie’s four cousins?
Video Solution
Problem 26.6.11 (AMC 10/12)
Let f be a function defined on the set of positive rational numbers with the property that f (a · b) = f (a) + f (b) for all positive rational numbers a and b. Suppose that f
as have the property that f (p) = p for every prime number p. For which of the following
numbers x is f (x) < 0?
Video Solution
Problem 26.6.12 (AMC 10)
Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to
the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to
make the same number of interior pieces as pieces along the perimeter of the pan. What
is the greatest possible number of brownies she can produce?
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Video Solution
Additional Problems
Problem 26.6.13 (AMC 10)
Consider the set of all fractions xy , where x and y are relatively prime positive integers. How many of these fractions have the property that if both numerator and
denominator are increased by 1, the value of the fraction is increased by 10%?
Problem 26.6.14 (AIME)
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas
from the pile, keeps three-fourths of them, and divides the rest equally between the
other two. The second monkey takes some bananas from the pile, keeps one-fourth of
them, and divides the rest equally between the other two. The third monkey takes the
remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally
between the other two. Given that each monkey receives a whole number of bananas
whenever the bananas are divided, and the numbers of bananas the first, second, and
third monkeys have at the end of the process are in the ratio 3 : 2 : 1,what is the least
possible total for the number of bananas?
Problem 26.6.15 (AMC 10)
There exists a unique strictly increasing sequence of nonnegative integers a1 < a2 < · · · <
ak such that
2289 + 1
= 2a1 + 2a2 + · · · + 2ak .
217 + 1
What is k?
Problem 26.6.16 (AMC 10)
For how many integers n is
n
the square of an integer?
20 − n
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Problem 26.6.17 (AMC 10)
The number 2013 is expressed in the form
2013 =
a1 !a2 !...am !
b1 !b2 !...bn !
,
where a1 ≥ a2 ≥ · · · ≥ am and b1 ≥ b2 ≥ · · · ≥ bn are positive integers and a1 + b1 is
as small as possible. What is |a1 − b1 |?
Problem 26.6.18 (AMC 10/12)
For k > 0, let Ik = 10 . . . 064, where there are k zeros between the 1 and the 6.
Let N (k) be the number of factors of 2 in the prime factorization of Ik . What is the
maximum value of N (k)?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Problem 26.6.19 (AMC 10)
Two fair dice, each with at least 6 faces are rolled. On each face of each dice is printed
a distinct integer from 1 to the number of faces on that die, inclusive. The probability
of rolling a sum of 7 is 34 of the probability of rolling a sum of 10, and the probability
1
of rolling a sum of 12 is 12
. What is the least possible number of faces on the two dice
combined?
Problem 26.6.20 (AMC 10)
How many ordered pairs (m, n) of positive integers, with m ≥ n, have the property that
their squares differ by 96?
Problem 26.6.21 (AMC 10)
For how many positive integers n less than or equal to 24 is n! evenly divisible by
1 + 2 + · · · + n?
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Problem 26.6.22 (AIME)
Let S be the set of all perfect squares whose rightmost three digits in base 10 are
256. Let T be the set of all numbers of the form x−256
, where x is in S. In other words,
1000
T is the set of numbers that result when the last three digits of each number in S are
truncated. Find the remainder when the tenth smallest element of T is divided by 1000.
Problem 26.6.23 (AMC 10/12)
For each integer n ≥ 2, let Sn be the sum of all products jk, where j and k are
integers and 1 ≤ j < k ≤ n. What is the sum of the 10 least values of n such that Sn is
divisible by 3?
Problem 26.6.24 (AIME)
Find the largest integer n satisfying the following conditions:
(i) n2 can be expressed as the difference of two consecutive cubes;
(ii) 2n + 79 is a perfect square.
Answers
26.1 200
26.2 33
26.3 239
26.4 2
26.6.1 10
26.6.2 (D)aperf ectsquare
26.6.3 1239
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26.6.4 996
26.6.5 75
26.6.6 201
26.6.7 373
26.6.8 (E, O, E)
26.6.9 3128 − 2128
26.6.10 22
26.6.11 25
11
26.6.12 60
26.6.13 1
26.6.14 408
26.6.15 137
26.6.16 4
26.6.17 2
26.6.18 7
26.6.19 17
26.6.20 4
26.6.21 16
26.6.22 170
26.6.23 197
26.6.24 181
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Chapter 27
Diophantine Equations
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Chapter 27. Diophantine Equations
Definition 27.0.1. A Diophantine equation is a polynomial equation such that the only
solutions are integers (i.e. all variables have integer values).
Concept 27.0.2 (Strategies to Solve Diophantine Equations)
Here are some ideas on how to solve diophantine equations
• Take mods of different numbers. This is generally useful when you
1. Show there are no solutions to a Diophantine equation
2. Show that there are only a specific type of solution
• Bound the possible values of different terms, generally useful when there are a
finite number of solutions to your Diophantine equations
• Factoring, using the various factorizations (see the algebra section on this), can
help find all the solutions
• Make substitutions to simplify your Diophantine equation
• Look for conditions on what must be multiples/divisors of your variables and
rewrite your Diophantine equation in terms of that (ex. if you know a is multiple
of 6, then can make substitution a = 6k)
Example 27.1 (AMC 10)
How many right triangles have integer leg lengths a and b and a hypotenuse of length
b + 1, where b < 100?
Video Solution
Example 27.2 (AIME)
There exist unique positive integers x and y that satisfy the equation x2 +84x+2008 = y 2 .
Find x + y.
Video Solution
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Chapter 27. Diophantine Equations
Example 27.3 (AIME)
Assume that a, b, c, and d are positive integers such that a5 = b4 , c3 = d2 , and c − a = 19.
Determine d − b.
Video Solution
27.1
Practice Problems
Problem 27.1.1 (AMC 12)
Integers x and y with x > y > 0 satisfy x + y + xy = 80. What is x?
Video Solution
Problem 27.1.2 (AMC 10)
How many ordered pairs of integers (x, y) satisfy the equation
x2020 + y 2 = 2y?
Video Solution
Problem 27.1.3 (AMC 10)
Let a and b be relatively prime positive integers with a > b > 0 and
What is a − b?
a3 − b 3
73
=
.
3
(a − b)
3
Video Solution
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Problem 27.1.4 (AIME)
Find 3x2 y 2 if x and y are integers such that y 2 + 3x2 y 2 = 30x2 + 517.
Video Solution
Problem 27.1.5 (AMC 12)
How many ordered pairs of positive integers (b, c) exist where both x2 + bx + c = 0 and
x2 + cx + b = 0 do not have distinct, real solutions?
Video Solution
Problem 27.1.6 (AIME)
There is a prime number p such that 16p + 1 is the cube of a positive integer. Find p.
Video Solution
Problem 27.1.7 (USAMO)
Determine all non-negative integral solutions (n1 , n2 , . . . , n14 ) if any, apart from permutations, of the Diophantine Equation n41 + n42 + · · · + n414 = 1599.
Video Solution
Problem 27.1.8 (AMC 10A)
A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence
of positive integers. The first three terms of the resulting four-term sequence are 57, 60,
and 91. What is the fourth term of this sequence?
(A) 190
(B) 194
(C) 198
(D) 202
(E) 206
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Chapter 27. Diophantine Equations
Video Solution
Problem 27.1.9 (AMC 12)
Two distinct numbers are selected from the set {1, 2, 3, 4, . . . , 36, 37} so that the sum of
the remaining 35 numbers is the product of these two numbers. What is the difference of
these two numbers?
Video Solution
Problem 27.1.10 (AMC 10)
Hiram’s algebra notes are 50 pages long and are printed on 25 sheets of paper; the
first sheet contains pages 1 and 2, the second sheet contains pages 3 and 4, and so on.
One day he leaves his notes on the table before leaving for lunch, and his roommate
decides to borrow some pages from the middle of the notes. When Hiram comes back, he
discovers that his roommate has taken a consecutive set of sheets from the notes and
that the average (mean) of the page numbers on all remaining sheets is exactly 19. How
many sheets were borrowed?
Video Solution
Additional Problems
Problem 27.1.11 (AMC 10)
Two farmers agree that pigs are worth 300 dollars and that goats are worth 210 dollars.
When one farmer owes the other money, he pays the debt in pigs or goats, with ”change”
received in the form of goats or pigs as necessary. (For example, a 390 dollar debt could
be paid with two pigs, with one goat received in change.) What is the amount of the
smallest positive debt that can be resolved in this way?
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Chapter 27. Diophantine Equations
Problem 27.1.12 (AHSME)
How many triples (a, b, c) of positive integers which satisfy the simultaneous equations
ab + bc = 44
ac + bc = 23?
Problem 27.1.13 (AIME)
One of Euler’s conjectures was disproved in the 1960s by three American mathematicians
when they showed there was a positive integer such that
1335 + 1105 + 845 + 275 = n5 .
Find the value of n.
Problem 27.1.14 (AIME)
Find the largest possible value of k for which 311 is expressible as the sum of k consecutive
positive integers.
Problem 27.1.15 (AIME)
For some integer m, the polynomial x3 − 2011x + m has the three integer roots a,
b, and c. Find |a| + |b| + |c|.
Problem 27.1.16 (AMC 12)
Let m ≥ 5 be an odd integer, and let D(m) denote the number of quadruples (a1 , a2 , a3 , a4 )
of distinct integers with 1 ≤ ai ≤ m for all i such that m divides a1 + a2 + a3 + a4 . There
is a polynomial
q(x) = c3 x3 + c2 x2 + c1 x + c0
such that D(m) = q(m) for all odd integers m ≥ 5. What is c1 ?
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Answers
27.1 6
27.2 80
27.3 757
27.1.1 26
27.1.2 4
27.1.3 3
27.1.4 588
27.1.5 6
27.1.6 307
27.1.7 N ointegralsolutions
27.1.8 206
27.1.9 10
27.1.10 13
27.1.11 30
27.1.12 2
27.1.13 144
27.1.14 486
27.1.15 98
27.1.16 11
300
Chapter 28
Bases
Video Lecture
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Chapter 28. Bases
Definition 28.0.1 (Bases). A number expressed in base-n is similar to base 10 except instead
of regrouping to a new place value every 10, we regroup every n.
Concept 28.0.2
A number in base n with digits am , am−1 , . . . , a2 , a1 , a0 can be written as:
am am−1 . . . a2 a1 a0
This number can be evaluated as
am nm + am−1 nm−1 + · · · + a2 n2 + a1 n1 + a0 n0
Example 28.1 (AMC 10/12)
The base-nine representation of the number N is 27,006,000,052nine . What is the remainder
when N is divided by 5?
Video Solution
Example 28.2 (AIME)
Find the three-digit positive integer a b c whose representation in base nine is b c a nine ,
where a, b, and c are (not necessarily distinct) digits.
Video Solution
Example 28.3 (AMC 10)
A base-10 three digit number n is selected at random. Which of the following is closest
to the probability that the base-9 representation and the base-11 representation of n are
both three-digit numerals?
Video Solution
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Chapter 28. Bases
Concept 28.0.3
We can also have bases with decimals (both repeating and terminal).
A number in base n with digits am , am−1 , . . . , a2 , a1 , a0 before the decimal point
and the digits a−1 , a−2 , . . . , a−q after the decimal point can be written as:
am am−1 . . . a1 a0 .a−1 a−2 . . . a−q
This number can be evaluated as:
am nm + a(m−1) n(m−1) + · · · + a1 n1 + a0 n0 + a−1 n−1 + a−2 n−2 + . . . a−q n−q
Example 28.4 (AMC 12)
For some positive integer k, the repeating base-k representation of the (base-ten) fraction
7
is 0.23k = 0.232323...k . What is k?
51
Video Solution
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28.1
Chapter 28. Bases
Practice Problems
Problem 28.1.1 (AMC 10)
In base 10, the number 2013 ends in the digit 3. In base 9, on the other hand, the same
number is written as (2676)9 and ends in the digit 6. For how many positive integers b
does the base-b-representation of 2013 end in the digit 3?
Video Solution
Problem 28.1.2 (AMC 10)
Hexadecimal (base-16) numbers are written using numeric digits 0 through 9 as well
as the letters A through F to represent 10 through 15. Among the first 1000 positive
integers, there are n whose hexadecimal representation contains only numeric digits.
What is the sum of the digits of n?
Video Solution
Problem 28.1.3 (AIME)
For each positive integer n, let f (n) be the sum of the digits in the base-four representation of n and let g(n) be the sum of the digits in the base-eight representation of f (n).
For example, f (2020) = f (1332104 ) = 10 = 128 , and g(2020) = the digit sum of 128 = 3.
Let N be the least value of n such that the base-sixteen representation of g(n) cannot be
expressed using only the digits 0 through 9. Find the remainder when N is divided by
1000.
Video Solution
Problem 28.1.4 (AIME)
A positive integer N has base-eleven representation abc and base-eight representation
1bca, where a, b, and c represent (not necessarily distinct) digits. Find the least such N
expressed in base ten.
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Video Solution
Problem 28.1.5 (AMC 10/12)
For which of the following integers b is the base-b number 2021b − 221b not divisible by 3?
Video Solution
Problem 28.1.6 (AMC 10)
Let n be a positive integer and d be a digit such that the value of the numeral 32d in
base n equals 263, and the value of the numeral 324 in base n equals the value ¯of¯¯the
¯¯¯
numeral 11d1 in base six. What is n + d?
¯¯¯¯
Video Solution
Additional Problems
Problem 28.1.7 (AMC 10)
Let n be a positive integer and d be a digit such that the value of the numeral 32d in
base n equals 263, and the value of the numeral 324 in base n equals the value of the
numeral 11d1 in base six. What is n + d?
Problem 28.1.8 (AMC 10)
For which of the following integers b is the base-b number 2021b − 221b not divisible by 3?
Problem 28.1.9 (AMC 10)
What is the greatest possible sum of the digits in the base-seven representation of
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a positive integer less than 2019?
Problem 28.1.10 (AMC 12)
In the equation below, A and B are consecutive positive integers, and A, B, and
A + B represent number bases:
132A + 43B = 69A+B .
What is A + B?
Problem 28.1.11 (AIME)
The number n can be written in base 14 as a b c, can be written in base 15 as a c b, and
can be written in base 6 as a c a c , where a > 0. Find the base-10 representation of n.
Problem 28.1.12 (AIME)
Call a positive integer N a 7-10 double if the digits of the base-7 representation of
N form a base-10 number that is twice N . For example, 51 is a 7-10 double because its
base-7 representation is 102. What is the largest 7-10 double?
Problem 28.1.13 (AIME)
Let N be the number of positive integers that are less than or equal to 2003 and
whose base-2 representation has more 1’s than 0’s. Find the remainder when N is divided
by 1000.
Problem 28.1.14 (AIME)
A rational number written in base eight is ab.cd, where all digits are nonzero. The
same number in base twelve is bb.ba. Find the base-ten number abc.
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Problem 28.1.15 (AIME)
Find the number of positive integers less than or equal to 2017 whose base-three representation contains no digit equal to 0.
Problem 28.1.16 (AIME)
The number n can be written in base 14 as a b c, can be written in base 15 as a c b, and
can be written in base 6 as a c a c , where a > 0. Find the base-10 representation of n.
Problem 28.1.17 (AIME)
Find the sum of all positive integers b < 1000 such that the base-b integer 36b is a
perfect square and the base-b integer 27b is a perfect cube.
Problem 28.1.18 (AIME)
A positive integer N has base-eleven representation abc and base-eight representation
1bca, where a, b, and c represent (not necessarily distinct) digits. Find the least such N
expressed in base ten.
Problem 28.1.19 (AIME)
There exist r unique nonnegative integers n1 > n2 > · · · > nr and r unique integers ak
(1 ≤ k ≤ r) with each ak either 1 or −1 such that
a1 3n1 + a2 3n2 + · · · + ar 3nr = 2008.
Find n1 + n2 + · · · + nr .
Problem 28.1.20 (AIME)
For each positive integer n, let f (n) be the sum of the digits in the base-four representation of n and let g(n) be the sum of the digits in the base-eight representation of f (n).
For example, f (2020) = f (1332104 ) = 10 = 128 , and g(2020) = the digit sum of 128 = 3.
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Chapter 28. Bases
Let N be the least value of n such that the base-sixteen representation of g(n) cannot be
expressed using only the digits 0 through 9. Find the remainder when N is divided by
1000.
Problem 28.1.21 (AMC 10)
Bernardo chooses a three-digit positive integer N and writes both its base-5 and base-6
representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written.
Treating the two numbers as base-10 integers, he adds them to obtain an integer S. For
example, if N = 749, Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains
the sum S = 13,689. For how many choices of N are the two rightmost digits of S, in
order, the same as those of 2N ?
Answers
28.1 3
28.2 227
28.3 0.7
28.4 16
28.1.1 13
28.1.2 21
28.1.3 151
28.1.4 621
28.1.5 8
28.1.6 11
28.1.7 11
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28.1.8 8
28.1.9 22
28.1.10 13
28.1.11 925
28.1.12 315
28.1.13 155
28.1.14 321
28.1.15 222
28.1.16 925
28.1.17 371
28.1.18 621
28.1.19 21
28.1.20 151
28.1.21 25
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Chapter 29
Miscellaneous Number Theory
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29.1
Chapter 29. Miscellaneous Number Theory
Palindromes
Definition 29.1.1. A palindrome is a number that reads the same forward and backward.
Example 29.1 (AIME)
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome
is a number that reads the same forward and backward, such as 777 or 383.)
29.2
Chicken McNugget Theorem
Theorem 29.2.1 (Chicken McNugget Theorem)
The maximum value that cannot be expressed as the sum of non-negative multiples of a
and b is ab − a − b if a and b are relatively prime.
For relatively prime positive integers a, b ,there are exactly
(a − 1)(b − 1)
2
positive integers which cannot be expressed in the form ma + nb where m and n are
positive integers.
Remark 29.2.2
This theorem is useful in finding solutions to problems like ”the maximum amount of
money that can’t be created with 3 cent and 5 cent coins”.
Example 29.2 (AMC 10)
The town of Hamlet has 3 people for each horse, 4 sheep for each cow, and 3 ducks for
each person. Which of the following could not possibly be the total number of people,
horses, sheep, cows, and ducks in Hamlet?
(A) 41
(B) 47
(C) 59
(D) 61
(E) 66
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Example 29.3 (AMC 10)
The product (8)(888 . . . 8), where the second factor has k digits, is an integer whose
digits have a sum of 1000. What is k?
(A) 901
(B) 911
(C) 919
(D) 991
(E) 999
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29.3
Chapter 29. Miscellaneous Number Theory
Practice Problems
Problem 29.3.1 (2021 AMC 12B Problem 3)
Suppose
2+
1
1 + 2+ 2
1
3+x
=
144
.
53
What is the value of x?
Video Solution
Problem 29.3.2 (AMC 10)
Driving along a highway, Megan noticed that her odometer showed 15951 (miles). This
number is a palindrome-it reads the same forward and backward. Then 2 hours later, the
odometer displayed the next higher palindrome. What was her average speed, in miles
per hour, during this 2-hour period?
Video Solution
Problem 29.3.3 (AMC 10)
A palindrome, such as 83438, is a number that remains the same when its digits are reversed. The numbers x and x + 32 are three-digit and four-digit palindromes, respectively.
What is the sum of the digits of x?
Video Solution
Problem 29.3.4 (AMC 10)
Mr. Zhou places all the integers from 1 to 225 into a 15 by 15 grid. He places 1
in the middle square (eighth row and eighth column) and places other numbers one by
one clockwise, as shown in part in the diagram below. What is the sum of the greatest
number and the least number that appear in the second row from the top?
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Video Solution
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Chapter 29. Miscellaneous Number Theory
Problem 29.3.5 (AMC 12)
Let d(n) denote the number of positive integers that divide n, including 1 and n. For
example, d(1) = 1, d(2) = 2, and d(12) = 6. (This function is known as the divisor
function.) Let
d(n)
f (n) = 3 √ .
n
There is a unique positive integer N such that f (N ) > f (n) for all positive integers
n ̸= N . What is the sum of the digits of N ?
Video Solution
Additional Problems
Problem 29.3.6 (AMC 10/12)
A palindrome between 1000 and 10, 000 is chosen at random. What is the probability
that it is divisible by 7?
Problem 29.3.7 (AMC 10)
How many 7-digit palindromes (numbers that read the same backward as forward)
can be formed using the digits 2, 2, 3, 3, 5, 5, 5?
Problem 29.3.8 (AMC 12)
A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 10 with no leading zeros. A 6-digit palindrome n is chosen
n
uniformly at random. What is the probability that 11
is also a palindrome?
Problem 29.3.9 (AMC 10)
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is 1 year
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Chapter 29. Miscellaneous Number Theory
older than Chloe, and Zoe is exactly 1 year old today. Today is the first of the 9 birthdays
on which Chloe’s age will be an integral multiple of Zoe’s age. What will be the sum of
the two digits of Joey’s age the next time his age is a multiple of Zoe’s age?
Video Solution
Problem 29.3.10 (AIME)
Ninety-four bricks, each measuring 4′′ × 10′′ × 19′′ , are to be stacked one on top of
another to form a tower 94 bricks tall. Each brick can be oriented so it contributes 4′′ or
10′′ or 19′′ to the total height of the tower. How many different tower heights can be
achieved using all ninety-four of the bricks?
Problem 29.3.11 (AMC 10)
Let N be the positive integer 7777 . . . 777, a 313-digit number where each digit is a
7. Let f (r) be the leading digit of the rth root of N . What is
f (2) + f (3) + f (4) + f (5) + f (6)?
Problem 29.3.12 (AIME)
Find the sum of all positive integers n such that, given an unlimited supply of stamps
of denominations 5, n, and n + 1 cents, 91 cents is the greatest postage that cannot be
formed.
Answers
29.1 550
29.2 47
29.3 991
29.3.1 34
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29.3.2 55
29.3.3 24
29.3.4 367
29.3.5 9
29.3.6 15
29.3.7 6
29.3.8 11
30
29.3.9 11
29.3.10 465
29.3.11 8
29.3.12 071
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Geometry
318
Chapter 30
Angle Chasing
Video Lecture
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30.1
Chapter 30. Angle Chasing
Basics of Angle Chasing
• Sum of angles in triangle is 180°
• Sum of angles in a line is 180°
• A triangle with 2 equal angles will have their corresponding sides equal and a triangle
with 2 sides equal will have their corresponding angles equal. Such a triangle is called
an isosceles triangle.
Concept 30.1.1 (Complementary Angle)
Complementary angles are a pair of angles with the sum of 90 degrees
Concept 30.1.2 (Supplementary Angle)
Supplementary angles are a pair of angles with the sum of 180 degrees
Concept 30.1.3 (Intersecting lines)
When two lines intersect, the vertical angles are equal. Vertical angles are each of the
pairs of opposite angles made by two intersecting lines. ”Vertical” in this case means
they share the same Vertex (corner point), not the usual meaning of up-down.
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Concept 30.1.4 (Parallel Lines)
Corresponding angles equal
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Chapter 30. Angle Chasing
Example 30.1 (AMC 10)
Triangle ABC has AB = 2 · AC. Let D and E be on AB and BC, respectively, such
that ∠BAE = ∠ACD. Let F be the intersection of segments AE and CD, and suppose
that △CF E is equilateral. What is ∠ACB?
A
D
C
E
B
Video Solution
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Chapter 30. Angle Chasing
Example 30.2 (AIME)
In △ABC with AB = AC, point D lies strictly between A and C on side AC, and point
E lies strictly between A and B on side AB such that AE = ED = DB = BC. The
, where m and n are relatively prime positive integers.
degree measure of ∠ABC is m
n
Find m + n.
A
E
D
C
B
Video Solution
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Chapter 30. Angle Chasing
Theorem 30.1.5
Sum of interior angle of a polygon = (n − 2) · 180
Interior angle of a regular polygon =
(n − 2)
· 180
n
Exterior angle of a regular polygon =
360
n
Fact 30.1.6. Important Interior Angles
Number of sides in regular polygon Interior Angle of regular polygon
3
60
4
90
5
108
6
120
8
135
9
140
10
144
Theorem 30.1.7 (Inscribed Arc Theorem)
The angle formed by an arc in the center or the arc angle is double of the angle formed
on the edge.
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Chapter 30. Angle Chasing
Corollary 30.1.8 (Inscribed Right Triangle)
Inscribed triangle with diameter as one side is always a right triangle.
Definition 30.1.9 (Chord). A Chord is a line segment between any two distinct points on
the circle. The diameter of the circle is the longest chord in the circle.
Theorem 30.1.10
The perpendicular bisector of any chord passes through the center. In the figure below,
the perpendicular bisectors of AB and CD intersect at the center O.
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Corollary 30.1.11
Chapter 30. Angle Chasing
• Congruent chords are equidistant from the center of a circle.
• If two chords in a circle are congruent, then their intercepted arcs are congruent.
• If two chords in a circle are congruent, then they determine two central angles that
are congruent.
Theorem 30.1.12
The angle marked in the diagram is half of the difference of the 2 red arcs.
⌢ ⌢
BD − AC
∠AP C =
2
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Chapter 30. Angle Chasing
Theorem 30.1.13
If two chords AB and CD intersect at P, then the ∠BP C and ∠AP D are equal to the
average of the two arcs.
∠BP C = ∠AP D =
⌢
⌢
BC + AD
2
Theorem 30.1.14
If a tangent R intersects the circle at Q, and a chord QP is drawn, then the ∠RQP is
equal to half the arc angle
Remark 30.1.15
Circles are really useful for angle chasing so keep an eye out for the inscribed arc theorem
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Chapter 30. Angle Chasing
that can be used in many angle chasing problems.
Remark 30.1.16
A useful trick to solving angle chasing problems with regular polygons is to draw a circle
around the polygon and use the inscribed arc theorem.
Theorem 30.1.17
Equal chords mark out equal arcs
This basically means that if you have 2 chords of the same length, the sector of the circle
they mark out will be equal
Definition 30.1.18 (Tangent). A tangent is any line from a point external to the circle that
just touches the circle.
Theorem 30.1.19 (Right Angle Tangency Point)
If you connect the center of a circle to the point where the circle and a line are tangent,
they will form a right angle.
Remark 30.1.20
This property is very useful in circle problems as it allows us to work with right angles.
In addition, another helpful technique is drawing useful radii to various points in your
diagram as that opens up new information to work with.
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Chapter 30. Angle Chasing
Theorem 30.1.21 (Properties of a Cyclic quadrilateral)
Sum of opposite angles = 180
Example 30.3 (AMC 10)
Let △ABC be an isosceles triangle with BC = AC and ∠ACB = 40◦ . Construct the
circle with diameter BC, and let D and E be the other intersection points of the circle
with the sides AC and AB, respectively. Let F be the intersection of the diagonals of
the quadrilateral BCDE. What is the degree measure of ∠BF C?
C
F
B
E
D
A
Video Solution
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30.2
Chapter 30. Angle Chasing
Practice Problems
Problem 30.2.1 (AMC 10)
Let ABCDEF be an equiangular
hexagon. The lines AB, CD, and EF determine
√
a triangle
with
√ with area 192 3, and the lines BC, DE, and F A determine a triangle
√
area 324 3. The perimeter of hexagon ABCDEF can be expressed as m + n p, where
m, n, and p are positive integers and p is not divisible by the square of any prime. What
is m + n + p?
Video Solution
Problem 30.2.2 (AMC)
Two congruent circles centered at points A and B each pass through the other circle’s center. The line containing both A and B is extended to intersect the circles at
points C and D. The circles intersect at two points, one of which is E. What is the degree
measure of ∠CED?
Video Solution
Problem 30.2.3 (MATHCOUNTS)
A square is located in the interior of a regular hexagon, and certain vertices are labeled as shown. What is the degree measure of ∠ABC?
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Chapter 30. Angle Chasing
Video Solution
Problem 30.2.4 (AMC 10)
In the given circle, the diameter EB is parallel to DC, and AB is parallel to ED.
The angles AEB and ABE are in the ratio 4 : 5. What is the degree measure of angle
BCD?
A
E
O
B
D C
Video Solution
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Chapter 30. Angle Chasing
Problem 30.2.5 (AMC 10)
The keystone arch is an ancient architectural feature. It is composed of congruent
isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom
sides of the two end trapezoids are horizontal. In an arch made with 9 trapezoids, let x
be the angle measure in degrees of the larger interior angle of the trapezoid. What is x?
Video Solution
Additional Problems
Problem 30.2.6 (AHSME)
△ABC is isosceles with base AC. Points P and Q are respectively in CB and AB
and such that AC = AP = P Q = QB. The number of degrees in ∠B is:
Problem 30.2.7 (AHSME)
In this diagram AB and AC are the equal sides of an isosceles △ABC, in which is
inscribed equilateral △DEF . Designate ∠BF D by a, ∠ADE by b, and ∠F EC by c.
Then:
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Chapter 30. Angle Chasing
A
E
c
D b
a
B
(A) b = a+c
2
(B) b = a−c
2
F
(C) a = b−c
2
C
(D) a = b+c
2
(E) none of these
Problem 30.2.8 (AMC 10)
How many non-similar triangles have angles whose degree measures are distinct positive
integers in arithmetic progression?
Problem 30.2.9 (MATHCOUNTS)
Concave quadrilateral ABCD is symmetric about the line AC. The measures of angles DAB and ABC are 84 degrees and 32 degrees, respectively. The dashed line segments
bisect angles ABC and ADC. What is the degree measure of the acute angle at which
the two dashed line segments intersect?
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Chapter 30. Angle Chasing
Problem 30.2.10 (AMC 10/12)
As shown in the figure below, point E lies on the opposite half-plane determined by line
CD from point A so that ∠CDE = 110◦ . Point F lies on AD so that DE = DF , and
ABCD is a square. What is the degree measure of ∠AF E?
E
A
F
D
B
110◦
C
Problem 30.2.11 (AHSME)
In the adjoining figure, ABCD is a square, ABE is an equilateral triangle and point E
is outside square ABCD. What is the measure of ∡AED in degrees?
D
A
E
C
B
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Chapter 30. Angle Chasing
Video Solution
Problem 30.2.12 (AMC 10)
The angles of quadrilateral ABCD satisfy ∠A = 2∠B = 3∠C = 4∠D. What is the
degree measure of ∠A, rounded to the nearest whole number?
Problem 30.2.13 (AMC 10)
The sum of two angles of a triangle is 65 of a right angle, and one of these two angles is 30◦ larger than the other. What is the degree measure of the largest angle in the
triangle?
Problem 30.2.14 (AMC 10)
Mary divides a circle into 12 sectors. The central angles of these sectors, measured
in degrees, are all integers and they form an arithmetic sequence. What is the degree
measure of the smallest possible sector angle?
Problem 30.2.15 (AHSME)
An acute isosceles triangle, ABC, is inscribed in a circle. Through B and C, tangents to the circle are drawn, meeting at point D. If ∠ABC = ∠ACB = 2∠D and x is
the radian measure of ∠A, then x =
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Chapter 30. Angle Chasing
Problem 30.2.16 (AIME)
Triangle ABC is isosceles with AC = BC and ∠ACB = 106◦ . Point M is in the
interior of the triangle so that ∠M AC = 7◦ and ∠M CA = 23◦ . Find the number of
degrees in ∠CM B.
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Chapter 30. Angle Chasing
Problem 30.2.17 (AMC 10)
Quadrilateral ABCD has AB = BC = CD, angle ABC = 70 and angle BCD = 170.
What is the measure of angle BAD?
Problem 30.2.18 (AIME)
A beam of light strikes BC at point C with angle of incidence α = 19.94◦ and reflects with an equal angle of reflection as shown. The light beam continues its path,
reflecting off line segments AB and BC according to the rule: angle of incidence equals
angle of reflection. Given that β = α/10 = 1.994◦ and AB = BC, determine the number
of times the light beam will bounce off the two line segments. Include the first reflection
at C in your count.
Answers
30.1 90°
30.2 547
30.3 110
30.2.1 55
30.2.2 120
30.2.3 45
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30.2.4 130
30.2.5 100°
30.2.6 25 57
30.2.7 a = b+c
2
30.2.8 59
30.2.9 64
30.2.10 170
30.2.11 15
30.2.12 173
30.2.13 72
30.2.14 8
30.2.15 3π
7
30.2.16 83
30.2.17 85
30.2.18 071
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Chapter 31
Triangle Area and Length
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31.1
Chapter 31. Triangle Area and Length
Area of a Triangle
There are many ways to calculate the area of a triangle. Here are some of the most useful
formulas for calculating the area of a triangle:
Theorem 31.1.1 (Using base and height)
A triangle with base b and height h has an area of
1
·b·h
2
Theorem 31.1.2 (Heron’s Formula)
A triangle with sides a, b, c and semiperimeter s has an area of
q
s(s − a)(s − b)(s − c)
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Chapter 31. Triangle Area and Length
Definition 31.1.3 (Incenter). The incenter of a triangle is the intersection of all the angle
bisectors. This point is also the center of the incircle, and equidistant from all the three sides.
Definition 31.1.4 (In-radius). The inradius of a triangle is the radius of the inscribed circle
in the triangle.
Theorem 31.1.5
Inradius r of a right triangle:
1
r = (a + b − c)
2
where a and b are the legs of the triangle, and c is the hypotenuse.
Theorem 31.1.6 (Using inradius)
A triangle with inradius r (the radius of the circle that can be inscribed in a triangle)
and semiperimeter s has an area of:
Area = inradius · semiperimeter
A = rs
Remark 31.1.7
Note that if we know the area of the triangle and it’s semi-perimeter, we can apply the
inradius formula to find the inradius of the triangle.
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Chapter 31. Triangle Area and Length
Definition 31.1.8 (Circumcenter). The circumcenter of a triangle is the intersection of all
3 perpendicular bisectors (the line that bisects a segment and is perpendicular to it). This
point is also the center of the circumcircle and equidistant from all the three vertices.
Definition 31.1.9 (Circumradius). The circum-radius of a triangle is the radius of circle
that a triangle is inscribed in.
Theorem 31.1.10 (Using circumradius)
A triangle with circumradius R (the radius of the circle that the triangle can be inscribed
in) and sides a, b, c has an area of
abc
A=
4R
Remark 31.1.11
Similar to the inradius problem, if we know all 3 sides of a triangle, we can apply Heron’s
and easily calculate the circumradius of the triangle.
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Chapter 31. Triangle Area and Length
Theorem 31.1.12 (Shoelace Theorem using coordinates)
Suppose the polygon P has vertices (a1 , b1 ), (a2 , b2 ), ... , (an , bn ), listed in clockwise
order. Then the area (A) of P is
A=
1
|(a1 b2 + a2 b3 + · · · + an b1 ) − (b1 a2 + b2 a3 + · · · + bn a1 )|
2
You can also go counterclockwise order, as long as you find the absolute value of the
answer.
The Shoelace Theorem gets its name because if one lists the coordinates in a column,
(a1 , b1 )
(a2 , b2 )
..
.
(an , bn )
(a1 , b1 )
and marks the pairs of coordinates to be multiplied,
Remark 31.1.13 (Intuitive Way of Thinking about Shoelace Theorem)
Steps to Shoelace Theorem
1. Line up all of your polygon’s coordinates in a vertical line
2. Repeat your first coordinate at the bottom of your line
3. Let the sum of products of all rightward diagonally pairs be A
4. Let the sum of products of all leftward diagonally pairs be B
5. Find
1
|A − B|
2
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Chapter 31. Triangle Area and Length
Theorem 31.1.14 (Pick’s Theorem)
If a polygon has vertices with integer coordinates (lattice points) then the area of the
polygon is i + 2b − 1 where i is the number of lattice points inside the polygon and b is
the number of lattice points on the boundary of the polygon.
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Chapter 31. Triangle Area and Length
Example 31.1 (AIME)
As shown in the figure, triangle ABC is divided into six smaller triangles by lines drawn
from the vertices through a common interior point. The areas of four of these triangles
are as indicated. Find the area of triangle ABC.
C
84
35
40
A
30
B
Video Solution
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31.2
Chapter 31. Triangle Area and Length
Triangle Properties
Definition 31.2.1 (Median). A median is a line connecting a point to the midpoint of the
opposite side.
Theorem 31.2.2 (Median in a Right Triangle)
In a right triangle ABC, let the median from point C intersect AB at a point M. Then
AM = BM = CM.
Basically, in a right triangle AB is the diameter of the circumcircle, and MA, MB, and
MC are radii of the circumcircle.
B
M
C
A
Definition 31.2.3 (Centroid). In a triangle, the intersection of all 3 medians in a triangle is
the centroid.
Theorem 31.2.4
The centroid of a triangle is on the median and it is 23 of the way from from one of
vertices to the midpoint of the opposite side.
Definition 31.2.5 (Cevian). A cevian is any line from any vertex of a triangle to the opposite
side. Medians and angle bisectors are special cases of cevians.
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Chapter 31. Triangle Area and Length
Example 31.2 (AMC 12)
Three congruent isosceles triangles are constructed with their bases on the sides of an
equilateral triangle of side length 1. The sum of the areas of the three isosceles triangles
is the same as the area of the equilateral triangle. What is the length of one of the two
congruent sides of one of the isosceles triangles?
Video Solution
Example 31.3 (AMC 10)
Triangle ABC has side-lengths AB = 12, BC = 24, and AC = 18. The line through
the incenter of △ABC parallel to BC intersects AB at M and AC at N. What is the
perimeter of △AM N ?
Video Solution
Theorem 31.2.6 (Stewart’s Theorem)
Given a triangle △ABC with sides of length a, b, c opposite vertices of A, B, C respectively.
If cevian AD is drawn so that BD = m, DC = n and AD = d, we have that
man + dad = bmb + cnc
.
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Chapter 31. Triangle Area and Length
Remark 31.2.7
A way to remember this is the saying ”A Man and his Dad put a Bomb in the Sink
Corollary 31.2.8 (Stewart’s For Angle Bisector)
If AD is an angle bisector, then d2 + mn = bc.
Note that this follows from Stewart’s Theorem and the Angle Bisector Theorem.
Corollary 31.2.9 (Stewart’s Theorem For Medians)
If AD is a median, then d2 = 12 (b2 + c2 ) − 41 a2
31.3
Practice Problems
Problem 31.3.1 (AMC 10)
Find the area of the shaded region.
1
7
1
4
4
1
7
1
Video Solution
Problem 31.3.2 (AMC 10)
Triangle ABC has a right angle at B. Point D is the foot of the altitude from B,
AD = 3, and DC = 4. What is the area of △ABC?
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A
3
D
4
B
C
Video Solution
Problem 31.3.3 (AMC 10)
The figure below shows a square and four equilateral triangles, with each triangle
having a side lying on a side of the square, such that each triangle has side length 2 and
the third vertices of the triangles meet at the center of the square. The region inside the
square but outside the triangles is shaded. What is the area of the shaded region?
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Video Solution
Problem 31.3.4 (AMC 10)
Triangle AM C is isosceles with AM = AC. Medians M V and CU are perpendicular to each other, and M V = CU = 12. What is the area of △AM C?
A
U
V
P
M
C
Video Solution
Problem 31.3.5 (AMC 10)
Points P and Q lie in a plane with P Q = 8. How many locations for point R in
this plane are there such that the triangle with vertices P , Q, and R is a right triangle
with area 12 square units?
Video Solution
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Chapter 31. Triangle Area and Length
Problem 31.3.6 (AMC 10)
Rectangle ABCD has AB = 5 and BC = 4. Point E lies on AB so that EB = 1,
point G lies on BC so that CG = 1, and point F lies on CD so that DF = 2. Segments
AG and AC intersect EF at Q and P , respectively. What is the value of PEFQ ?
E
A
B
Q
P
G
D
F
C
Video Solution
Additional Problems
Problem 31.3.7 (AMC 10)
Given a triangle with side lengths 15, 20, and 25, find the triangle’s shortest altitude.
Problem 31.3.8 (AMC 10/12)
A triangle with side lengths in the ratio 3 : 4 : 5 is inscribed in a circle with radius 3.
What is the area of the triangle?
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Chapter 31. Triangle Area and Length
Problem 31.3.9 (AMC 10)
√
The two legs of a right triangle, which are altitudes, have lengths 2 3 and 6. How
long is the third altitude of the triangle?
Problem 31.3.10 (AMC 12)
Triangle ABC has AB = 13 and AC = 15, and the altitude to BC has length 12.
What is the sum of the two possible values of BC?
Problem 31.3.11 (AMC 10)
A triangle has side lengths 10, 10, and 12. A rectangle has width 4 and area equal to the
area of the triangle. What is the perimeter of this rectangle?
Problem 31.3.12 (AMC 12)
How many noncongruent integer-sided triangles with positive area and perimeter less
than 15 are neither equilateral, isosceles, nor right triangles?
Problem 31.3.13 (AMC 10)
Right triangles T1 and T2 , have areas of 1 and 2, respectively. A side of T1 is congruent to a side of T2 , and a different side of T1 is congruent to a different side of T2 .
What is the square of the product of the lengths of the other (third) side of T1 and T2 ?
Problem 31.3.14 (AMC 12)
Isosceles triangles T and T ′ are not congruent but have the same area and the same
perimeter. The sides of T have lengths 5, 5, and 8, while those of T ′ have lengths a, a,
and b. Which of the following numbers is closest to b?
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Chapter 31. Triangle Area and Length
Problem 31.3.15 (AMC 12)
Triangle ABC is equilateral with AB = 1. Points E and G are on AC and points
D and F are on AB such that both DE and F G are parallel to BC. Furthermore,
triangle ADE and trapezoids DF GE and F BCG all have the same perimeter. What is
DE + F G?
C
G
E
A
D
F
B
Problem 31.3.16 (AMC 10)
Points P and Q lie in a plane with P Q = 8. How many locations for point R in
this plane are there such that the triangle with vertices P , Q, and R is a right triangle
with area 12 square units?
Problem 31.3.17 (AIME)
Triangle ABC has AB = 9 and BC : AC = 40 : 41. What’s the largest area that
this triangle can have?
Problem 31.3.18 (AMC 10)
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to
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Chapter 31. Triangle Area and Length
the legs of the triangle so that the triangle is divided into a square and two smaller right
triangles. The area of one of the two small right triangles is m times the area of the
square. The ratio of the area of the other small right triangle to the area of the square is
1
(A) 2m+1
(C) 1 − m
(B) m
1
(D) 4m
(E) 8m1 2
Problem 31.3.19 (AMC 10)
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines
from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as
shown. What is the area of the shaded quadrilateral?
7
3
7
Problem 31.3.20 (AMC 10)
In ABC we have AB = 25, BC = 39, and AC = 42. Points D and E are on AB
and AC respectively, with AD = 19 and AE = 14. What is the ratio of the area of
triangle ADE to the area of the quadrilateral BCED?
A
14
E
19
28
D
6
B
C
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Chapter 31. Triangle Area and Length
Problem 31.3.21 (AMC 12)
Let ABCD be a unit square. Let Q1 be the midpoint of CD. For i = 1, 2, . . . , let
Pi be the intersection of AQi and BD, and let Qi+1 be the foot of the perpendicular
from Pi to CD. What is
∞
X
Area of △DQi Pi ?
i=1
Answers
31.1 315
√
31.2
3
3
31.3 30
31.3.1 6 12
√
31.3.2 7 3
√
31.3.3 12 − 4 3
31.3.4 96
31.3.5 8
10
31.3.6 91
31.3.7 12
31.3.8 8.64
31.3.9 3
31.3.10 18
31.3.11 32
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Chapter 31. Triangle Area and Length
31.3.12 5
31.3.13 28
3
31.3.14 3
31.3.15
21
13
31.3.16 8
31.3.17 420
1
31.3.18 4m
31.3.19 18
31.3.20 19
56
31.3.21 14
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Chapter 32
Special Triangles
Video Lecture
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32.1
Chapter 32. Special Triangles
Equilateral Triangle
Theorem 32.1.1
If the side length of an equilateral triangle is a
√
3
Height of the triangle =
a
2
This follows directly from the 30 − 60 − 90 triangle.
√
Area of the equilateral triangle =
3 2
a
4
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32.2
Chapter 32. Special Triangles
Right Triangles
Theorem 32.2.1 (Pythagorean Theorem)
A right triangle with legs a and b and hypotenuse c satisfies the following relation:
c 2 = a2 + b 2
B
a
c
C
b
A
Fact 32.2.2. Important Pythagorean Triples
3, 4, 5
5, 12, 13
7, 24, 25
8, 15, 17
9, 40, 41
20, 21, 29
If all numbers in a pythagorean triple are multiplied by a constant, the resulting numbers still form a pythagorean triple.
For example: These are all pythagorean triples:
3, 4, 5
6, 8, 10
9, 12, 15
12, 16, 20
15, 20, 25
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Chapter 32. Special Triangles
Example 32.1 (AMC 10/12)
Let △XOY be a right-angled triangle with m∠XOY = 90◦ . Let M and N be the
midpoints of legs OX and OY , respectively. Given that XN = 19 and Y M = 22, find
XY .
Video Solution
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32.2.1
Chapter 32. Special Triangles
45-45-90 Triangle
Theorem 32.2.3
If the side length of a 45-45-90 triangle is a
hypotenuse of the triangle =
Area of the triangle =
√
√
2 × side length = 2a
1
1
× side length2 = a2
2
2
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32.2.2
Chapter 32. Special Triangles
30-60-90 Triangle
Theorem 32.2.4
If the short leg length of a 30-60-90 triangle is a
Long Leg of the triangle =
√
3 × short leg =
√
3a
hypotenuse of the triangle = 2 × short leg = 2a
√
√
3
3 2
Area =
× short leg2 =
a
2
2
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Chapter 32. Special Triangles
Example 32.2 (AIME)
In triangle ABC, angles A and B measure 60 degrees and 45 degrees, respectively. The
bisector of angle A intersects
√ BC at T , and AT = 24. The area of triangle ABC can be
written in the form a + b c, where a, b, and c are positive integers, and c is not divisible
by the square of any prime. Find a + b + c.
C
T
24
30◦
45◦
30◦
A
D
B
Video Solution
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32.2.3
Chapter 32. Special Triangles
13-14-15 Triangle
Theorem 32.2.5
If the three sides of a triangle are 13, 14, and 15, it can be divided into two right triangles
with side lengths:
5, 12, 13 and 9, 12, 15
Area of this triangle = 12 × 14 × 12 = 84
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32.3
Chapter 32. Special Triangles
Practice Problems
Problem 32.3.1 (AMC 10)
Let ABC be an equilateral triangle. Extend side AB beyond B to a point B ′ so that
BB ′ = 3 · AB. Similarly, extend side BC beyond C to a point C ′ so that CC ′ = 3 · BC,
and extend side CA beyond A to a point A′ so that AA′ = 3 · CA. What is the ratio of
the area of △A′ B ′ C ′ to the area of △ABC?
Video Solution
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Chapter 32. Special Triangles
Problem 32.3.2 (AMC 10)
Triangle ABC has a right angle at B, AB = 1, and BC = 2. The bisector of ∠BAC
meets BC at D. What is BD?
A
1
B
D
C
Video Solution
Additional Problems
Problem 32.3.3 (AIME)
In quadrilateral ABCD, ∠B is a right angle, diagonal AC is perpendicular to CD,
AB = 18, BC = 21, and CD = 14. Find the perimeter of ABCD.
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Chapter 32. Special Triangles
B
A
C
D
Problem 32.3.4 (AMC 10)
In rectangle ABCD, AB = 20 and BC = 10. Let E be a point on CD such that
∠CBE = 15◦ . What is AE?
Problem 32.3.5 (AMC 12)
How many non-congruent right triangles with positive integer leg lengths have areas that
are numerically equal to 3 times their perimeters?
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Chapter 32. Special Triangles
Problem 32.3.6 (AMC 10)
Points A, B, C and D lie on a line, in that order, with AB = CD and BC = 12.
Point E is not on the line, and BE = CE = 10. The perimeter of △AED is twice the
perimeter of △BEC. Find AB.
Problem 32.3.7 (AMC 10)
A large equilateral triangle is constructed by using toothpicks to create rows of small
equilateral triangles. For example, in the figure, we have 3 rows of small congruent
equilateral triangles, with 5 small triangles in the base row. How many toothpicks would
be needed to construct a large equilateral triangle if the base row of the triangle consists
of 2003 small equilateral triangles?
2
1
4
3
5
Problem 32.3.8 (AMC 12)
Let A0 = (0, 0). Distinct √
points A1 , A2 , . . . lie on the x-axis, and distinct points B1 , B2 , . . .
lie on the graph of y = x. For every positive integer n, An−1 Bn An is an equilateral
triangle. What is the least n for which the length A0 An ≥ 100?
Problem 32.3.9 (AMC 10)
Suppose that △ABC is an equilateral triangle of side length s, with the property
√
that there is a unique point P inside the triangle such that AP = 1, BP = 3, and
CP = 2. What is s?
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Chapter 32. Special Triangles
Answers
32.1 26
32.2 291
32.3.1 37 : 1
√
32.3.2
5−1
2
32.3.3 084
32.3.4 20
32.3.5 6
32.3.6 9
32.3.7 1, 507, 509
32.3.8 17
32.3.9
√
7
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Chapter 33
Similar Triangles
Video Lecture
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Chapter 33. Similar Triangles
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Chapter 33. Similar Triangles
Theorem 33.0.1
For similar triangles:
1. All the angles of the triangles are same
2. All corresponding sides have same ratio
3. Area ratio is the square of side length ratio
Concept 33.0.2 (Similarity Test)
Two triangles are similar if the three angles in the triangle are the same. In other words,
the triangles are the same shape multiplied by a scale factor. In general, triangles are
similar if:
• AA similarity: Two angles of the triangles are same, which basically means that
the third angle will be equal)
• SAS similarity (Side Angle Side): Two sides are proportional and the angle between
the sides is equal
• SSS similarity (Side Side Side): All three sides are proportional
• HL similarity (Hypotenuse Leg): In a right triangle, hypotenuse and leg are
proportional
• LL similarity (LL Leg): In a right triangle, the two legs are proportional
Warning: SSA does not mean triangles are similar
An easy way to detect similar triangles is if bases of triangles are parallel and the
sides of the triangles are collinear (see figure below)
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Chapter 33. Similar Triangles
Example 33.1 (AIME)
A point P is chosen in the interior of △ABC such that when lines are drawn through P
parallel to the sides of △ABC, the resulting smaller triangles t1 , t2 , and t3 in the figure,
have areas 4, 9, and 49, respectively. Find the area of △ABC.
A
t2
t1
t3
B
C
Video Solution
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Chapter 33. Similar Triangles
Theorem 33.0.3 (Special Properties of right triangles)
In a right triangle ABC where B is the right angle, the following triangles are similar
△ABC ∼ △ADB ∼ △BDC
Length of the perpendicular to the hypotenuse (BD) =
Also note that:
q
AB·BC
AC
AD · CD = BD2
AD · AC = AB2
CD · CA = CB2
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Chapter 33. Similar Triangles
Example 33.2 (AMC 10)
In square ABCD, points P and Q lie on AD and AB, respectively. Segments BP and
CQ intersect at right angles at R, with BR = 6 and P R = 7. What is the area of the
square?
D
C
P
7
R
6
A
Q
B
Video Solution
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Chapter 33. Similar Triangles
Example 33.3 (AMC 10/12)
Quadrilateral ABCD satisfies ∠ABC = ∠ACD = 90◦ , AC = 20, and CD = 30.
Diagonals AC and BD intersect at point E, and AE = 5. What is the area of quadrilateral
ABCD?
D
A
30
F
x
E
20
B
C
Video Solution
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33.1
Chapter 33. Similar Triangles
Angle Bisector Theorem
Theorem 33.1.1 (Angle Bisector Theorem)
If the line AD bisects angle A, then
AB
AC
=
BD
CD
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Chapter 33. Similar Triangles
Example 33.4 (AMC 12)
In △ABC, AB = 6, BC = 7, and CA = 8. Point D lies on BC, and AD bisects ∠BAC.
Point E lies on AC, and BE bisects ∠ABC. The bisectors intersect at F . What is the
ratio AF : F D?
C
E
D
F
A
B
Video Solution
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Chapter 33. Similar Triangles
Example 33.5 (AIME)
In triangle ABC, AB = 20 and AC = 11. The angle bisector of ∠A intersects BC at
point D, and point M is the midpoint of AD. Let P be the point of the intersection of
m
AC and BM . The ratio of CP to P A can be expressed in the form , where m and n
n
are relatively prime positive integers. Find m + n.
B
20
D
M
A
P
11
D′
C
Video Solution
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33.2
Chapter 33. Similar Triangles
Practice Problems
Problem 33.2.1 (AMC 10)
A paper triangle with sides of lengths 3, 4, and 5 inches, as shown, is folded so that point
A falls on point B. What is the length in inches of the crease?
B
5
A
4
3
C
Video Solution
Problem 33.2.2 (AMC 10)
Three unit squares and two line segments connecting two pairs of vertices are shown.
What is the area of △ABC?
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Chapter 33. Similar Triangles
A
B
C
Video Solution
Problem 33.2.3 (AMC 10/12)
Trapezoid ABCD has AB ∥ CD, BC = CD = 43, and AD ⊥ BD. Let O be the
intersection of the diagonals AC and BD, and let P be the
√ midpoint of BD. Given that
OP = 11, the length of AD can be written in the form m n, where m and n are positive
integers and n is not divisible by the square of any prime. What is m + n?
A
B
P
43
11
O
D
43
C
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Chapter 33. Similar Triangles
Video Solution
Problem 33.2.4 (AMC 10)
Right triangle ABC has leg lengths AB = 20 and BC = 21. Including AB and
BC, how many line segments with integer length can be drawn from vertex B to a point
on hypotenuse AC?
C
P
A
B
Video Solution
Problem 33.2.5 (AMC 10)
In quadrilateral ABCD, ∠BAD ∼
= ∠ADC and ∠ABD ∼
= ∠BCD, AB = 8, BD = 10,
m
and BC = 6. The length CD may be written in the form n , where m and n are relatively
prime positive integers. Find m + n.
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Chapter 33. Similar Triangles
Video Solution
Problem 33.2.6 (AMC 10)
In △ABC with a right angle at C, point D lies in the interior of AB and point E
lies in the interior of BC so that AC = CD, DE = EB, and the ratio AC : DE = 4 : 3.
What is the ratio AD : DB?
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Chapter 33. Similar Triangles
B
E
D
C
A
Video Solution
Problem 33.2.7 (AMC 10)
Triangle ABC with AB = 50 and AC = 10 has area 120. Let D be the midpoint
of AB, and let E be the midpoint of AC. The angle bisector of ∠BAC intersects DE
and BC at F and G, respectively. What is the area of quadrilateral F DBG?
Video Solution
Problem 33.2.8 (AMC 10)
In rectangle ABCD, AB = 6 and BC = 3. Point E between B and C, and point
F between E and C are such that BE = EF = F C. Segments AE and AF intersect
BD at P and Q, respectively. The ratio BP : P Q : QD can be written as r : s : t where
the greatest common factor of r, s, and t is 1. What is r + s + t?
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Chapter 33. Similar Triangles
A
B
P
E
Q
F
D
C
Video Solution
Problem 33.2.9 (AMC 12)
Triangle ABC has AB = 13, BC = 14 and AC = 15. Let P be the point on AC
such that P C = 10. There are exactly two points D and E on line BP such that
quadrilaterals ABCD and ABCE are trapezoids. What is the distance DE?
Video Solution
Problem 33.2.10 (AMC 10)
Triangle ABC with AB = 50 and AC = 10 has area 120. Let D be the midpoint
of AB, and let E be the midpoint of AC. The angle bisector of ∠BAC intersects DE
and BC at F and G, respectively. What is the area of quadrilateral F DBG?
Video Solution
Problem 33.2.11 (AMC 10A)
Let △ABC be a scalene triangle. Point P lies on BC so that AP bisects ∠BAC.
The line through B perpendicular to AP intersects the line through A parallel to BC at
point D. Suppose BP = 2 and P C = 3. What is AD ?
(A) 8
(B) 9
(C) 10
(D) 11
(E) 12
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Chapter 33. Similar Triangles
Video Solution
Problem 33.2.12 (AMC 10/12)
The diagram below shows a rectangle with side lengths 4 and 8 and a square with
side length 5. Three vertices of the square lie on three different sides of the rectangle, as
shown. What is the area of the region inside both the square and the rectangle?
5
4
8
Video Solution
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Chapter 33. Similar Triangles
Problem 33.2.13 (AMC 10B #20)
Let ABCD be a rhombus with ∠ADC = 46◦ . Let E be the midpoint of CD, and
let F be the point on BE such that AF is perpendicular to BE. What is the degree
measure of ∠BF C?
Video Solution
Additional Problems
Problem 33.2.14 (AMC 10)
Rectangle ABCD has AB = 6 and BC = 3. Point M is chosen on side AB so
that ∠AM D = ∠CM D. What is the degree measure of ∠AM D?
A
M
B
6
D
6
3
C
Problem 33.2.15 (AIME)
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of
the area of shaded region C to the area of shaded region B is 11/5. Find the ratio of
shaded region D to the area of shaded region A.
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Chapter 33. Similar Triangles
A
B
C
D
Problem 33.2.16 (AIME)
Let P be an interior point of triangle ABC and extend lines from the vertices through P
to the opposite sides. Let a, b, c, and d denote the lengths of the segments indicated in
the figure. Find the product abc if a + b + c = 43 and d = 3.
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Chapter 33. Similar Triangles
Problem 33.2.17 (AMC 12)
In triangle ABC, AB = 13, BC = 14, and CA = 15. Distinct points D, E, and
F lie on segments BC, CA, and DE, respectively, such that AD ⊥ BC, DE ⊥ AC, and
AF ⊥ BF . The length of segment DF can be written as m
, where m and n are relatively
n
prime positive integers. What is m + n?
B
5
13
D
F
A
E
9
C
Problem 33.2.18 (AMC 10)
A square with side length x is inscribed in a right triangle with sides of length 3,
4, and 5 so that one vertex of the square coincides with the right-angle vertex of the
triangle. A square with side length y is inscribed in another right triangle with sides of
length 3, 4, and 5 so that one side of the square lies on the hypotenuse of the triangle.
x
What is ?
y
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Chapter 33. Similar Triangles
C
E
D
x
A
B
F
C′
S
R
T
y
A′
Q
B′
Problem 33.2.19 (AIME)
Trapezoid ABCD has sides AB = 92, BC = 50, CD = 19, and AD = 70, with
AB parallel to CD. A circle with center P on AB is drawn tangent to BC and AD.
Given that AP = m
, where m and n are relatively prime positive integers, find m + n.
n
Problem 33.2.20 (AIME)
In trapezoid ABCD, leg BC is perpendicular to√bases AB and √
CD, and diagonals
AC and BD are perpendicular. Given that AB = 11 and AD = 1001, find BC 2 .
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Chapter 33. Similar Triangles
B
√
11
A
√
x
1001
y
C
D
Problem 33.2.21 (AMC 10)
In triangle ABC, AB = 13, BC = 14, and CA = 15. Distinct points D, E, and
F lie on segments BC, CA, and DE, respectively, such that AD ⊥ BC, DE ⊥ AC, and
AF ⊥ BF . The length of segment DF can be written as m
, where m and n are relatively
n
prime positive integers. What is m + n?
(A) 18
(B) 21
(C) 24
(D) 27
(E) 30
B
5
13
D
F
A
E
9
C
Problem 33.2.22 (AIME)
Rhombus P QRS is inscribed in rectangle ABCD so that vertices P , Q, R, and S are
interior points on sides AB, BC, CD, and DA, respectively. It is given that P B = 15,
BQ = 20, P R = 30, and QS = 40. Let m
, in lowest terms, denote the perimeter of
n
ABCD. Find m + n.
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Chapter 33. Similar Triangles
P
A
15
B
S
20
O
Q
D
R
C
Problem 33.2.23 (AMC 10)
In the right triangle △ACE, we have AC = 12, CE = 16, and EA = 20. Points
B, D, and F are located on AC, CE, and EA, respectively, so that AB = 3, CD = 4,
and EF = 5. What is the ratio of the area of △DBF to that of △ACE?
A
3
B
15
9
F
5
C
4
D
12
E
Problem 33.2.24 (AMC 10)
Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpendicular from B to diagonal AC. What is the area of △AED?
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Chapter 33. Similar Triangles
A
B
3
E
4
D
C
Problem 33.2.25 (AMC 10)
In rectangle ABCD, AB = 1, BC = 2, and points E, F , and G are midpoints of
BC, CD, and AD, respectively. Point H is the midpoint of GE. What is the area of the
shaded region?
A
B
1
H
G
E
1
D
1
2
F
1
2
C
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Chapter 33. Similar Triangles
Problem 33.2.26 (AMC 10)
Rectangle ABCD has AB = 6 and BC = 3. Point M is chosen on side AB so
that ∠AM D = ∠CM D. What is the degree measure of ∠AM D?
A
M
B
3
6
D
C
6
Problem 33.2.27 (AMC 10)
Rectangle ABCD has AB = 8 and BC = 6. Point M is the midpoint of diagonal
AC, and E is on AB with M E ⊥ AC. What is the area of △AM E?
D
C
M
A
E
B
Problem 33.2.28 (AMC 10)
Rectangle ABCD has AB = 4 and BC = 3. Segment EF is constructed through
B so that EF is perpendicular to DB, and A and C lie on DE and DF , respectively.
What is EF ?
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Chapter 33. Similar Triangles
C
4
D
F
3
A
B
E
Problem 33.2.29 (AMC 10)
In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that
DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of △AEB.
E
D
1
2
F
C
G
3
A
5
B
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Chapter 33. Similar Triangles
Problem 33.2.30 (AMC 10)
In rectangle ABCD, we have AB = 8, BC = 9, H is on BC with BH = 6, E is
on AD with DE = 4, line EC intersects line AH at G, and F is on line AD with
GF ⊥ AF . Find the length of GF .
G
F
D
6
H
C
4
E
B
A
Problem 33.2.31 (AIME)
Square ABCD has sides of length 1. Points E and F are on BC and CD, respectively, so that △AEF is equilateral. A square with vertex B has sides that are parallel
to √those of ABCD and a vertex on AE. The length of a side of this smaller square is
a− b
, where a, b, and c are positive integers and b is not divisible by the square of any
c
prime. Find a + b + c.
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Chapter 33. Similar Triangles
1
A
D
F
A′
D′
s
B
C′
E
C
Problem 33.2.32 (AMC 12)
Let ABC be a triangle where M is the midpoint of AC, and CN is the angle bisector of ∠ACB with N on AB. Let X be the intersection of the median BM and the
bisector CN . In addition △BXN is equilateral with AC = 2. What is BX 2 ?
Answers
33.1 144
33.2 117
33.3 360
33.4 2 : 1
33.5 51
33.2.1 15
8
33.2.2 15
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Chapter 33. Similar Triangles
33.2.3 194
33.2.4 13
33.2.5 069
33.2.6 2 : 3
33.2.7 75
33.2.8 20
√
33.2.9 12 2
33.2.10 75
33.2.11 10
33.2.12 15
5
8
33.2.13 113
33.2.14 75
33.2.15 408
33.2.16 441
33.2.17 21
33.2.18 37
35
33.2.19 164
33.2.20 110
33.2.21 21
33.2.22 677
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Chapter 33. Similar Triangles
7
33.2.23 16
33.2.24 54
25
33.2.25
1
6
33.2.26 75
33.2.27 75
8
33.2.28 125
12
33.2.29 25
2
33.2.30 20
33.2.31 12
√
2
33.2.32 10−6
7
399
Chapter 34
Quadrilaterals
Video Lecture
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34.1
Chapter 34. Quadrilaterals
Square
Theorem 34.1.1 (Area of a Square)
Any square with side length s has an area of
s2
and a perimeter of
34.2
4s
Rectangle
Theorem 34.2.1 (Area of a Rectangle)
Any rectangle with base b and height h has an area of
bh
and a perimeter of
2b + 2h
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Chapter 34. Quadrilaterals
Example 34.1 (AIME)
Let ABCD be a square, and let E and F be points on AB and BC, respectively. The
line through E parallel to BC and the line through F parallel to AB divide ABCD into
two squares and two nonsquare rectangles. The sum of the areas of the two squares is
AE
9
of the area of square ABCD. Find EB
+ EB
.
10
AE
Video Solution
34.3
Rhombus
Theorem 34.3.1 (Area of a Rhombus)
A rhombus with diagonals d1 and d2 has an area of
1
d1 d2
2
and a perimeter of
2×
Concept 34.3.2 (Rhombus)
q
d21 + d22
1. All sides equal
2. Opposite angles congruent
3. Diagonals are perpendicular bisectors
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34.4
Chapter 34. Quadrilaterals
Parallelogram
Theorem 34.4.1 (Area of a Parallelogram)
A parallelogram with base b and height h has an area of
bh
A parallelogram with diagonals d1 and d2 has an area of
1
d1 d2 ∗ sin(θ)
2
where θ is the central angle of the parallelogram
Concept 34.4.2 (Parallelogram Properties)
1. Diagonals bisect each other
2. Opposite sides congruent
3. Opposite angles congruent
4. Base angles supplementary
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Chapter 34. Quadrilaterals
Example 34.2 (AIME)
In △ABC, AB = 425, BC = 450, and AC = 510. An interior point P is then drawn,
and segments are drawn through P parallel to the sides of the triangle. If these three
segments are of an equal length d, find d.
Video Solution
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34.5
Chapter 34. Quadrilaterals
Trapezoid
Theorem 34.5.1 (Area of a Trapezoid)
A trapezoid with 2 bases b1 and b2 and a height h has an area of
b1 + b2
·h
2
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Chapter 34. Quadrilaterals
Example 34.3 (AMC 12)
Let ABCD be an isosceles trapezoid with BC ∥ AD and AB = CD. Points X and
Y lie on diagonal AC with X between A and Y , as shown in the figure. Suppose
∠AXD = ∠BY C = 90◦ , AX = 3, XY = 1, and Y C = 2. What is the area of ABCD?
Video Solution
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34.6
Chapter 34. Quadrilaterals
Practice Problems
Problem 34.6.1 (AMC 12)
Convex quadrilateral ABCD has AB = 3, BC = 4, CD = 13, AD = 12, and
∠ABC = 90◦ , as shown. What is the area of the quadrilateral?
B
A
C
D
Video Solution
Problem 34.6.2 (AMC 10)
The five small shaded squares inside this unit square are congruent and have disjoint
interiors. The midpoint of each side of the middle square coincides with one
√ of the
vertices of the other four small squares as shown. The common side length is a−b 2 , where
a and b are positive integers. What is a + b ?
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Chapter 34. Quadrilaterals
Video Solution
Problem 34.6.3 (AMC 10)
Square EF GH has one vertex on each side of square ABCD. Point E is on AB
with AE = 7 · EB. What is the ratio of the area of EF GH to the area of ABCD?
Video Solution
Problem 34.6.4 (AMC 10)
Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpendicular from B to diagonal AC. What is the area of △AED?
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Chapter 34. Quadrilaterals
A
3
B
E
4
D
C
Video Solution
Problem 34.6.5 (AMC 10)
A square piece of paper has side length 1 and vertices A, B, C, andD in that order.
As shown in the figure the paper is folded so that vertex C meets edge AD at point
C ′ , and edge BC intersects edge AB at point E. Suppose that C ′ D = 13 . What is the
perimeter of △AEC ′ ?
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Chapter 34. Quadrilaterals
A
C’
D
E
B
C
Video Solution
Problem 34.6.6 (AMC 12)
Let ABCD be a square. Let E, F, G and H be the centers, respectively, of equilateral triangles with bases AB, BC, CD, and DA, each exterior to the square. What is
the ratio of the area of square EF GH to the area of square ABCD?
Problem 34.6.7 (AMC 10/12)
Trapezoid ABCD has AB ∥ CD, BC = CD = 43, and AD ⊥ BD. Let O be the
intersection of the diagonals AC and BD, and let P be √
the midpoint of BD. GIven that
OP = 11, the length AD can be written in the form m n, where m and n are positive
integers and n is not divisible by the square of any prime. What is m + n?
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Chapter 34. Quadrilaterals
A
B
P
43
11
O
D
43
C
Video Solution
Problem 34.6.8 (AMC 12)
Let ABCD be an isosceles trapezoid having parallel bases AB and CD with AB > CD.
Line segments from a point inside ABCD to the vertices divide the trapezoid into four
triangles whose areas are 2, 3, 4, and 5 starting with the triangle with base CD and
AB
moving clockwise as shown in the diagram below. What is the ratio CD
?
Video Solution
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Chapter 34. Quadrilaterals
Problem 34.6.9 (AMC 12)
Let ABCD be a parallelogram with area 15. Points P and Q are the projections
of A and C, respectively, onto the line BD; and points R and S are the projections of
B and D, respectively, onto the line AC. See the figure, which also shows the relative
locations of these points.
Suppose P Q = 6 and RS = 8, and let d denote the length of BD, the longer diagonal
√
of ABCD. Then d2 can be written in the form m + n p, where m, n, and p are positive
integers and p is not divisible by the square of any prime. What is m + n + p?
Video Solution
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Chapter 34. Quadrilaterals
Problem 34.6.10 (AMC 10/12)
Isosceles trapezoid ABCD has parallel sides AD and BC, with BC < AD and AB = CD.
There is a point P in the plane such that P A = 1, P B = 2, P C = 3, and P D = 4. What
BC
is AD
?
(A) 14
(B) 13
(C) 12
(D) 23
(E) 34
Video Solution
Additional Problems
Problem 34.6.11 (AMC 10/12)
Four congruent rectangles are placed as shown. The area of the outer square is 4
times that of the inner square. What is the ratio of the length of the longer side of each
rectangle to the length of its shorter side?
413
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Chapter 34. Quadrilaterals
Problem 34.6.12 (AMC 10)
In square ABCD, points E and H lie on AB and DA, respectively, so that AE = AH.
Points F and G lie on BC and CD, respectively, and points I and J lie on EH so that
F I ⊥ EH and GJ ⊥ EH. See the figure below. Triangle AEH, quadrilateral BF IE,
quadrilateral DHJG, and pentagon F CGJI each has area 1. What is F I 2 ?
D
G
C
F
H
J
I
A
E
B
Problem 34.6.13 (AMC 10)
A closed box with a square base is to be wrapped with a square sheet of wrapping
paper. The box is centered on the wrapping paper with the vertices of the base lying on
the midlines of the square sheet of paper, as shown in the figure on the left. The four
corners of the wrapping paper are to be folded up over the sides and brought together to
meet at the center of the top of the box, point A in the figure on the right. The box has
base length w and height h. What is the area of the sheet of wrapping paper?
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Chapter 34. Quadrilaterals
A
A
w
w
Problem 34.6.14 (AMC 10)
Rectangle ABCD has AB = 3 and BC = 4. Point E is the foot of the perpendicular from B to diagonal AC. What is the area of △AED?
A
3
B
E
4
D
C
Problem 34.6.15 (AMC 12)
The internal angles of quadrilateral ABCD form an arithmetic progression. Triangles
ABD and DCB are similar with ∠DBA = ∠DCB and ∠ADB = ∠CBD. Moreover,
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Chapter 34. Quadrilaterals
the angles in each of these two triangles also form an arithmetic progression. In degrees,
what is the largest possible sum of the two largest angles of ABCD?
Problem 34.6.16 (AMC 10)
Consider the 12-sided polygon ABCDEF GHIJKL, as shown. Each of its sides has
length 4, and each two consecutive sides form a right angle. Suppose that AG and CH
meet at M . What is the area of quadrilateral ABCM ?
K
A
B
L
C
D
I
F
E
H
G
M
J
Problem 34.6.17 (AIME)
Let ABCD be a square, and let E and F be points on AB and BC, respectively.
The line through E parallel to BC and the line through F parallel to AB divide ABCD
into two squares and two nonsquare rectangles. The sum of the areas of the two squares
9
AE
is 10
of the area of square ABCD. Find EB
+ EB
.
AE
Problem 34.6.18 (AMC 10)
Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD
intersect at E, AC = 14, and △AED and △BEC have equal areas. What is AE?
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A
Chapter 34. Quadrilaterals
9
B
E
D
12
C
Problem 34.6.19 (AMC 10)
In rectangle P QRS, P Q = 8 and QR = 6. Points A and B lie on P Q, points C
and D lie on QR, points E and F lie on RS, and points G and H lie on SP so that
AP = BQ < 4 and the convex octagon ABCDEF GH is √equilateral. The length of
a side of this octagon can be expressed in the form k + m n, where k, m, and n are
integers and n is not divisible by the square of any prime. What is k + m + n?
Problem 34.6.20 (AMC 10)
In rectangle ABCD, DC = 2 · CB and points E and F lie on AB so that ED and F D
trisect ∠ADC as shown. What is the ratio of the area of △DEF to the area of rectangle
ABCD?
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Chapter 34. Quadrilaterals
A
E
F
D
B
C
Problem 34.6.21 (AMC 10)
Trapezoid ABCD has parallel sides AB of length 33 and CD of length 21. The other
two sides are of lengths 10 and 14. The angles A and B are acute. What is the length of
the shorter diagonal of ABCD?
E
A
33
F
B
10
14
D
C
21
Problem 34.6.22 (AMC 10)
In rectangle ABCD, AB = 6, AD = 30, and G is the midpoint of AD. Segment
AB is extended 2 units beyond B to point E, and F is the intersection of ED and BC.
What is the area of quadrilateral BF DG?
A
15
G
15
D
6
6
B
2
F
C
E
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Chapter 34. Quadrilaterals
Problem 34.6.23 (AMC 10)
Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose
that AB = 9, DC = 12, and the area of △AKD is 24. What is the area of trapezoid
ABCD?
Problem 34.6.24 (AIME)
In the diagram below, ABCD is a rectangle with side lengths AB = 3 and BC = 11,
and AECF is a rectangle with side lengths AF = 7 and F C = 9, as shown. The area of
the shaded region common to the interiors of both rectangles is m
, where m and n are
n
relatively prime positive integers. Find m + n.
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Chapter 34. Quadrilaterals
F
A
D
B
C
E
Problem 34.6.25 (AMC 10)
A rectangular floor measures a by b feet, where a and b are positive integers and
b > a. An artist paints a rectangle on the floor with the sides of the rectangle parallel
to the floor. The unpainted part of the floor forms a border of width 1 foot around the
painted rectangle and occupies half the area of the whole floor. How many possibilities
are there for the ordered pair (a, b)?
Problem 34.6.26 (AMC 10)
In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, and F
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Chapter 34. Quadrilaterals
as the midpoint of DA. The area of ABEF is twice the area of F ECD. What is
AB/DC?
Problem 34.6.27 (AIME)
ABCD is a rectangular sheet of paper that has been folded so that corner B is matched
with point B ′ on edge AD. The crease is EF, where E is on AB and F is on CD. The
dimensions AE = 8, BE = 17, and CF = 3 are given. The perimeter of rectangle ABCD
is m/n, where m and n are relatively prime positive integers. Find m + n.
C′
F
D
C
B′
A
E
B
Problem 34.6.28 (AIME)
In convex quadrilateral KLM N side M N is perpendicular to diagonal KM , side KL
is perpendicular to diagonal LN , M N = 65, and KL = 28. The line through L
perpendicular to side KN intersects diagonal KM at O with KO = 8. Find M O.
Problem 34.6.29 (AMC 12)
In the figure, ABCD is a square of side length 1. The rectangles JKHG and EBCF
are congruent. What is BE?
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C
Chapter 34. Quadrilaterals
F
H
D
G
K
B
E
J
A
Answers
34.1 18
34.2 306
√
34.3 3 35
34.6.1 36
34.6.2 11
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Chapter 34. Quadrilaterals
34.6.3 25
32
34.6.4 54
25
34.6.5 2
√
34.6.6 2+3 3
34.6.7 194
34.6.8 2 +
√
2
34.6.9 81
34.6.10 31
34.6.11 3
√
34.6.12 8 − 4 2
34.6.13 2(w + h)2
54
34.6.14 25
34.6.15 240
34.6.16 88
5
34.6.17 018
34.6.18 6
34.6.19 7
√
34.6.20
3
6
34.6.21 25
34.6.22 135
2
34.6.23 98
34.6.24 109
34.6.25 2
34.6.26 5
34.6.27 293
34.6.28 090
34.6.29 2 −
√
3
423
Chapter 35
Circles
Video Lecture
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Chapter 35. Circles
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Chapter 35. Circles
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35.1
Chapter 35. Circles
Circle Area
Theorem 35.1.1 (Area and Circumference)
A circle with radius r has
Area = πr2
Circumference = 2πr
Theorem 35.1.2 (Arcs of a circle)
An arc of a circle with radius r and angle a°
Area of a sector = πr2 ×
a°
= π × radius2 × fraction of circle in sector
360
Length of the arc = 2πr ×
a°
= 2π × radius × fraction of circle in sector
360
Definition 35.1.3 (Angle of an arc). This is the angle that the arc makes at the center of
the circle.
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Chapter 35. Circles
Example 35.1 (AMC 10)
Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1.
8
The area of the shaded region in the diagram is 13
of the area of the unshaded region.
What is the radian measure of the acute angle formed by the two lines? (Note: π radians
is 180 degrees.)
Video Solution
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Chapter 35. Circles
Example 35.2 (AMC 10)
As shown in the figure, line segment AD is trisected by points B and C so that
AB = BC = CD = 2. Three semicircles of radius 1, AEB, BFC, and CGD, have their
diameters on AD, and are tangent to line EG at E, F, and G, respectively. A circle of
radius 2 has its center on F. The area of the region inside the circle but outside the three
semicircles, shaded in the figure, can be expressed in the form
⌢ ⌢
⌢
√
a
· π − c + d,
b
where a, b, c, and d are positive integers and a and b are relatively prime. What is
a + b + c + d?
F
E
A
B
G
C
D
Video Solution
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35.2
Chapter 35. Circles
Circle Length
Concept 35.2.1
In geometry problems with circles, always try to draw all important radii
(i.e. connect center to all points on a circle)
Example 35.3 (AIME)
A circle with diameter P Q of length 10 is internally tangent at P to a circle of radius
20. Square ABCD is constructed with A and B on the larger circle, CD tangent at Q
to the smaller circle, and√the smaller circle outside ABCD. The length of AB can be
written in the form m + n, where m and n are integers. Find m + n.
D
P
A
Q
C
B
Video Solution
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Chapter 35. Circles
Example 35.4 (AIME)
A machine-shop
√ cutting tool has the shape of a notched circle, as shown. The radius of
the circle is 50 cm, the length of AB is 6 cm and that of BC is 2 cm. The angle ABC
is a right angle. Find the square of the distance (in centimeters) from B to the center of
the circle.
A
B
C
Video Solution
35.3
Power of a Point
Theorem 35.3.1 (Power of Point For 2 Tangents)
From a given point P external to a circle, the two tangents to the circle are equal.
PS = PT
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Chapter 35. Circles
Theorem 35.3.2 (Power of Point Inside Circle)
If AB and CD are two secants in a circle with Center O, which intersect at a point P,
the line segments satisfy the following property:
P A · P B = P C · P D = r2 − OP 2
Theorem 35.3.3 (Power of Point Outside Circle)
If AB and CD are two secants in a circle, which intersect at a point P outside the circle
with Center O, the line segments satisfy the following property:
P A · P B = P C · P D = OP 2 − r2
where r is the radius of the circle
Remark 35.3.4
Power of a point is useful when dealing with circles and chord lengths.
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Chapter 35. Circles
Example 35.5 (AMC 12)
√
Let AB be a diameter in a circle of radius 5√ 2. Let CD be a chord in the circle that
intersects AB at a point E such that BE = 2 5 and ∠AEC = 45◦ . What is CE 2 +DE 2 ?
C
A
√
5 2
45◦ E
√
2 5
B
D
Video Solution
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Chapter 35. Circles
Example 35.6 (AIME)
Let ABCD be a parallelogram with ∠BAD < 90◦ . A circle tangent to sides DA, AB,
and BC intersects diagonal AC at points P and Q with AP < AQ, as shown. Suppose
that AP√= 3, P Q = 9, and QC = 16. Then the area of ABCD can be expressed in the
form m n, where m and n are positive integers, and n is not divisible by the square of
any prime. Find m + n.
B
C
Q
P
A
D
Video Solution
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35.4
Chapter 35. Circles
Cyclic Quadrilaterals
Theorem 35.4.1 (Properties of a Cyclic quadrilateral)
Sum of opposite angles = 180
You can draw a circle around quadrilateral and use angle chasing properties for circles
like the inscribed angle theorem.
Concept 35.4.2
Sometimes, there may be quadrilaterals not explicitly stated to be in a circle. Then, if
the sum of opposite angles is 180° or if the angles satisfy properties from the inscribed
angle theorem, then you can identify them to be a cyclic quadrilateral, and you can
use any of the properties of cyclic quadrilaterals or even circles because you know the
quadrilateral can be inscribed in a circle.
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Chapter 35. Circles
Theorem 35.4.3 (Ptolemy’s Theorem)
In a cyclic quadrilateral ABCD
Product of diagonals = Sum of the product of both pairs of opposite sides
AC · BD = AB · CD + AD · BC
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Chapter 35. Circles
Theorem 35.4.4 (Brahmagupta’s Formula)
In a cyclic quadrilateral with side lengths a, b, c, d, the area of the quadrilateral can be
found as:
A=
q
(s − a)(s − b)(s − c)(s − d)
where s is the semiperimeter of the quadrilateral and can be calculated as
s=
a+b+c+d
2
Basically, to find the area of a cyclic quadrilateral
1. Find the perimeter and divide by 2
2. Subtract each of the side lengths from it to get 4 values
3. Multiply your 4 values
4. Take the square root of your product
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Chapter 35. Circles
Example 35.7 (AMC 10)
√
√
Isosceles triangle ABC has AB = AC = 3 6, and a circle with radius 5 2 is tangent
to line AB at B and to line AC at C. What is the area of the circle that passes through
vertices A, B, and C?
A
√
3 6
√
3 6
O2
B
C
√
5 2
√
5 2
O1
Video Solution
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Chapter 35. Circles
Example 35.8 (AIME)
A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted
by AB, has length 31. Find the sum of the lengths of the three diagonals that can be
drawn from A.
E
D
81
81
81
z
y
F
C
x
81
81
31
A
B
Video Solution
35.5
Practice Problems
Problem 35.5.1 (AMC 10)
Externally tangent circles with centers at points A and B have radii of lengths 5 and
3, respectively. A line externally tangent to both circles intersects ray AB at point C.
What is BC?
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Chapter 35. Circles
A
8
B
x
C
3
5
E
D
Video Solution
Problem 35.5.2 (AMC 10)
In the figure below, N congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap.
Let A be the combined area of the small semicircles and B be the area of the region
inside the large semicircle but outside the semicircles. The ratio A : B is 1 : 18. What is
N?
...
Video Solution
Problem 35.5.3 (AMC 10)
Seven cookies of radius 1 inch are cut from a circle of cookie dough, as shown. Neighboring
cookies are tangent, and all except the center cookie are tangent to the edge of the dough.
The leftover scrap is reshaped to form another cookie of the same thickness. What is the
radius in inches of the scrap cookie?
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Chapter 35. Circles
Video Solution
Problem 35.5.4 (AMC 10)
Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points A and B, as shown in the diagram. The distance
AB can be written in the form m
, where m and n are relatively prime positive integers.
n
What is m + n?
A
B
Video Solution
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Chapter 35. Circles
Problem 35.5.5 (AMC 10)
Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?
Video Solution
Problem 35.5.6 (AMC 10)
The figure below shows 13 circles of radius 1 within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure,
inside the larger circle but outside all the circles of radius 1?
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Chapter 35. Circles
Video Solution
Problem 35.5.7 (AMC 12)
The diameter AB of a circle of radius 2 is extended to a point D outside the circle
so that BD = 3. Point E is chosen so that ED = 5 and line ED is perpendicular to line
AD. Segment AE intersects the circle at a point C between A and E. What is the area
of △ABC?
E
C
A
O
B
D
Video Solution
Problem 35.5.8 (AMC 10)
In △ABC, AB = 86, and AC = 97. A circle with center A and radius AB intersects BC at points B and X. Moreover BX and CX have integer lengths. What is
BC?
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Chapter 35. Circles
86
B
k
E
k
90◦ D
h
m
86
A
97
C
Video Solution
Problem 35.5.9 (AMC 12)
If circular arcs AC and BC have centers at B and A, respectively, then there ex⌢
⌢
⌢
ists a circle tangent to both AC and BC, and to AB. If the length of BC is 12, then the
circumference of the circle is
Video Solution
444
OmegaLearn.org
Chapter 35. Circles
Problem 35.5.10 (AMC 10)
Triangle ABC is an isosceles right triangle with AB = AC = 3. Let M be the midpoint of
hypotenuse BC. Points I and E lie on sides AC and AB, respectively, so that AI > AE
and AIM E is a cyclic√ quadrilateral. Given that triangle EM I has area 2, the length CI
can be written as a−c b , where a, b, and c are positive integers and b is not divisible by
the square of any prime. What is the value of a + b + c?
C
I
M
3
E
A
3
B
Video Solution
Problem 35.5.11 (AMC 10)
√
A quadrilateral is inscribed in a circle of radius 200 2. Three of the sides of this
quadrilateral have length 200. What is the length of the fourth side?
445
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Chapter 35. Circles
B
C
E
A
F
D
θ
O
Video Solution
Problem 35.5.12 (AMC 12)
Semicircle Γ has diameter AB of length 14. Circle
√ Ω lies tangent to◦ AB at a point P
and intersects
Γ at points Q and R. If QR = 3 3 and ∠QP R = 60 , then the area of
√
a b
△P QR is c , where a and c are relatively prime positive integers, and b is a positive
integer not divisible by the square of any prime. What is a + b + c?
Video Solution
Problem 35.5.13 (AMC 10/12)
Let S be the set of circles in the coordinate plane that are tangent to each of the
three circles with equations x2 + y 2 = 4, x2 + y 2 = 64, and (x − 5)2 + y 2 = 3. What is
the sum of the areas of all circles in S?
(A) 48π
(B) 68π
(C) 96π
(D) 102π
(E) 136π
446
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Chapter 35. Circles
Video Solution
Additional Problems
Problem 35.5.14 (AMC 10)
Mary divides a circle into 12 sectors. The central angles of these sectors, measured
in degrees, are all integers and they form an arithmetic sequence. What is the degree
measure of the smallest possible sector angle?
Problem 35.5.15 (AMC 10)
A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an
outside diameter of 20 cm. The outside diameter of each of the outer rings is 1 cm less
than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What
is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
447
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Chapter 35. Circles
20
..
.
3
Problem 35.5.16 (AMC 10)
A circle of radius 1 is tangent to a circle of radius 2. The sides of △ABC are tangent to the circles as shown, and the sides AB and AC are congruent. What is the area
of △ABC?
A
1
2
B
C
448
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Chapter 35. Circles
Problem 35.5.17 (AIME)
The diagram shows twenty congruent circles arranged in three rows and enclosed in a
rectangle. The circles are tangent to one another and to the sides of the rectangle as
shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter
1 √
dimension can be written as ( p − q) where p and q are positive integers. Find p + q.
2
Problem 35.5.18 (AMC 10)
Let ABCD be a cyclic quadrilateral. The side lengths of ABCD are distinct integers less than 15 such that BC · CD = AB · DA. What is the largest possible value of
BD?
Problem 35.5.19 (AMC 10)
In the figure below, semicircles with centers at A and B and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter
JK. The two smaller semicircles are externally tangent to each other and internally
tangent to the largest semicircle. A circle centered at P is drawn externally tangent to
the two smaller semicircles and internally tangent to the largest semicircle. What is the
radius of the circle centered at P ?
449
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Chapter 35. Circles
Problem 35.5.20 (AMC 10)
In △ABC, AB = 6, AC = 8, BC = 10, and D is the midpoint of BC. What is
the sum of the radii of the circles inscribed in △ADB and △ADC?
Problem 35.5.21 (AMC 10)
Triangle ABC is equilateral with side length 6. Suppose that O is the center of the
inscribed circle of this triangle. What is the area of the circle passing through A, O, and
C?
Problem 35.5.22 (AMC 12)
Triangle ABC has AB = 27, AC = 26, and BC = 25. Let I be the intersection
of the internal angle bisectors of △ABC. What is BI?
Problem 35.5.23 (AMC 12)
Circles with radii 1, 2, and 3 are mutually externally tangent. What is the area of
the triangle determined by the points of tangency?
(A) 35
(B) 54
(C) 1
(D) 65
(E) 34
450
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Chapter 35. Circles
Problem 35.5.24 (AMC 10)
√
√
Isosceles triangle ABC has AB = AC = 3 6, and a circle with radius 5 2 is tangent to line AB at B and to line AC at C. What is the area of the circle that passes
through vertices A, B, and C?
Problem 35.5.25 (AMC 10)
A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed
circle. What is the distance between the centers of those circles?
A
E
F
I
B
D
O
C
451
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Chapter 35. Circles
Problem 35.5.26 (AIME)
Let △P QR be a right triangle with P Q = 90, P R = 120, and QR = 150. Let C1
be the inscribed circle. Construct ST with S on P R and T on QR, such that ST is
perpendicular to P R and tangent to C1 . Construct U V with U on P Q and V on QR
such that U V is perpendicular to P Q and tangent to C1 . Let C2 be the inscribed circle
of △RST and C3 the inscribed
circle of △QU V . The distance between the centers of C2
√
and C3 can be written as 10n. What is n?
R
S
T
V
P
U
Q
Problem 35.5.27 (AIME)
Circles C1 and C2 are externally tangent, and they are both internally tangent to
circle C3 . The radii of C1 and C2 are 4 and 10, respectively, and the centers of the three
circles are all collinear. A chord of C3 √is also a common external tangent of C1 and C2 .
Given that the length of the chord is mp n where m, n, and p are positive integers, m and
p are relatively prime, and n is not divisible by the square of any prime, find m + n + p.
452
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Chapter 35. Circles
B
T2
T
A
H
T1
O1
O3
O2
Problem 35.5.28 (AMC 10)
Right triangle ABC has side lengths BC = 6, AC = 8, and AB = 10.
A circle centered at O is tangent to line BC at B and passes through A. A circle centered
at P is tangent to line AC at A and passes through B. What is OP ?
P
θ
B
6
D
O
θ
M
θ
C
8
A
453
OmegaLearn.org
Chapter 35. Circles
Problem 35.5.29 (AMC 10)
Points A and B lie on a circle centered at O, and ∠AOB = 60◦ . A second circle
is internally tangent to the first and tangent to both OA and OB. What is the ratio of
the area of the smaller circle to that of the larger circle?
B
C
r
P
2r
r
◦
30
O
Q
A
R
Problem 35.5.30 (AMC 10)
Rectangle P QRS lies in a plane with P Q = RS = 2 and QR = SP = 6. The
rectangle is rotated 90◦ clockwise about R, then rotated 90◦ clockwise about the point S
moved to after the first rotation. What is the length of the path traveled by point P ?
P
Q′
P′
S
R′′
Q′′
R
S′
P ′′
2
Q
6
454
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Chapter 35. Circles
Problem 35.5.31 (AMC 10)
Circles with centers O and P have radii 2 and 4, respectively, and are externally tangent.
Points A and B on the circle with center O and points C and D on the circle with center
P are such that AD and BC are common external tangents to the circles. What is the
area of the concave hexagon AOBCP D?
D
A
4
2
P
O
B
C
Problem 35.5.32 (AMC 10)
Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal
length, as shown. What is the area of the shaded region that lies within the large
semicircle but outside the smaller semicircles?
A
B
1
2
1
Problem 35.5.33 (AIME)
In the adjoining figure, two circles with radii 8 and 6 are drawn with their centers
455
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Chapter 35. Circles
12 units apart. At P , one of the points of intersection, a line is drawn in such a way that
the chords QP and P R have equal length. Find the square of the length of QP .
Q
P
12
R
Problem 35.5.34 (AIME)
Triangle ABC is inscribed in circle ω. Points P and Q are on side AB with AP <
AQ. Rays CP and CQ meet ω again at S and T (other than C), respectively. If
AP = 4, P Q = 3, QB = 6, BT = 5, and AS = 7, then ST = m
, where m and n are
n
relatively prime positive integers. Find m + n.
456
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Chapter 35. Circles
Problem 35.5.35 (AMC 10)
Let AB be a diameter of a circle and let C be a point on AB with 2 · AC = BC.
Let D and E be points on the circle such that DC ⊥ AB and DE is a second diameter.
What is the ratio of the area of △DCE to the area of △ABD?
457
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Chapter 35. Circles
D
A
C
B
E
Problem 35.5.36 (AMC 10)
Circles A, B and C are externally tangent to each other, and internally tangent to
circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the
center of D. What is the radius of circle B?
D
B
A
C
Problem 35.5.37 (AMC 10)
Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of AB. What is the area
inside circle C but outside circle A and circle B?
458
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Chapter 35. Circles
M
A
C
B
D
Problem 35.5.38 (AMC 12)
A collection of circles in the upper half-plane, all tangent to the x-axis, is constructed in
layers as follows. Layer L0 consists of two circles of radii 702 and 732 that are externally
S
tangent. For k ≥ 1, the circles in k−1
j=0 Lj are ordered according to their points of
tangency with the x-axis. For every pair of consecutive circles in this order, a new circle
is constructed externally tangent to each of the two circles in the pair. Layer Lk consists
S
of the 2k−1 circles constructed in this way. Let S = 6j=0 Lj , and for every circle C denote
by r(C) its radius. What is
X
1
q
?
r(C)
C∈S
459
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Chapter 35. Circles
Answers
35.1 π7
35.2 17
35.3 312
35.4 026
35.5 100
35.6 150
35.7 26π
35.8 384
35.5.1 12
35.5.2 19
35.5.3
√
2
35.5.4 69
√
35.5.5 10π + 4 3
√
35.5.6 4π 3
35.5.7 140
37
35.5.8 61
35.5.9 27
35.5.10 12
35.5.11 500
460
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Chapter 35. Circles
35.5.12 122
35.5.13 136π
35.5.14 8
35.5.15 173
√
35.5.16 16 2
35.5.17 154
35.5.18
q
425
2
35.5.19 67
35.5.20 17
6
35.5.21 12π
35.5.22 15
35.5.23 65
35.5.24 26π
√
65
2
35.5.25
35.5.26 725
35.5.27 405
35.5.28 35
12
35.5.29 19
35.5.30 3 +
√ 10 π
√
35.5.31 24 2
461
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Chapter 35. Circles
√
35.5.32 76 π − 23
35.5.33 130
35.5.34 43
35.5.35 13
35.5.36 89
35.5.37 2
35.5.38 143
14
462
Chapter 36
Polygons
Video Lecture
463
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Chapter 36. Polygons
464
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Chapter 36. Polygons
Theorem 36.0.1
Sum of interior angle of a polygon = (n − 2) · 180
Interior angle of a regular polygon =
(n − 2)
· 180
n
Exterior angle of a regular polygon =
360
n
Fact 36.0.2. Important Interior Angles
Number of sides in regular polygon Interior Angle of regular polygon
3
60
4
90
5
108
6
120
8
135
9
140
10
144
465
OmegaLearn.org
36.1
Chapter 36. Polygons
Hexagon
Theorem 36.1.1
Sum of interior angle of a regular hexagon = (6 − 2) · 180 = 720
Interior angle of a regular hexagon =
(6 − 2)
· 180 = 120
6
Exterior angle of a regular hexagon =
360
= 60
6
√
3 2
Area of a regular hexagon = 6 ·
s
4
Length of the diagonal of a regular hexagon = 2s
466
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Chapter 36. Polygons
Example 36.1 (AMC 10)
Six regular hexagons surround a regular hexagon of side length 1 as shown. What is the
area of △ABC?
B
A
C
Video Solution
467
OmegaLearn.org
Chapter 36. Polygons
Example 36.2 (AIME)
In convex hexagon ABCDEF , all six sides are congruent, ∠A and ∠D are right angles,
√
and ∠B, ∠C, ∠E, and ∠F are congruent. The area of the hexagonal region is 2116( 2 +
1). Find AB.
E
D
C
F
A
B
Video Solution
468
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Chapter 36. Polygons
Example 36.3 (AIME)
A hexagon that is inscribed in a circle has side lengths 22, 22, 20, 22, 22, and 20 in that
√
order. The radius of the circle can be written as p + q, where p and q are positive
integers. Find p + q.
B
C
A
D
F
E
Video Solution
469
OmegaLearn.org
36.2
Chapter 36. Polygons
Octagon
Theorem 36.2.1
Sum of interior angle of a regular octagon = (8 − 2) · 180 = 1080
Interior angle of a regular octagon =
(8 − 2)
· 180 = 135
6
Exterior angle of a regular octagon =
Area of a regular octagon = 2(1 +
360
= 45
8
√
2)s2
Remark 36.2.2
A regular hexagon can be divided into 6 congruent equilateral triangles.
470
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Chapter 36. Polygons
Example 36.4 (AMC 10)
A regular octagon ABCDEF GH has an area of one square unit. What is the area of
the rectangle ABEF ?
A
B
H
C
G
D
F
E
Video Solution
471
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36.3
Chapter 36. Polygons
Pentagons
Example 36.5 (AMC 10)
In the figure shown below, ABCDE is a regular pentagon and AG = 1. What is
F G + JH + CD?
A
F
E
G
J
B
H
I
D
C
Video Solution
36.4
Practice Problems
Problem 36.4.1 (AMC 10)
A wire is cut into two pieces, one of length a and the other of length b. The piece
of length a is bent to form an equilateral triangle, and the piece of length b is bent to
form a regular hexagon. The triangle and the hexagon have equal area. What is ab ?
Problem 36.4.2 (AMC 10)
Let ABCDEF be an equiangular hexagon. The lines AB, CD, and EF determine
472
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Chapter 36. Polygons
√
a triangle
with
√ with area 192 3, and the lines BC, DE, and F A determine a triangle
√
area 324 3. The perimeter of hexagon ABCDEF can be expressed as m + n p, where
m, n, and p are positive integers and p is not divisible by the square of any prime. What
is m + n + p?
Video Solution
Problem 36.4.3 (AMC 10)
Two circles lie outside regular hexagon ABCDEF . The first is tangent to AB, and the
second is tangent to DE. Both are tangent to lines BC and F A. What is the ratio of
the area of the second circle to that of the first circle?
(A) 18
(B) 27
(C) 36
(D) 81
(E) 108
473
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Chapter 36. Polygons
Example 36.6 (AMC 12)
Regular octagon ABCDEF GH has area n. Let m be the area of quadrilateral ACEG.
?
What is m
n
A
B
H
C
G
D
F
E
Video Solution
Problem 36.4.4 (AMC 10/12)
The figure below is constructed from 11 line segments,
√
√ each of which has length 2.
The area of pentagon ABCDE can be written as m + n, where m and n are positive
integers. What is m + n?
474
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Chapter 36. Polygons
Video Solution
Problem 36.4.5 (AMC 10)
In regular hexagon ABCDEF , points W , X, Y , and Z are chosen on sides BC, CD, EF ,
and F A respectively, so lines AB, ZW , Y X, and ED are parallel and equally spaced.
What is the ratio of the area of hexagon W CXY F Z to the area of hexagon ABCDEF ?
E
D
Z
W
F
C
Y
X
A
B
Video Solution
475
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Chapter 36. Polygons
Additional Problems
Problem 36.4.6 (AMC 10)
A regular octagon is formed by cutting an isosceles right triangle from each of the
corners of a square with sides of length 2000. What is the length of each side of the
octagon?
Problem 36.4.7 (AMC 10)
Spot’s doghouse has a regular hexagonal base that measures one yard on each side.
He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of
the region outside of the doghouse that Spot can reach?
Problem 36.4.8 (AMC 10)
Equilateral △ABC has side length 1, and squares ABDE, BCHI, CAF G lie outside
the triangle. What is the area of hexagon DEF GHI?
476
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Chapter 36. Polygons
F
E
A
G
D
B
C
I
H
Problem 36.4.9 (AMC 12)
Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the
pentagon, and calculated the area of the region between the two circles. Bethany did
the same with a regular heptagon (7 sides). The areas of the two regions were A and B,
respectively. Each polygon had a side length of 2. Which of the following is true?
25
(A) A = 49
B
(B) A = 57 B
(C) A = B
(D) A = 57 B
(E) A = 49
B
25
Problem 36.4.10 (AMC 10/12)
Equiangular hexagon ABCDEF has side lengths AB = CD = EF = 1 and BC =
DE = F A = r. The area of △ACE is 70% of the area of the hexagon. What is the sum
of all possible values of r?
Problem 36.4.11 (AIME)
Let △P QR be a triangle with ∠P = 75◦ and ∠Q = 60◦ . A regular hexagon ABCDEF
with side length 1 is drawn inside △P QR so that side AB lies on P Q, side CD lies on
QR, and one of the remaining vertices lies on RP . There are positive
√ integers a, b, c, and
a+b c
d such that the area of △P QR can be expressed in the form d , where a and d are
relatively prime, and c is not divisible by the square of any prime. Find a + b + c + d.
477
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Chapter 36. Polygons
Problem 36.4.12 (AIME)
Let ABCDEF be an equiangular hexagon such that AB = 6, BC = 8, CD = 10,
and DE = 12. Denote by d the diameter of the largest circle that fits inside the hexagon.
Find d2 .
D
C
E
F
B
A
Answers
√
36.1 3 3
36.2 046
36.3 272
36.4 21
36.5 1 +
√
5
√
36.4.1
6
2
36.4.2 55
36.4.3 81
478
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Chapter 36. Polygons
√
36.6
2
2
36.4.4 23
36.4.5 11
27
√
36.4.6 2000( 2 − 1)
36.4.7 3π
36.4.8 3 +
√
3
36.4.9 A = B
36.4.10 6
36.4.11 021
36.4.12 147
479
Chapter 37
3-D Geometry
Video Lecture
480
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Chapter 37. 3-D Geometry
481
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37.1
Chapter 37. 3-D Geometry
Cube
Theorem 37.1.1 (Volume and Surface Area of a cube)
Volume of a cube = (side length)3 = a3
Surface area of a cube = 6 × (side length)2 = 6a2
Length of space diagonal of a cube =
√
3 × side length =
√
3a
482
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37.2
Chapter 37. 3-D Geometry
Rectangular Prism
Theorem 37.2.1 (Volume and Surface Area of a rectangular prism)
Volume of a rectangular prism = l × b × h = product of all three dimensions
Surface area of a rectangular prism = 2(lb + bh + lh)
Length of space diagonal of a rectangular prism =
√
l2 + b2 + h2
483
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37.3
Chapter 37. 3-D Geometry
Cylinder
Theorem 37.3.1 (Volume and Surface Area of a cylinder)
Volume of a cylinder = πr2 h
Surface area of a cylinder = 2πr2 + 2πrh
= 2πr(r + h)
484
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37.4
Chapter 37. 3-D Geometry
Cone
Theorem 37.4.1 (Volume and Surface Area of a cone)
1
Volume of a cone = πr2 h
3
which basically means
1
Volume of a Cone = π · radius2 · height
3
Surface area of a cone = πr2 + πrs = πr(r + s)
where s is the lateral or slant height
which can also be written as
π · radius2 + π · radius × slant height
Remark 37.4.2
The slant height s can be calculated by the following formula
√
s = r2 + h2
or
slant height =
q
radius2 + height2
485
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37.5
Chapter 37. 3-D Geometry
Sphere
Theorem 37.5.1 (Volume and Surface Area of a sphere)
4
Volume of a sphere = πr3
3
Surface area of a sphere = 4πr2
486
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37.6
Chapter 37. 3-D Geometry
Tetrahedron
Theorem 37.6.1 (Volume of a tetrahedron)
Volume of any tetrahedron =
1
· base area · height
3
Theorem 37.6.2 (Volume of a regular tetrahedron (all sides equal))
√
Volume of a regular tetrahedron =
2 3
s
12
487
OmegaLearn.org
37.7
Chapter 37. 3-D Geometry
Pyramid
Theorem 37.7.1 (Volume of a pyramid)
Volume of any pyramid =
1
· base area · height
3
Theorem 37.7.2 (Volume of a regular pyramid (all sides equal))
When the pyramid has a square base, and all the sides are equal
√
2 3
s
Volume of a regular pyramid =
6
488
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37.8
Chapter 37. 3-D Geometry
Volume of Complex Polyhedrons
Example 37.1 (AMC 10)
A regular hexagon with sides of length 6 has an isosceles triangle attached to each side.
Each of these triangles has two sides of length 8. The isosceles triangles are folded to
make a pyramid with the hexagon as the base of the pyramid. What is the volume of
the pyramid?
Video Solution
37.9
Cross Sections
Example 37.2 (AIME)
Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent
to two others and their centers are the vertices of a regular octagon. A ninth sphere is
placed on the flat surface so that√it is tangent to each of the other eight spheres. The
radius of this last sphere is a + b c, where a, b, and c are positive integers, and c is not
divisible by the square of any prime. Find a + b + c.
Video Solution
Example 37.3 (AIME)
A right prism with height h has bases that are regular hexagons with sides of length 12.
A vertex A of the prism and its three adjacent vertices are the vertices of a triangular
pyramid. The dihedral angle (the angle between the two planes) formed by the face of
the pyramid that lies in a base of the prism and the face of the pyramid that does not
contain A measures 60 degrees. Find h2 .
Video Solution
489
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Chapter 37. 3-D Geometry
Example 37.4 (AMC 10)
A sphere with center O has radius 6. A triangle with sides of length 15, 15, and 24 is
situated in space so that each of its sides are tangent to the sphere. What is the distance
between O and the plane determined by the triangle?
Video Solution
Example 37.5 (AMC 10)
Inside a right circular cone with base radius 5 and height 12 are three congruent spheres
with radius r. Each sphere is tangent to the other two spheres and also tangent to the
base and side of the cone. What is r?
Video Solution
490
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Chapter 37. 3-D Geometry
Example 37.6 (AMC 12)
A 4 × 4 × h rectangular box contains a sphere of radius 2 and eight smaller spheres of
radius 1. The smaller spheres are each tangent to three sides of the box, and the larger
sphere is tangent to each of the smaller spheres. What is h?
Video Solution
491
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37.10
Chapter 37. 3-D Geometry
Practice Problems
Problem 37.10.1 (AMC 10A)
Seven cubes, whose volumes are 1, 8, 27, 64, 125, 216, and 343 cubic units, are stacked
vertically to form a tower in which the volumes of the cubes decrease from bottom to
top. Except for the bottom cube, the bottom face of each cube lies completely on top of
the cube below it. What is the total surface area of the tower (including the bottom) in
square units?
Video Solution
Problem 37.10.2 (AMC 8)
Jerry cuts a wedge from a 6-cm cylinder of bologna as shown by the dashed curve.
Which answer choice is closest to the volume of his wedge in cubic centimeters?
6 cm
8 cm
Video Solution
492
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Chapter 37. 3-D Geometry
Problem 37.10.3 (AMC 8)
In the cube ABCDEF GH with opposite vertices C and E, J and I are the midpoints of segments F B and HD, respectively. Let R be the ratio of the area of the
cross-section EJCI to the area of one of the faces of the cube. What is R2 ?
E
F
H
J
G
A
I
B
D
C
Video Solution
Problem 37.10.4 (AMC 8)
Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long,
10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How
many blocks does the fort contain?
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Video Solution
Problem 37.10.5 (MATHCOUNTS)
The two cones shown have parallel bases and common apex T. TW = 32 m, WV
= 8 m, and ZY = 5 m. What is the volume of the frustum with circle W and circle Z as
its bases? Express your answer in terms of π.
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Video Solution
Problem 37.10.6 (AMC 10/12)
An inverted cone with base radius 12 cm and height 18 cm is full of water. The
water is poured into a tall cylinder whose horizontal base has a radius of 24 cm. What is
the height in centimeters of the water in the cylinder?
Video Solution
Problem 37.10.7 (AMC 10/12)
Two right circular cones with vertices facing down as shown in the figure below contain
the same amount of liquid. The radii of the tops of the liquid surfaces are 3 cm and 6 cm.
Into each cone is dropped a spherical marble of radius 1 cm, which sinks to the bottom
and is completely submerged without spilling any liquid. What is the ratio of the rise of
the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
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Video Solution
Problem 37.10.8 (AMC 10A)
A bowl is formed by attaching four regular hexagons of side 1 to a square of side
1. The edges of adjacent hexagons coincide, as shown in the figure. What is the area of
the octagon obtained by joining the top eight vertices of the four hexagons, situated on
the rim of the bowl?
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(A) 6
(B) 7
Chapter 37. 3-D Geometry
√
(C) 5 + 2 2
(D) 8
(E) 9
Video Solution
Problem 37.10.9 (AMC 10)
What is√the volume of
√ tetrahedron ABCD with edge lengths AB = 2, AC = 3, AD = 4,
BC = 13, BD = 2 5, and CD = 5 ?
Video Solution
Problem 37.10.10 (AMC 12)
Let ABCD be a rectangle and let DM be a segment perpendicular to the plane of
ABCD. Suppose that DM has integer length, and the lengths of M A, M C, and M B
are consecutive odd positive integers (in this order). What is the volume of pyramid
M ABCD?
Video Solution
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Additional Problems
Problem 37.10.11 (MATHCOUNTS)
A right circular cone is sliced into four pieces by planes parallel to its base, as shown in
the figure. All of these pieces have the same height. What is the ratio of the volume of
the second-largest piece to the volume of the largest piece? Express your answer as a
common fraction.
Problem 37.10.12 (AMC 10)
The centers of the faces of the right rectangular prism shown below are joined to
create an octahedron. What is the volume of this octahedron?
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Problem 37.10.13 (AMC 10)
A pyramid with a square base is cut by a plane that is parallel to its base and 2
units from the base. The surface area of the smaller pyramid that is cut from the top
is half the surface area of the original pyramid. What is the altitude of the original
pyramid?
Problem 37.10.14 (AMC 10A)
A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red
stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two
complete revolutions around it. What is the area of the stripe in square feet?
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A
3
80
B
30
Problem 37.10.15 (AMC 10)
A right circular cylinder with its diameter equal to its height is inscribed in a right
circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder
and cone coincide. Find the radius of the cylinder.
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A
12 − 2r
D
2r
E
2r
B
10
C
Problem 37.10.16 (AMC 12)
A regular hexagon with sides of length 6 has an isosceles triangle attached to each
side. Each of these triangles has two sides of length 8. The isosceles triangles are folded
to make a pyramid with the hexagon as the base of the pyramid. What is the volume of
the pyramid?
Problem 37.10.17 (AMC 10)
An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that
has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone.
Assume that the melted ice cream occupies 75% of the volume of the frozen ice cream.
What is the ratio of the cone’s height to its radius? (Note: a cone with radius r and
height h has volume πr2 h/3 and a sphere with radius r has volume 4πr3 /3.)
Problem 37.10.18 (AMC 12)
Triangle ABC, with sides of length 5, 6, and 7, has one vertex on the positive x-
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axis, one on the positive y-axis, and one on the positive z-axis. Let O be the origin.
What is the volume of tetrahedron OABC?
B
5
b
6
a
O
c
A
7
C
Problem 37.10.19 (AMC 10)
A cube with side length 1 is sliced by a plane that passes through two diagonally opposite
vertices A and C and the midpoints B and D of two opposite edges not containing A or
C, as shown. What is the area of quadrilateral ABCD?
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Problem 37.10.20 (AMC 10)
The centers of the faces of the right rectangular prism shown below are joined to
create an octahedron. What is the volume of this octahedron?
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Problem 37.10.21 (AMC 12)
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a
right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter
base at the top, packed solid with ice cream, together with a solid cone of ice cream of
height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the
total volume of the ice cream, in cubic inches?
Problem 37.10.22 (AMC 12)
A pyramid has a square base with side of length 1 and has lateral faces that are
equilateral triangles. A cube is placed within the pyramid so that one face is on the base
of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid.
What is the volume of this cube?
Problem 37.10.23 (AMC 10)
A pyramid has a square base with sides of length 1 and has lateral faces that are
equilateral triangles. A cube is placed within the pyramid so that one face is on the base
of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid.
What is the volume of this cube?
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Chapter 37. 3-D Geometry
Problem 37.10.24 (AMC 10)
Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the
smaller having a central angle of 120 degrees. He makes two circular cones, using each
sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller
cone to that of the larger?
(A) 18
(B) 14
√
(C) 1010
√
(D) 65
√
(E) 55
Problem 37.10.25 (AMC 10/12)
Six spheres of radius 1 are positioned so that their centers are at the vertices of a
regular hexagon of side length 2. The six spheres are internally tangent to a larger sphere
whose center is the center of the hexagon. An eighth sphere is externally tangent to the
six smaller spheres and internally tangent to the larger sphere. What is the radius of
this eighth sphere?
Problem 37.10.26 (AMC 10)
A sphere is inscribed in a truncated right circular cone as shown. The volume of
the truncated cone is twice that of the sphere. What is the ratio of the radius of the
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bottom base of the truncated cone to the radius of the top base of the truncated cone?
1
1
s
r
s
r−1
Problem 37.10.27 (AIME)
A cylindrical log has diameter 12 inches. A wedge is cut from the log by making
two planar cuts that go entirely through the log. The first is perpendicular to the axis of
the cylinder, and the plane of the second cut forms a 45◦ angle with the plane of the first
cut. The intersection of these two planes has exactly one point in common with the log.
The number of cubic inches in the wedge can be expressed as nπ, where n is a positive
integer. Find n.
Problem 37.10.28 (AMC 10)
In the rectangular parallelepiped shown, AB = 3, BC = 1, and CG = 2. Point M is the
midpoint of F G. What is the volume of the rectangular pyramid with base BCHE and
apex M ?
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H
G
M
E
F
2
D
C
1
A
3
B
Problem 37.10.29 (AMC 10)
Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is
the volume of this octahedron?
Problem 37.10.30 (AMC 10)
A sphere is inscribed in a cube that has a surface area of 24 square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters
of the inner cube?
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Problem 37.10.31 (AMC 10)
Logan is constructing a scaled model of his town. The city’s water tower stands 40 meters
high, and the top portion is a sphere that holds 100,000 liters of water. Logan’s miniature
water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?
Problem 37.10.32 (AIME)
When a right triangle is rotated about one leg, the volume of the cone produced is
800π cm3 . When the triangle is rotated about the other leg, the volume of the cone
produced is 1920π cm3 . What is the length (in cm) of the hypotenuse of the triangle?
Problem 37.10.33 (AIME)
The solid shown has a square base of side length s. The upper edge is√parallel to
the base and has length 2s. All other edges have length s. Given that s = 6 2, what is
the volume of the solid?
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Problem 37.10.34 (AIME)
A container in the shape of a right circular cone is 12 inches tall and its base has
a 5-inch radius. The liquid that is sealed inside is 9 inches deep when the cone is held
with its point down and its base horizontal. When the liquid is held with its point up
√
and its base horizontal, the height of the liquid is m − n 3 p, from the base where m, n,
and p are positive integers and p is not divisible by the cube of any prime number. Find
m + n + p.
Problem 37.10.35 (AIME)
A rectangular box has width 12 inches, length 16 inches, and height m
inches, where m
n
and n are relatively prime positive integers. Three faces of the box meet at a corner of
the box. The center points of those three faces are the vertices of a triangle with an area
of 30 square inches. Find m + n.
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Chapter 37. 3-D Geometry
Problem 37.10.36 (AMC 10)
Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of
radius 2 rests on them. What is the distance from the plane to the top of the larger
sphere?
Problem 37.10.37 (AIME)
In a regular tetrahedron the centers of the four faces are the vertices of a smaller
tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is
m/n, where m and n are relatively prime positive integers. Find m + n.
Problem 37.10.38 (AIME)
A cylindrical barrel with radius 4 feet and height 10 feet is full of water. A solid
cube with side length 8 feet is set into the barrel so that the diagonal of the cube is
vertical. The volume of water thus displaced is v cubic feet. Find v 2 .
Problem 37.10.39 (AIME)
Cube ABCDEF GH, labeled as shown below, has edge length 1 and is cut by a plane
passing through vertex D and the midpoints M and N of AB and CG respectively. The
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plane divides the cube into two solids. The volume of the larger of the two solids can be
written in the form pq , where p and q are relatively prime positive integers. Find p + q.
H
E
G
F
N
D
C
A
M
B
Answers
√
37.1 36 21
37.2 152
37.3 108
√
37.4 2 5
√
3
37.5 90−40
11
√
37.6 2 + 2 7
37.10.1 658
37.10.2 151
37.10.3 32
37.10.4 280
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37.10.5 516π
37.10.6 1.5
37.10.7 4 : 1
37.10.8 7
37.10.9 4
√
37.10.10 24 5
37.10.11 <>
37.10.12 10
√
37.10.13 4 + 2 2
37.10.14 240
37.10.15 30
11
√
37.10.16 36 21
37.10.17 3 : 1
37.10.18
√
95
√
37.10.19
6
2
37.10.20 10
37.10.21 44π
3
√
37.10.22 5 2 − 7
√
37.10.23 5 2 − 7
37.10.24
√
10
10
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Chapter 37. 3-D Geometry
37.10.25 32
√
3+ 5
37.10.26
2
37.10.27 216
37.10.28 2
37.10.29 16
37.10.30 8
37.10.31 0.4
37.10.32 026
37.10.33 288
37.10.34 052
37.10.35 041
37.10.36 3 +
√
69
3
37.10.37 028
37.10.38 384
37.10.39 089
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Chapter 38
Area and Length of Complex Shapes
Video Lecture
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Chapter 38. Area and Length of Complex Shapes
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38.1
Chapter 38. Area and Length of Complex Shapes
Area of Complex Shapes
Concept 38.1.1
Tricks to finding the area of complex shapes
• Divide the shape into “nicer” areas which are easier to calculate
• Extend Lines
– You generally want to extend lines when they form nicer shapes/areas to work
with, such as triangles
• Break up areas
– A common way to do so is to drop altitudes as doing so generally allows you
to form right triangles
Remark 38.1.2
A common technique is to find the area of shapes and then find the area of a shape in
terms of a variable (like altitude, inradius, circumradius, etc.) and then solve for that
variable.
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Chapter 38. Area and Length of Complex Shapes
Example 38.1 (AIME)
The two squares shown share the same center O and have sides of length 1. The length
of AB is 43/99 and the area of octagon ABCDEF GH is m/n, where m and n are
relatively prime positive integers. Find m + n.
Video Solution
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Chapter 38. Area and Length of Complex Shapes
Example 38.2 (AMC 10)
Circles centered at A
√ and B each have radius 2, as shown. Point O is the midpoint
of AB, and OA = 2 2. Segments OC and OD are tangent to the circles centered at
A and B, respectively, and EF is a common tangent. What is the area of the shaded
region ECODF ?
E
F
C
A
D
B
O
2
2
Video Solution
Example 38.3 (AIME)
√
In equiangular octagon CAROLIN E, CA = RO = LI = N E = 2 and AR = OL =
IN = EC = 1. The self-intersecting octagon CORN ELIA encloses six non-overlapping
triangular regions. Let K be the area enclosed by CORN ELIA, that is, the total area
a
of the six triangular regions. Then K = , where a and b are relatively prime positive
b
integers. Find a + b.
Video Solution
38.2
Length of complex shapes
Concept 38.2.1
Finding Length of Complex Shapes
• Having equal angles means equal lengths and vice versa
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• Be on the lookout for 90 degree angles, as you can use Pythagorean theorem
• Split the length into multiple components by using some of these techniques
– Drawing extra lines
– Dropping Altitudes
• Extending lines to create similar triangles, special triangles, etc. and then subtracting the extra length
Example 38.4 (AMC 12)
In △ABC, ∠C = 90◦ and AB = 12. Squares ABXY and CBW Z are constructed
outside of the triangle. The points X, Y , Z, and W lie on a circle. What is the perimeter
of the triangle?
X
W
B
Y
M
E
Z
C
A
Video Solution
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Chapter 38. Area and Length of Complex Shapes
Example 38.5 (AMC 12)
A round table has radius 4. Six rectangular place mats are placed on the table. Each
place mat has width 1 and length x as shown. They are positioned so that each mat has
two corners on the edge of the table, these two corners being end points of the same side
of length x. Further, the mats are positioned so that the inner corners each touch an
inner corner of an adjacent mat. What is x?
1
x
Video Solution
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38.3
Chapter 38. Area and Length of Complex Shapes
Practice Problems
Problem 38.3.1 (AMC 12)
Convex quadrilateral ABCD has AB = 3, BC = 4, CD = 13, AD = 12, and
∠ABC = 90◦ , as shown. What is the area of the quadrilateral?
A
B
C
D
Video Solution
Problem 38.3.2 (AMC 12)
Regular octagon ABCDEF GH has area n. Let m be the area of quadrilateral ACEG.
What is m
?
n
A
B
H
C
G
D
F
E
Video Solution
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Chapter 38. Area and Length of Complex Shapes
Problem 38.3.3 (AMC 10)
As shown in the figure below, six semicircles lie in the interior of a regular hexagon
with side length 2 so that the diameters of the semicircles coincide with the sides of the
hexagon. What is the area of the shaded region —- inside the hexagon but outside all of
the semicircles?
2
Video Solution
38.4
Additional Problems
Problem 38.4.1 (AMC 12)
A circle of radius 2 is centered at A. An equilateral triangle with side 4 has a vertex at A. What is the difference between the area of the region that lies inside the circle
but outside the triangle and the area of the region that lies inside the triangle but outside
the circle?
Problem 38.4.2 (AIME)
A rectangle has sides of length a and 36. A hinge is installed at each vertex of the
rectangle, and at the midpoint of each side of length 36. The sides of length a can be
pressed toward each other keeping those two sides parallel so the rectangle becomes
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Chapter 38. Area and Length of Complex Shapes
a convex hexagon as shown. When the figure is a hexagon with the sides of length a
parallel and separated by a distance of 24, the hexagon has the same area as the original
rectangle. Find a2 .
Problem 38.4.3 (AIME)
ABCD is a rectangular sheet of paper that has been folded so that corner B is matched
with point B ′ on edge AD. The crease is EF, where E is on AB and F is on CD. The
dimensions AE = 8, BE = 17, and CF = 3 are given. The perimeter of rectangle ABCD
is m/n, where m and n are relatively prime positive integers. Find m + n.
C′
F
D
C
B′
A
E
B
Problem 38.4.4 (AMC 10)
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B
and C respectively. What fraction of the area of △ABC lies outside the circle?
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.5 (AMC 12)
On each side of a unit square, an equilateral triangle of side length 1 is constructed. On
each new side of each equilateral triangle, another equilateral triangle of side length 1 is
constructed. The interiors of the square and the 12 triangles have no points in common.
Let R be the region formed by the union of the square and all the triangles, and S be the
smallest convex polygon that contains R. What is the area of the region that is inside S
but outside R?
Problem 38.4.6 (AMC 12)
Vertex E of equilateral △ABE is in the interior of unit square ABCD. Let R be
the region consisting of all points inside ABCD and outside △ABE whose distance from
AD is between 13 and 23 . What is the area of R?
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A
B
E
D
C
Problem 38.4.7 (AMC 12)
Triangle ABC has AC = 3, BC = 4, and AB = 5. Point D is on AB, and CD
bisects the right angle. The inscribed circles of △ADC and △BCD have radii ra and rb ,
respectively. What is ra /rb ?
A
15
7
D
OA
3
20
7
OB
C
sin θ = 35
45◦
4
B
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.8 (AMC 10)
A circle with center O√ has area 156π. Triangle ABC is equilateral, BC is a chord
on the circle, OA = 4 3, and point O is outside △ABC. What is the side length of
△ABC?
B
s
2
X
C
√
s 3
2
s
√
156
A
√
4 3
O
Problem 38.4.9 (AMC 10)
The figure shown is called a trefoil and is constructed by drawing circular sectors
about the sides of the congruent equilateral triangles. What is the area of a trefoil whose
horizontal base has length 2?
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2
Problem 38.4.10 (AMC 10)
Two circles lie outside regular hexagon ABCDEF . The first is tangent to AB, and the
second is tangent to DE. Both are tangent to lines BC and F A. What is the ratio of
the area of the second circle to that of the first circle?
Problem 38.4.11 (AMC 10)
√
3
A square of side length 1 and a circle of radius
share the same center. What
3
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Chapter 38. Area and Length of Complex Shapes
is the area inside the circle, but outside the square?
A
X
B
a
1
2
√
3
3
O
Problem 38.4.12 (AMC 10)
An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point of the
triangle?
3
6
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.13 (AMC 10)
The closed curve in the figure is made up of 9 congruent circular arcs each of length 2π
,
3
where each of the centers of the corresponding circles is among the vertices of a regular
hexagon of side 2. What is the area enclosed by the curve?
◦
◦
◦
◦
◦
◦
Problem 38.4.14 (AMC 10)
A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with
the center at each of the vertices, creating circular sectors as shown. The region inside
the hexagon but outside the sectors is shaded as shown What is the area of the shaded
region?
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.15 (AMC 10/12)
A semicircle of diameter 1 sits at the top of a semicircle of diameter 2, as shown.
The shaded area inside the smaller semicircle and outside the larger semicircle is called a
lune. Determine the area of this lune.
1
2
Problem 38.4.16 (AMC 10/12)
Square ABCD has side length 2. A semicircle with diameter AB is constructed inside the square, and the tangent to the semicircle from C intersects side AD at E. What
is the length of CE?
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Chapter 38. Area and Length of Complex Shapes
D
C
E
A
B
Problem 38.4.17 (AMC 10)
A circle of radius 1 is internally tangent to two circles of radius 2 at points A and
B, where AB is a diameter of the smaller circle. What is the area of the region, shaded
in the picture, that is outside the smaller circle and inside each of the two larger circles?
2
A
1
B
2
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.18 (AMC 10)
A circle of radius 2 is centered at O. Square OABC has side length 1. Sides AB
and CB are extended past B to meet the circle at D and E, respectively. What is the
area of the shaded region in the figure, which is bounded by BD, BE, and the minor arc
connecting D and E?
D
C
O
B
E
A
Problem 38.4.19 (AMC 10)
A paint brush is swept along both diagonals of a square to produce the symmetric
painted area, as shown. Half the area of the square is painted. What is the ratio of the
side length of the square to the brush width?
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.20 (AMC 10)
Circles centered at
√ A and B each have radius 2, as shown. Point O is the midpoint of
AB, and OA = 2 2. Segments OC and OD are tangent to the circles centered at A and
B, respectively, and EF is a common tangent. What is the area of the shaded region
ECODF ?
E
F
C
A
D
B
O
2
2
Problem 38.4.21 (AMC 10)
A unit square is rotated 45◦ about its center. What is the area of the region swept out
by the interior of the square?
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.22 (AMC 10)
Sides AB and AC of equilateral triangle ABC are tangent to a circle at points B
and C respectively. What fraction of the area of △ABC lies outside the circle?
A
B
C
O
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Chapter 38. Area and Length of Complex Shapes
Problem 38.4.23 (AMC 12)
Circles ω1 , ω2 , and ω3 each have radius 4 and are placed in the plane so that each
circle is externally tangent to the other two. Points P1 , P2 , and P3 lie on ω1 , ω2 , and
ω3 respectively such that P1 P2 = P2 P3 = P3 P1 and line Pi Pi+1 is tangent to ωi for each
i = 1, 2, 3,√where√P4 = P1 . See the figure below. The area of △P1 P2 P3 can be written in
the form a + b for positive integers a and b. What is a + b?
ω1
P1
P3
ω3
ω2
P2
Problem 38.4.24 (AIME)
Squares S1 and S2 are inscribed in right triangle ABC, as shown in the figures below. Find AC + CB if area (S1 ) = 441 and area (S2 ) = 440.
A
E
D
S1
C
F
B
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A
X
Y
S2
W
C
Z
B
Answers
38.1 185
√
38.2 8 2 − 4 − π
38.3 023
√
38.4 12 + 12 2
√
√
3
38.5 3 7−
2
38.3.1 36
√
38.3.2
2
2
√
38.3.3 3 3 − π
38.4.1 4(π −
√
3)
38.4.2 720
38.4.3 293
√
38.4.4 34 − 4 273π
38.4.5 1
√
3
38.4.6 12−5
36
3
38.4.7 28
10 −
√ 2
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Chapter 38. Area and Length of Complex Shapes
38.4.8 6
38.4.9 23 π
38.4.10 81
√
2π
3
38.4.11
−
9
3
38.4.12 54 + 9π
√
38.4.13 π + 6 3
√
38.4.14 54 3 − 18π
√
38.4.15
3
1
− 24
π
4
38.4.16 25
√
38.4.17 53 π − 2 3
38.4.18 π3 + 1 −
√
3
√
38.4.19 2 2 + 2
√
38.4.20 8 2 − 4 − π
38.4.21 2 −
√
2 + π4
√
38.4.22 43 − 4 273π
38.4.23 552
38.4.24 462
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Chapter 39
Coordinate Geometry
Video Lecture
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39.1
Chapter 39. Coordinate Geometry
Line in Coordinate Plane
Definition 39.1.1 (Equation of a Line). The equation of a line is ax + by + c = 0.
Definition 39.1.2 (Slope-Intercept Form). An equation of a line in slope-intercept form is
y = mx + b
Theorem 39.1.3 (Slope of a line through 2 points)
The slope of a line passing through 2 points (x1 , y1 ) and (x2 , y2 ) is
y2 − y1
x2 − x1
Theorem 39.1.4 (Slope of a line through angle)
The slope of a line with an angle of θ above the x-axis is tan(θ)
Theorem 39.1.5 (Distance between 2 points)
The distance between 2 points (x1 , y1 ) and (x2 , y2 ) is
q
(x1 − x2 )2 + (y1 − y2 )2
Theorem 39.1.6 (Point to line formula)
The distance between a point (x0 , y0 ) and a line ax + by + c = 0 is
|a · x0 + b · y0 + c|
√
a2 + b 2
.
Remark 39.1.7
Be careful not to get the equation of the line confused with ax + by = c
Remark 39.1.8
Note that this distance represents the shortest possible distance which would be length
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Chapter 39. Coordinate Geometry
of the perpendicular line.
Remark 39.1.9
This formula is a bit confusing so an easy way to remember the numerator is that it’s
just the equation of the line with the values of the point plugged in as the x and y values
39.1.1
Circle in Coordinate Plane
Theorem 39.1.10 (Equation of a Circle)
A circle with center (a, b) and radius r has equation
(x − a)2 + (y − b)2 = r2
Example 39.1 (AMC 10)
Let points A = (0, 0), B = (1, 2), C = (3, 3), and D = (4, 0). Quadrilateral ABCD is
cut
into
equal area pieces by a line passing through A. This line intersects CD at point
p r
, , where these fractions are in lowest terms. What is p + q + r + s?
q s
C
B
E
15
8
A
27
8
4
D
Video Solution
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Chapter 39. Coordinate Geometry
Example 39.2 (AMC 10)
A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle.
What is the distance between the centers of those circles?
Video Solution
Example 39.3 (AMC 10)
In rectangle ABCD, AB = 6 and BC = 3. Point E between B and C, and point F
between E and C are such that BE = EF = F C. Segments AE and AF intersect BD
at P and Q, respectively. The ratio BP : P Q : QD can be written as r : s : t where the
greatest common factor of r, s, and t is 1. What is r + s + t?
A
B
P
E
Q
F
D
C
Video Solution
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Chapter 39. Coordinate Geometry
Example 39.4 (AMC 10)
In the right triangle △ACE, we have AC = 12, CE = 16, and EA = 20. Points B, D,
and F are located on AC, CE, and EA, respectively, so that AB = 3, CD = 4, and
EF = 5. What is the ratio of the area of △DBF to that of △ACE?
A
3
B
15
9
F
5
C
4
D
12
E
Video Solution
Example 39.5 (PUMAC)
Let ABCD be a square with side length 8. Let M be the midpoint of BC and let ω be
the circle passing through M , A, and D. Let O be the center of ω, X be the intersection
point (besides A) of ω with AB, and Y be the intersection point of OX and AM . Find
the length OY .
Video Solution
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39.2
Chapter 39. Coordinate Geometry
Practice Problems
Problem 39.2.1 (AMC 10)
The interior of a quadrilateral is bounded by the graphs of (x + ay)2 = 4a2 and
(ax − y)2 = a2 , where a a positive real number. What is the area of this region in
terms of a, valid for all a > 0?
Video Solution
Problem 39.2.2 (AMC 12)
A laser is placed at the point (3,5). The laser beam travels in a straight line. Larry wants
the beam to hit and bounce off the y-axis, then hit and bounce off the x-axis, then hit
the point (7, 5). What is the total distance the beam will travel along this path?
Video Solution
Problem 39.2.3 (AMC 10/12)
The point P (a, b) in the xy-plane is first rotated counterclockwise by 90° around the
point (1, 5) and then reflected about the line y = −x. The image of P after these two
transformations is at (−6, 3). What is b − a?
Video Solution
Problem 39.2.4 (AMC 10)
The area of the region bounded by the graph of
x2 + y 2 = 3|x − y| + 3|x + y|
is m + nπ, where m and n are integers. What is m + n?
Video Solution
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Chapter 39. Coordinate Geometry
Additional Problems
Problem 39.2.5 (AMC 10)
The line 12x + 5y = 60 forms a triangle with the coordinate axes. What is the sum of
the lengths of the altitudes of this triangle?
Problem 39.2.6 (AMC 12)
Triangle ABC has vertices A = (3, 0), B = (0, 3), and C, where C is on the line
x + y = 7. What is the area of △ABC?
Problem 39.2.7 (AMC 10)
The lines with equations ax − 2y = c and 2x + by = −c are perpendicular and intersect at (1, −5). What is c?
Problem 39.2.8 (AMC 12)
The point P = (1, 2, 3) is reflected in the xy-plane, then its image Q is rotated by
180◦ about the x-axis to produce R, and finally, R is translated by 5 units in the
positive-y direction to produce S. What are the coordinates of S?
Problem 39.2.9 (AMC 10)
Line segment AB is a diameter of a circle with AB = 24. Point C, not equal to
A or B, lies on the circle. As point C moves around the circle, the centroid (center of
mass) of △ABC traces out a closed curve missing two points. To the nearest positive
integer, what is the area of the region bounded by this curve?
Problem 39.2.10 (AMC 12)
Right triangle ABC has side lengths BC = 6, AC = 8, and AB = 10.
A circle centered at O is tangent to line BC at B and passes through A. A circle
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Chapter 39. Coordinate Geometry
centered at P is tangent to line AC at A and passes through B. What is OP ?
Problem 39.2.11 (AMC 12)
1
Two lines with slopes
and 2 intersect at (2, 2). What is the area of the triangle
2
enclosed by these two lines and the line x + y = 10?
Problem 39.2.12 (AMC 12)
A triangle has vertices (0, 0), (1, 1), and (6m, 0), and the line y = mx divides the
triangle into two triangles of equal area. What is the sum of all possible values of m?
Problem 39.2.13 (AMC 12)
Let A, B and C be three distinct points on the graph of y = x2 such that line AB is
parallel to the x-axis and △ABC is a right triangle with area 2008. What is the sum of
the digits of the y-coordinate of C?
Problem 39.2.14 (AMC 10)
In rectangle ABCD, we have A = (6, −22), B = (2006, 178), D = (8, y), for some
integer y. What is the area of rectangle ABCD?
Problem 39.2.15 (AIME)
Find the area of the region enclosed by the graph of |x − 60| + |y| = x4 .
Problem 39.2.16 (AIME)
Find the number of ordered pairs (x, y) of positive integers that satisfy x ≤ 2y ≤ 60 and
y ≤ 2x ≤ 60.
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Chapter 39. Coordinate Geometry
Problem 39.2.17 (AIME)
Consider the parallelogram with vertices (10, 45), (10, 114), (28, 153), and (28, 84). A line
through the origin cuts this figure into two congruent polygons. The slope of the line is
m/n, where m and n are relatively prime positive integers. Find m + n.
Problem 39.2.18 (AIME)
5
Let L be the line with slope 12
that contains the point A = (24, −1), and let M
be the line perpendicular to line L that contains the point B = (5, 6). The original
coordinate axes are erased, and line L is made the x-axis and line M the y-axis. In the
new coordinate system, point A is on the positive x-axis, and point B is on the positive
y-axis. The point P with coordinates (−14, 27) in the original system has coordinates
(α, β) in the new coordinate system. Find α + β.
Problem 39.2.19 (AIME)
Triangle ABC lies in the cartesian plane and has an area of 70. The coordinates
of B and C are (12, 19) and (23, 20), respectively, and the coordinates of A are (p, q).
The line containing the median to side BC has slope −5. Find the largest possible value
of p + q.
A (p,q)
(17,22)
B (12,19)
M
C (23,20)
Problem 39.2.20 (AMC 10)
In rectangle ABCD, we have AB = 8, BC = 9, H is on BC with BH = 6, E is
on AD with DE = 4, line EC intersects line AH at G, and F is on line AD with
GF ⊥ AF . Find the length of GF .
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Chapter 39. Coordinate Geometry
G
H
C
F
D
4
E
6
B
A
Answers
39.1 58
√
39.2
65
2
39.3 20
7
39.4 16
39.5 25
11
2
39.2.1 a8a
2 +1
√
39.2.2 10 2
39.2.3 7
39.2.4 54
39.2.5
281
13
39.2.6 6
39.2.7 13
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Chapter 39. Coordinate Geometry
39.2.8 (1, 3, 3)
39.2.9 50
39.2.10 35
12
39.2.11 6
39.2.12 − 16
39.2.13 18
39.2.14 40, 400
39.2.15 480
39.2.16 480
39.2.17 118
39.2.18 031
39.2.19 047
39.2.20 20
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Advanced Topics
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Chapter 40
Floor and Ceiling Functions
Video Lecture
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Chapter 40. Floor and Ceiling Functions
Definition 40.0.1 (Floor, Ceiling, and Fractional Part Functions).
⌊x⌋ = Greatest integer less than or equal to x
⌈x⌉ = Smallest integer greater than or equal to x
{x} = Fractional part of x (the value after the decimal point)
Concept 40.0.2 (Common Floor and Ceiling Problems Techniques)
Most floor and ceiling problems can be solved using these techniques.
1. Make the substitution x = ⌊x⌋ + {x}
2. Use the floor or ceiling function to find an inequality
For example, if you know that y = ⌊x⌋, then y ≤ x < y + 1
3. Graph your equations and look for intersection points (we recommend using graph
paper)
Example 40.1 (AIME)
How many of the first 1000 positive integers can be expressed in the form
⌊2x⌋ + ⌊4x⌋ + ⌊6x⌋ + ⌊8x⌋,
where x is a real number, and ⌊z⌋ denotes the greatest integer less than or equal to z?
Video Solution
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Chapter 40. Floor and Ceiling Functions
Example 40.2 (AMC 10)
Let ⌊x⌋ denote the greatest integer less than or equal to x. How many real numbers x
satisfy the equation x2 + 10, 000⌊x⌋ = 10, 000x?
Video Solution
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Chapter 40. Floor and Ceiling Functions
Example 40.3 (AMC 10)
How many positive integers n satisfy
√
n + 1000
= ⌊ n⌋?
70
(Recall that ⌊x⌋ is the greatest integer not exceeding x.)
Video Solution
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40.1
Chapter 40. Floor and Ceiling Functions
Practice Problems
Additional Problems
Problem 40.1.1 (AMC 10)
The graph of
f (x) = |⌊x⌋| − |⌊1 − x⌋|
is symmetric about which of the following? (Here ⌊x⌋ is the greatest integer not exceeding
x.)
(A) the y-axis (B)
the line x = 1
(C) the origin
1
,0
(E) the point (1, 0)
(D) the point
2
Problem 40.1.2 (AMC 10)
For how many positive integers n ≤ 1000 is
998
999
1000
+
+
n
n
n
not divisible by 3? (Recall that ⌊x⌋ is the greatest integer less than or equal to x.)
Problem 40.1.3 (AMC 10)
Cozy the Cat and Dash the Dog are going up a staircase with a certain number of
steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump.
Cozy goes two steps up with each jump (though if necessary, he will just jump the last
step). Dash goes five steps up with each jump (though if necessary, he will just jump the
last steps if there are fewer than 5 steps left). Suppose Dash takes 19 fewer jumps than
Cozy to reach the top of the staircase. Let s denote the sum of all possible numbers of
steps this staircase can have. What is the sum of the digits of s?
Problem 40.1.4 (AIME)
Suppose r is a real number for which
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j
k
Chapter 40. Floor and Ceiling Functions
j
k
j
k
j
k
19
20
21
91
r + 100
+ r + 100
+ r + 100
+ · · · + r + 100
= 546. Find ⌊100r⌋. (For real x,
⌊x⌋ is the greatest integer less than or equal to x.)
Problem 40.1.5 (AIME)
How many positive integers N less than 1000 are there such that the equation x⌊x⌋ = N
has a solution for x?
Problem 40.1.6 (AIME)
Given a nonnegative real number x, let ⟨x⟩ denote the fractional part of x; that is,
⟨x⟩ = x − ⌊x⌋, where ⌊x⌋ denotes the greatest integer less than or equal to x. Suppose
that a is positive, ⟨a−1 ⟩ = ⟨a2 ⟩, and 2 < a2 < 3. Find the value of a12 − 144a−1 .
Problem 40.1.7 (AIME)
A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let D be the
difference between the mode and the arithmetic mean of the sample. What is the largest
possible value of ⌊D⌋? (For real x, ⌊x⌋ is the greatest integer less than or equal to x.)
Problem 40.1.8 (AIME)
j
k
Find the least positive integer k for which the equation 2002
= k has no integer
n
solutions for n. (The notation ⌊x⌋ means the greatest integer less than or equal to x.)
Problem 40.1.9 (AIME)
Find the number of positive integers n less than 1000 for which there exists a positive real number x such that n = x⌊x⌋.
Note: ⌊x⌋ is the greatest integer less than or equal to x.
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Chapter 40. Floor and Ceiling Functions
Problem 40.1.10 (AIME)
Given a real number x, let ⌊x⌋ denote the greatest integer less than or equal to x.
For a certain integer k, there are exactly 70 positive integers n1 , n2 , . . . , n70 such that
√
√
√
k = ⌊ 3 n1 ⌋ = ⌊ 3 n2 ⌋ = · · · = ⌊ 3 n70 ⌋ and k divides ni for all i such that 1 ≤ i ≤ 70.
Find the maximum value of nki for 1 ≤ i ≤ 70.
Problem 40.1.11 (AMC 12)
How many positive integers n satisfy
√
n + 1000
= ⌊ n⌋?
70
(Recall that ⌊x⌋ is the greatest integer not exceeding x.)
Problem 40.1.12 (AIME)
For a real number a, let ⌊a⌋ denote the greatest integer less than or equal to a. Let R denote the region in the coordinate plane consisting of points (x, y) such that ⌊x⌋2 +⌊y⌋2 = 25.
The region R is completely contained in a disk of radius r (a disk
is the union of a circle
√
m
and its interior). The minimum value of r can be written as n , where m and n are
integers and m is not divisible by the square of any prime. Find m + n.
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Chapter 40. Floor and Ceiling Functions
Problem 40.1.13 (AIME)
Find the number ofj positive
k j k integers
j k n ≤ 600 whose value can be uniquely determined
when the values of n4 , n5 , and n6 are given, where ⌊x⌋ denotes the greatest integer
less than or equal to the real number x.
Answers
40.1 600
40.2 199
40.3 6
40.1.1 D
40.1.2 22
40.1.3 13
40.1.4 743
40.1.5 412
40.1.6 233
40.1.7 947
40.1.8 049
40.1.9 496
40.1.10 553
40.1.11 6
40.1.12 132
40.1.13 80or81
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Chapter 41
Inequalities
Video Lecture
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41.1
Chapter 41. Inequalities
Trivial Inequality
Theorem 41.1.1 (Trivial Inequality)
For real x, x2 ≥ 0
This means all perfect squares are 0 or more.
Corollary 41.1.2 (Completing the Square)
In a quadratic Q(x) = ax2 + bx + c,
2
b
b
and occurs when x = − 2a
• If a > 0, then the minimum value of Q(x) is c − 4a
2
b
b
• If a < 0, then the maximum value of Q(x) is c − 4a
and occurs when x = − 2a
Remark 41.1.3
Simple, yet powerful. This is the core of all inequalities and how more advanced
inequalities are derived.
The rest of the inequalities are optional for the AMC 10 but are still good to know.
41.2
AM-GM
Theorem 41.2.1 (AM-GM Inequality For 2 variables)
For non-negative reals a and b,
a+b √
≥ ab
2
Basically, this means the average of 2 non-negative numbers (arithmetic mean) is always
at least as big as the square root of the product of the 2 numbers (the geometric mean).
Note that equality in this expression occurs when a = b.
Corollary 41.2.2
The minimum value of x + x1 is 2 and occurs when x = 1
• The minimum value of a + b (if ab remains constant) occurs when a = b
• The maximum value of ab (if a + b remains constant) occurs when a = b
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Chapter 41. Inequalities
Example 41.1 (AMC 12)
Let A, M, and C be non-negative integers such that A + M + C = 12. What is the
maximum value of A · M · C + A · M + M · C + A · C?
Video Solution
Theorem 41.2.3 (AM-GM Inequality For More Variables)
For non-negative reals a1 , a2 , . . . an ,
√
a1 + a2 + · · · + an
≥ n a1 · a2 · a3 · · · · an
n
Note that equality occurs when a1 = a2 · · · = an . (essentially all the variables are equal).
Another way to say this is
q
Average of n numbers = n product of all n numbers
Remark 41.2.4
This means in general,
√
min(a1 + a2 + a3 + · · · + an ) = n · n a1 · a2 · a3 · · · · an
max(a1 · a2 · a3 · · · · an ) =
a1 + a2 + · · · + an
n
n
Essentially,
q
min(sum of all numbers) = n · n product of all numbers)
max(product of all numbers) = (average of all numbers)n
Remark 41.2.5
Generally, we use AM-GM to maximize products or minimize sums.
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Chapter 41. Inequalities
Theorem 41.2.6 (Weighted AM-GM Inequality)
For non-negative reals, ai , ci ,
q
c 1 · a1 + c 2 · a2 + · · · + c n · an
≥ c1 +c2 +···+cn ac11 · ac22 · ac33 · · · · · acnn
c1 + c2 + · · · + cn
Remark 41.2.7
Weighted AM-GM is very similar to AM-GM. One way to visualize weighted AM-GM
is that there are ck number of terms which are all equal to ak . So instead of writing
ak + ak + · · · + ak ck times in our sum we simply write ak · ck , and instead of writing
ak · ak · . . . ck times in our product we simply write ackk .
Remark 41.2.8
We use weighted AM-GM when we are trying to make the sum of all terms a constant
by multiplying weights to all (or some) the terms. Remember to divide by the weights
you multiplied at the end.
Example 41.2
Find the maximum value of x3 (4 − x).
Video Solution
Example 41.3 (AMC 12)
t
What is the maximum value of (2 −3t)t
for real values of t?
4t
Video Solution
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41.3
Chapter 41. Inequalities
Cauchy Schwarz
Theorem 41.3.1 (Cauchy Schwarz)
For reals ai and bi ,
(a1 · b1 + a2 · b2 + · · · + an · bn )2 ≤ (a21 + a22 + . . . a2n )(b21 + b22 + . . . b2n )
This means
(the sum of the products of all ak and bk )2 ≤
product of the sum of squares of all ak and bk
Equality holds when the ratio of
ai
bi
for all i is the same.
Remark 41.3.2
If you ever forget which side the ≥ sign faces, just try a small example like a1 = 1,
a2 = 2, b1 = 3, and b2 = 4.
Remark 41.3.3
You generally want to apply Cauchy Schwarz when you are dealing with sums of squares.
Corollary 41.3.4 (Titu’s Lemma)
For reals ai and bi ,
a2
(a1 + a2 + · · · + an )2
a21 a22
+
+ ··· + n ≥
b1
b2
bn
b1 + b2 + b3 + · · · + b n
Alternately,
√
√
√
( a1 + a2 + · · · + an )2
a1 a2
an
+
+ ··· +
≥
b1
b2
bn
b1 + b2 + b3 + · · · + b n
Note that this is a direct consequence of Cauchy Schwartz.
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Chapter 41. Inequalities
Example 41.4 (AMC 12)
Let a, b, and c be real numbers such that
a + b + c = 2, and
a2 + b2 + c2 = 12
What is the difference between the maximum and minimum possible values of c?
Video Solution
Concept 41.3.5 (Techniques For Optimization Problems)
Steps to find maximum/minimum of expressions
1. Try to find another simple expression for maximization is greater than or equal to
the expression you are given OR minimization is less than the expression you are
given by using 1 (or possibly even more) of the inequalities
(a) Trivial Inequality
(b) AM-GM
(c) Weighted AM-GM (AM-GM weighted and unweighted are useful for maximizing products and minimizing sums)
(d) Cauchy Schwartz (Cauchy Schwartz is useful when dealing with sums of
squares)
2. Verify that the equality case of your inequality holds true with your problem
conditions
3. Simplify your equality case and solve for the answer
41.4
Practice Problems
Problem 41.4.1 (AMC 10)
Define
P (x) = (x − 12 )(x − 22 ) · · · (x − 1002 ).
How many integers n are there such that P (n) ≤ 0?
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Video Solution
Problem 41.4.2 (AMC 10)
What is the least possible value of
(x + 1)(x + 2)(x + 3)(x + 4) + 2019
where x is a real number?
Video Solution
Video Solution (Meta-solving)
Problem 41.4.3 (AMC 10/12)
What is the least possible value of (xy − 1)2 + (x + y)2 for real numbers x and y?
Video Solution
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Chapter 41. Inequalities
Problem 41.4.4 (AMC 10)
Consider functions f that satisfy |f (x) − f (y)| ≤ 21 |x − y| for all real numbers x and y.
Of all such functions that also satisfy the equation f (300) = f (900), what is the greatest
possible value of
f (f (800)) − f (f (400))?
(A) 25
(B) 50
(C) 100
(D) 150
(E) 200
Video Solution
41.5
Additional Problems
Problem 41.5.1 (AMC 10/12)
What is the least possible value of (xy − 1)2 + (x + y)2 for real numbers x and y?
Problem 41.5.2 (AIME)
Suppose that |xi | < 1 for i = 1, 2, . . . , n. Suppose further that |x1 | + |x2 | + · · · + |xn | =
19 + |x1 + x2 + · · · + xn |. What is the smallest possible value of n?
Problem 41.5.3 (AIME)
Let x1 ≤ x2 ≤ · · · ≤ x100 be real numbers such that |x1 | + |x2 | + · · · + |x100 | = 1
and x1 + x2 + · · · + x100 = 0. Among all such 100-tuples of numbers, the greatest value
that x76 − x16 can achieve is m
, where m and n are relatively prime positive integers.
n
Find m + n.
Problem 41.5.4 (AMC 12)
There is a smallest positive real number a such that there exists a positive real number
b such that all the roots of the polynomial x3 − ax2 + bx − a are real. In fact, for this
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Chapter 41. Inequalities
value of a the value of b is unique. What is this value of b?
Problem 41.5.5
Let x1 , x2 , ..., x6 be non-negative real numbers such that x1 + x2 + x3 + x4 + x5 + x6 = 1,
1
and x1 x3 x5 + x2 x4 x6 ≥ 540
. Let p and q be positive relatively prime integers such that pq
is the maximum possible value of x1 x2 x3 + x2 x3 x4 + x3 x4 x5 + x4 x5 x6 + x5 x6 x1 + x6 x1 x2 .
Find p + q.
Problem 41.5.6 (AIME)
Alpha and Beta both took part in a two-day problem-solving competition. At the
end of the second day, each had attempted questions worth a total of 500 points. Alpha
scored 160 points out of 300 points attempted on the first day, and scored 140 points out
of 200 points attempted on the second day. Beta who did not attempt 300 points on the
first day, had a positive integer score on each of the two days, and Beta’s daily success
rate (points scored divided by points attempted) on each day was less than Alpha’s on
that day. Alpha’s two-day success ratio was 300/500 = 3/5. The largest possible two-day
success ratio that Beta could achieve is m/n, where m and n are relatively prime positive
integers. What is m + n?
Answers
41.1 112
41.2 ¡¿
1
41.3 12
41.4 16
3
41.4.1 5100
41.4.2 2018
41.4.3 1
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Chapter 41. Inequalities
41.4.4 50
41.5.1 1
41.5.2 20
41.5.3 841
19
41.5.5 540
41.5.6 849
567
Chapter 42
Logarithms
Note: This topic is mainly relevant for AMC 12.
Video Lecture
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42.1
Chapter 42. Logarithms
Basic Definitions
Definition 42.1.1 (Logarithm). A logarithm is the power to which a number must be raised
in order to get some other number.
Logarithms are expressed as
a = logb n
where b is the base and n is the number.
Basically, we are trying to calculate how many times we need to multiply the base to get the
number a, or what power do we need to raise the base to get the number a.
Theorem 42.1.2 (Converting to Logarithm and Exponents)
logx y = a =⇒ xa = y
.
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42.2
Chapter 42. Logarithms
Logarithmic Formulas
Theorem 42.2.1 (Important Formulas)
loga ar = r
loga bc = loga b + loga c
loga
b
= loga b − loga c
c
loga bc = c loga b
loga b · logb c =
logc
logb logc
·
=
= loga c
loga logb
loga
logb a =
1
loga b
logb a =
logd a
logd b
Remark 42.2.2
This last formula is known as the ”Base Change Formula” and is the most useful of them
all. Often times in logarithm problems you can just expand out your expression in terms
of this formula and simplify the expression to get the answer.
Remark 42.2.3
These formulas are extremely important for working with logarithms and should definitely
be memorized.
Remark 42.2.4
If you ever forget which way the sign of these logarithms are, you can just try a small
example like log10 100 + log10 1000 = log10 100, 000 so from here for example you could
figure out the sum of logarithms identity.
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Chapter 42. Logarithms
Remark 42.2.5
Natural Logarithm (Base e) where e is the Euler’s Number = 2.71828
A logarithm with a base of e is a called a natural log and is written as ln
Theorem 42.2.6 (Advanced Formulas)
logam an =
loga
n
m
1
= − loga b
b
log 1 b = − loga b
a
Remark 42.2.7
These formulas are less important and aren’t necessary for most logarithm problems,
but still good to know.
Concept 42.2.8
When you take a logarithm of numbers which form a geometric progression, the logarithms
of those numbers form an arithmetic progression.
Example 42.1 (AMC 12)
The solutions to the equation log3x 4 = log2x 8, where x is a positive real number other
than 13 or 12 , can be written as pq where p and q are relatively prime positive integers.
What is p + q?
Video Solution
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Chapter 42. Logarithms
Example 42.2 (AIME)
In a Martian civilization, all logarithms whose bases are not specified are assumed to be
base b, for some fixed b ≥ 2. A Martian student writes down
√
3 log( x log x) = 56
loglog x (x) = 54
and finds that this system of equations has a single real number solution x > 1. Find b.
Video Solution
Example 42.3 (AIME)
Let x, y, and z be real numbers satisfying the system
log2 (xyz − 3 + log5 x) = 5,
log3 (xyz − 3 + log5 y) = 4,
log4 (xyz − 3 + log5 z) = 4.
Find the value of | log5 x| + | log5 y| + | log5 z|.
Video Solution
Example 42.4 (AIME)
Let x, y, and z be positive real numbers that satisfy
2 logx (2y) = 2 log2x (4z) = log2x4 (8yz) ̸= 0.
1
The value of xy 5 z can be expressed in the form 2p/q
, where p and q are relatively prime
positive integers. Find p + q.
Video Solution
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42.3
Chapter 42. Logarithms
Practice Problems
Problem 42.3.1 (AMC 12)
What is the value of a for which log1 a + log1 a + log1 a = 1?
2
3
4
Video Solution
Problem 42.3.2 (AMC 12)
What is the value of
log3 7 · log5 9 · log7 11 · log9 13 · · · log21 25 · log23 27?
Video Solution
Problem 42.3.3 (AMC 12)
1
What is the value of (625log5 2015 ) 4 ?
Video Solution
Problem 42.3.4 (AMC 12)
There is a unique positive integer n such that
log2 (log16 n) = log4 (log4 n).
What is the sum of the digits of n?
Video Solution
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Chapter 42. Logarithms
Problem 42.3.5 (AMC 12)
There is a unique positive integer n such that
log2 (log16 n) = log4 (log4 n).
What is the sum of the digits of n?
Video Solution
Problem 42.3.6 (AMC 12)
The sequence
log12 162, log12 x, log12 y, log12 z, log12 1250
is an arithmetic progression. What is x?
Video Solution
Problem 42.3.7 (AMC 12)
For how many positive integers x is log10 (x − 40) + log10 (60 − x) < 2 ?
Video Solution
Problem 42.3.8 (AMC 12)
Positive real numbers a and b have the property that
q
log a +
q
log b + log
√
√
a + log b = 100
and all four terms on the left are positive integers, where log denotes the base 10 logarithm.
What is ab?
(A) 1052
(B) 10100
(C) 10144
(D) 10164
(E) 10200
Video Solution
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Chapter 42. Logarithms
Problem 42.3.9 (AMC 12)
Which of the following is the value of
q
log2 6 + log3 6?
Video Solution
Video Solution
Problem 42.3.10 (AIME)
The value of x that satisfies log2x 320 = log2x+3 32020 can be written as m
, where m
n
and n are relatively prime positive integers. Find m + n.
Video Solution
Problem 42.3.11 (AMC 12)
Positive real numbers x ̸= 1 and y ̸= 1 satisfy log2 x = logy 16 and xy = 64. What is
(log2 xy )2 ?
(A) 25
2
(B) 20
(C) 45
2
(D) 25
(E) 32
Video Solution
Problem 42.3.12 (AMC 12)
What is the value of
20
X
k2
log5k 3
k=1
!
·
100
X
!
log9k 25
k
?
k=1
Video Solution
Problem 42.3.13 (AMC 12)
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What is the value of
Chapter 42. Logarithms
log2 80 log2 160
−
?
log40 2
log20 2
Video Solution
Problem 42.3.14 (AMC 12)
Let S be the sum of all positive real numbers x for which
√
2
x2
=
√ 2x
2 .
Which of the following statements is true?
Video Solution
Additional Problems
Problem 42.3.15 (AMC 10)
Suppose that 4a = 5, 5b = 6, 6c = 7, and 7d = 8. What is a · b · c · d?
Problem 42.3.16 (AMC 12)
The sum of the base-10 logarithms of the divisors of 10n is 792. What is n?
Problem 42.3.17 (AIME)
In a Martian civilization, all logarithms whose bases are not specified as assumed
to be base b, for some fixed b ≥ 2. A Martian student writes down
√
3 log( x log x) = 56
loglog x (x) = 54
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Chapter 42. Logarithms
and finds that this system of equations has a single real number solution x > 1. Find b.
Problem 42.3.18 (AIME)
Let S be the sum of the base 10 logarithms of all the proper divisors (all divisors
of a number excluding itself) of 1000000. What is the integer nearest to S?
Problem 42.3.19 (AIME)
Positive integers a and b satisfy the condition
log2 (log2a (log2b (21000 ))) = 0.
Find the sum of all possible values of a + b.
Problem 42.3.20 (AIME)
The sequence a1 , a2 , . . . is geometric with a1 = a and common ratio r, where a and
r are positive integers. Given that log8 a1 + log8 a2 + · · · + log8 a12 = 2006, find the
number of possible ordered pairs (a, r).
Problem 42.3.21 (AMC 12)
If log(xy 3 ) = 1 and log(x2 y) = 1, what is log(xy)?
Problem 42.3.22 (AIME)
The increasing geometric sequence x0 , x1 , x2 , . . . consists entirely of integral powers
of 3. Given that
7
X
log3 (xn ) = 308
n=0
and
56 ≤ log3
7
X
!
xn ≤ 57,
n=0
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Chapter 42. Logarithms
find log3 (x14 ).
Problem 42.3.23 (AMC 12)
The solution of the equation 7x+7 = 8x can be expressed in the form x = logb 77 .
What is b?
Problem 42.3.24 (AMC 12)
For all integers n greater than 1, define an = log 12002 . Let b = a2 + a3 + a4 + a5
n
and c = a10 + a11 + a12 + a13 + a14 . Then b − c equals
(A) − 2
(B) − 1
1
(C) 2002
1
(D) 1001
(E) 21
Problem 42.3.25 (AMC 12)
The numbers log(a3 b7 ), log(a5 b12 ), and log(a8 b15 ) are the first three terms of an arithmetic
sequence, and the 12th term of the sequence is log bn . What is n?
Problem 42.3.26 (AIME)
Determine the value of ab if log8 a + log4 b2 = 5 and log8 b + log4 a2 = 7.
Problem 42.3.27 (AIME)
The number
2
3
+ log 2000
6
log4 20006
5
integers. Find m + n.
can be written as m
where m and n are relatively prime positive
n
Problem 42.3.28 (AMC 12)
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Chapter 42. Logarithms
The domain of the function f (x) = log 1 (log4 (log 1 (log16 (log 1 x)))) is an interval of
2
4
16
length m
,
where
m
and
n
are
relatively
prime
positive
integers.
What is m + n?
n
Problem 42.3.29 (AMC 12)
The solutions to the equation log3x 4 = log2x 8, where x is a positive real number
other than 13 or 12 , can be written as pq where p and q are relatively prime positive integers.
What is p + q?
Problem 42.3.30 (AMC 12)
Let m > 1 and n > 1 be integers. Suppose that the product of the solutions for x
of the equation
8(logn x)(logm x) − 7 logn x − 6 logm x − 2013 = 0
is the smallest possible integer. What is m + n?
Problem 42.3.31 (AMC 12)
Let S be the set of ordered triples (x, y, z) of real numbers for which
log10 (x + y) = z and log10 (x2 + y 2 ) = z + 1.
There are real numbers a and b such that for all ordered triples (x, y.z) in S we have
x3 + y 3 = a · 103z + b · 102z . What is the value of a + b?
Answers
42.1 31
42.2 216
42.3 265
42.4 49
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Chapter 42. Logarithms
42.3.1 24
42.3.2 6
42.3.3 2015
42.3.4 13
42.3.5 13
42.3.6 270
42.3.7 18
42.3.8 10164
42.3.9
q
log2 3 +
q
log3 2
42.3.10 103
42.3.11 20
42.3.12 21, 000
42.3.13 2
42.3.14 2 ≤ S < 6
42.3.15 32
42.3.16 11
42.3.17 216
42.3.18 141
42.3.19 881
42.3.20 046
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Chapter 42. Logarithms
42.3.21 35
42.3.22 091
42.3.23 87
42.3.24 −1
42.3.25 112
42.3.26 512
42.3.27 007
42.3.28 271
42.3.29 31
42.3.30 12
42.3.31 29
2
581
Chapter 43
Algebraic Trigonometry
Note: This topic is mainly relevant for AMC 12, but knowing some concepts can help make
some AMC 10 problems easier to solve.
Video Lecture
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43.1
Chapter 43. Algebraic Trigonometry
Trigonometric Identities
Theorem 43.1.1 (Trigonometric Identities)
Sine: sin(a) =
Opposite
Hypotenuse
Cosine: cos(a) =
Tangent: tan(a) =
Adjacent
Hypotenuse
Opposite
sin(a)
=
Adjacent
cos(a)
Remark 43.1.2
To remember the relationships, just use the mnemonics SOH, CAH, TOA:
SOH = Sin is Opposite over Hypotenuse
CAH = Cos is Adjacent over Hypotenuse
TOA = Tan is Opposite over Adjacent
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43.2
Chapter 43. Algebraic Trigonometry
More Trigonometric Identities
Theorem 43.2.1 (More Trigonometric Identities)
Cosecant: csc(a) =
Secant: sec(a) =
Cotangent: cot(a) =
43.3
Hypotenuse
1
=
Opposite
sin(a)
1
Hypotenuse
=
Adjacent
cos(a)
Adjacent
1
cos(a)
=
=
Opposite
tan(a)
sin(a)
Important Trigonometric Values
cos 0° = 1 √
√
cos 15° = 23+1
2
sin 0° = 0 √
√
sin 15° = 23−1
2
cos 30° = √23
cos 45° = 22
cos 60° = 1√
2
√
cos 75° = 23−1
2
cos 0° = 1
cos 90° = 0
cos 120° = − 1√
2
cos 135° = − √22
cos 150° = − 23
cos 180° = 1
sin 30° = 12√
sin 45° = √22
sin 60° = √23
√
sin 75° = 23+1
2
sin 0° = 0
sin 90° = 1 √
sin 120° = √23
sin 135° = 22
sin 150° = 12
sin 0° = 0
√
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43.4
Chapter 43. Algebraic Trigonometry
Unit Circle Identities
Theorem 43.4.1 (Unit Circle Identities)
sin(−a) = − sin(a)
sin(a) = sin(180 − a)
cos(a) = cos(−a)
cos(a) = − cos(180 − a)
tan(a) = − tan(180 − a)
tan(−a) = − tan(a)
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43.5
Chapter 43. Algebraic Trigonometry
Pythagorean Identities
Theorem 43.5.1 (Pythagorean Identities)
sin2 (a) + cos2 (a) = 1
tan2 (a) + 1 = sec2 (a)
cot2 (a) + 1 = csc2 (a)
43.6
Double Angle Identities
Theorem 43.6.1 (Double Angle Identities)
sin(2a) = 2 sin(a) cos(a)
cos(2a) = cos2 (a) − sin2 (a) = 2 cos2 (a) − 1 = 1 − 2 sin2 (a)
tan(2a) =
2 tan(a)
1 − tan2 (a)
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43.7
Chapter 43. Algebraic Trigonometry
Addition and Subtraction Identities
Theorem 43.7.1 (Addition and Subtraction Identities)
sin(a + b) = sin(a) cos(b) + sin(b) cos(a)
sin(a − b) = sin(a) cos(b) − sin(b) cos(a)
cos(a + b) = cos(a) cos(b) − sin(a) sin(b)
cos(a − b) = cos(a) cos(b) + sin(a) sin(b)
tan(a + b) =
tan(a) + tan(b)
1 − tan a tan b
tan(a − b) =
tan(a) − tan(b)
1 + tan a tan b
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43.8
Chapter 43. Algebraic Trigonometry
Half Angle Identities
Theorem 43.8.1 (Half Angle Identities)
s
a
1 − cos(a)
sin
=±
2
2
s
a
1 + cos(a)
=±
cos
2
2
43.9
Sum to Product Identities
Theorem 43.9.1 (Sum to Product Identities)
a+b
a−b
sin(a) + sin(b) = 2 sin
cos
2
2
!
a−b
a+b
sin(a) − sin(b) = 2 sin
cos
2
2
!
!
!
a−b
a+b
cos(a) + cos(b) = 2 cos
cos
2
2
!
!
a−b
a+b
cos(a) − cos(b) = −2 sin
sin
2
2
!
!
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43.10
Chapter 43. Algebraic Trigonometry
Product to Sum Identities
Theorem 43.10.1 (Product to Sum Identities)
1
sin(a) sin(b) = (cos(a − b) − cos(a + b))
2
1
cos(a) cos(b) = (cos(a − b) + cos(a + b))
2
1
sin(a) cos(b) = (sin(a + b) + sin(a − b))
2
Example 43.1 (AMC 12)
Suppose that sin a + sin b =
q
5
and cos a + cos b = 1. What is cos(a − b)?
3
Video Solution
Example 43.2 (AIME)
Let x be a real number such that sin10 x + cos10 x = 11
. Then sin12 x + cos12 x = m
36
n
where m and n are relatively prime positive integers. Find m + n.
Video Solution
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43.11
Chapter 43. Algebraic Trigonometry
Graphing to Solve Trigonometry Problems
Definition 43.11.1. A periodic function is a trigonometric function which repeats a pattern
of y-values at regular intervals. One complete repetition of the pattern is called a cycle. The
period of a function is the horizontal length of one complete cycle.
Period of sin, cos, and tan is 2π
43.12
Periods and Graphs of Trigonometric Functions
Concept 43.12.1 (Sine Graph)
Concept 43.12.2 (Cosine Graph)
Concept 43.12.3 (Tan Graph)
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Chapter 43. Algebraic Trigonometry
Remark 43.12.4
Long trigonometric expressions can be evaluated by telescoping, using identities in clever
ways, complex number substitutions (see complex numbers section below).
Example 43.3 (AMC 12)
For each integer n > 1, let F (n) be the number of solutions to the equation sin x = sin (nx)
P
on the interval [0, π]. What is 2007
n=2 F (n)?
Video Solution
43.13
Practice Problems
Problem 43.13.1 (AMC 12)
How many solutions does the equation sin
interval [0, π]?
π
cos x
2
= cos
π
sin x
2
have in the closed
Video Solution
Problem 43.13.2 (AMC 12)
Suppose that the roots of the polynomial P (x) = x3 + ax2 + bx + c are cos 2π
, cos 4π
, and
7
7
6π
cos 7 , where angles are in radians. What is abc?
Video Solution
Problem 43.13.3 (AMC 12)
How many values of θ in the interval 0 < θ ≤ 2π satisfy
1 − 3 sin θ + 5 cos 3θ = 0?
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Chapter 43. Algebraic Trigonometry
Video Solution
Additional Problems
Problem 43.13.4 (AMC 12)
The functions sin(x) and cos(x) are periodic with least period 2π. What is the least
period of the function cos(sin(x))?
Problem 43.13.5 (AMC 12)
The first two terms of a sequence are a1 = 1 and a2 = √13 . For n ≥ 1,
an+2 =
an + an+1
.
1 − an an+1
What is |a2009 |?
Problem 43.13.6 (AMC 12)
An object moves 8 cm in a straight line from A to B, turns at an angle α, measured in
radians and chosen at random from the interval (0, π), and moves 5 cm in a straight line
to C. What is the probability that AC < 7?
Problem 43.13.7 (AMC 12)
The number of x-intercepts on the graph of y = sin(1/x) in the interval (0.0001, 0.001) is
closest to
(A) 2900
(B) 3000
(C) 3100
(D) 3200
(E) 3300
Problem 43.13.8 (AMC 12)
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If
Chapter 43. Algebraic Trigonometry
2n
n=0 cos θ = 5, what is the value of cos 2θ?
P∞
Problem 43.13.9 (AMC 12)
How many solutions does the equation tan(2x) = cos( x2 ) have on the interval [0, 2π]?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Problem 43.13.10 (AMC 12)
Which of the following describes the largest subset of values of y within the closed
interval [0, π] for which
sin(x + y) ≤ sin(x) + sin(y)
for every x between 0 and π, inclusive?
Problem 43.13.11 (AMC 12)
Given that Ak = k(k−1)
cos k(k−1)π
, find |A19 + A20 + · · · + A98 |.
2
2
Problem 43.13.12 (AMC 12)
Given that (1 + sin t)(1 + cos
√t) = 5/4 and
m
(1 − sin t)(1 − cos t) = n − k, where k, m, and n are positive integers with m and n
relatively prime, find k + m + n.
Problem 43.13.13 (AMC 12)
Suppose a is a real number such that the equation
a · (sin x + sin (2x)) = sin (3x)
has more than one solution in the interval (0, π). The set of all such a that can be written
in the form
(p, q) ∪ (q, r),
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Chapter 43. Algebraic Trigonometry
where p, q, and r are real numbers with p < q < r. What is p + q + r?
Answers
43.1 13
43.2 067
43.3 2016532
43.13.1 2
1
43.13.2 32
43.13.3 6
43.13.4 π
43.13.5 0
43.13.6 13
43.13.7 2900
43.13.8 35
43.13.9 5
43.13.10 0 ≤ y ≤ π
43.13.11 040
43.13.12 027
43.13.13 −4
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Chapter 44
Geometric Trigonometry
Note: This topic is mainly relevant for AMC 12, but knowing some concepts like Law of
Sines and Law of Cosines can help make some AMC 10 problems easier to solve.
Video Lecture
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44.1
Chapter 44. Geometric Trigonometry
Area of a Triangle using trigonometry
Theorem 44.1.1 (Area of a Triangle using trigonometry)
In a triangle with side lengths, a, b, c, where the angle between sides a and b is denoted
by C
Area of the triangle =
44.2
1
· ab · sin(C)
2
Law of Sines
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Chapter 44. Geometric Trigonometry
Theorem 44.2.1 (Law of Sines)
In a triangle with sides a, b, c and angles A, B, and C where the side a is opposite to
the angle A, the side b is opposite to the angle B, and the side c is opposite the angle C,
we have
b
c
a
=
=
= 2R
sin A
sin B
sin C
where R is the circumradius of the triangle.
44.3
Law of Cosines
Theorem 44.3.1 (Law of Cosines)
In a triangle with side lengths, a, b, c, where the angle between sides a and b is denoted
by C
c2 = a2 + b2 − 2ab · cos C
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Chapter 44. Geometric Trigonometry
Video Lectures
Solving area problems using trigonometry
Example 44.1 (AIME)
In △P QR, P R = 15, QR = 20, and P Q = 25. Points A and B lie on P Q, points C and D
lie on QR, and points E and F lie on P R, with P A = QB = QC = RD = RE = P F = 5.
Find the area of hexagon ABCDEF .
Video Solution
Example 44.2 (AIME)
Triangle ABC has right angle at B, and contains a point P for which P A = 10, P B = 6,
and ∠AP B = ∠BP C = ∠CP A. Find P C.
A
P
B
C
Video Solution
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Chapter 44. Geometric Trigonometry
Example 44.3 (AMC 12)
Circles ω1 , ω2 , and ω3 each have radius 4 and are placed in the plane so that each circle
is externally tangent to the other two. Points P1 , P2 , and P3 lie on ω1 , ω2 , and ω3
respectively such that P1 P2 = P2 P3 = P3 P1 and line Pi Pi+1 is tangent to ωi for each
i = 1, 2, 3,√where√P4 = P1 . See the figure below. The area of △P1 P2 P3 can be written in
the form a + b for positive integers a and b. What is a + b?
ω1
P1
P3
ω3
ω2
P2
Video Solution
44.4
Practice Problems
Problem 44.4.1 (AMC 12)
Let ABC be an equilateral triangle. Extend side AB beyond B to a point B ′ so that
BB ′ = 3 · AB. Similarly, extend side BC beyond C to a point C ′ so that CC ′ = 3 · BC,
and extend side CA beyond A to a point A′ so that AA′ = 3 · CA. What is the ratio of
the area of △A′ B ′ C ′ to the area of △ABC?
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Chapter 44. Geometric Trigonometry
Video Solution
Problem 44.4.2 (AMC 12)
Regular octagon ABCDEF GH has area n. Let m be the area of quadrilateral ACEG.
What is m
?
n
Video Solution
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Chapter 44. Geometric Trigonometry
Additional Problems
Problem 44.4.3 (AHSME)
√
√
In △ABC, side a = 3, side b = 3, and side c > 3. Let x be the largest number such that the magnitude, in degrees, of the angle opposite side c exceeds x. Then x
equals:
Problem 44.4.4 (AHSME)
In triangle ABC, 3 sin A + 4 cos B = 6 and 4 sin B + 3 cos A = 1. Then ∠C in degrees is
Problem 44.4.5 (AMC 10)
In rectangle ABCD, AB = 20 and BC = 10. Let E be a point on CD such that
∠CBE = 15◦ . What is AE?
Problem 44.4.6 (AMC 12)
In the figure, equilateral hexagon ABCDEF has three nonadjacent
acute interior angles
√
◦
that each measure 30 . The enclosed area of the hexagon is 6 3. What is the perimeter
of the hexagon?
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A
B
F
D
C
E
Problem 44.4.7 (AIME)
Let ABCDEF be a regular hexagon. Let G, H, I, J, K, and L be the midpoints
of sides AB, BC, CD, DE, EF , and AF , respectively. The segments AH, BI, CJ,
DK, EL, and F G bound a smaller regular hexagon. Let the ratio of the area of the
smaller hexagon to the area of ABCDEF be expressed as a fraction m
where m and n
n
are relatively prime positive integers. Find m + n.
C
H
B
M
I
G
N
O
D
A
J
L
E
K
F
Problem 44.4.8 (AMC 12)
A circle centered at O has radius 1 and contains the point A. The segment AB is
tangent to the circle at A and ∠AOB = θ. If point C lies on OA and BC bisects ∠ABO,
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then OC =?
B
θ
O
C
A
Problem 44.4.9 (AIME)
Hexagon ABCDEF is divided into five rhombuses, P, Q, R,
√ S, and T , as shown. Rhombuses P, Q, R, and S are congruent, and each has area 2006. Let K be the area of
rhombus T . Given that K is a positive integer, find the number of possible values for K.
Problem 44.4.10 (AMC 12)
Alice and Bob live 10 miles apart. One day Alice looks due north from her house
and sees an airplane. At the same time Bob looks due west from his house and sees the
same airplane. The angle of elevation of the airplane is 30◦ from Alice’s position and 60◦
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from Bob’s position. Which of the following is closest to the airplane’s altitude, in miles?
(A) 3.5
(B) 4
(C) 4.5
(D) 5
(E) 5.5
Problem 44.4.11 (AIME)
In triangle ABC, tan ∠CAB = 22/7, and the altitude from A divides BC into segments of length 3 and 17. What is the area of triangle ABC?
Problem 44.4.12 (AIME)
In convex quadrilateral KLM N side M N is perpendicular to diagonal KM , side KL
is perpendicular to diagonal LN , M N = 65, and KL = 28. The line through L
perpendicular to side KN intersects diagonal KM at O with KO = 8. Find M O.
Problem 44.4.13 (AIME)
In △P QR, P R = 15, QR = 20, and P Q = 25. Points A and B lie on P Q, points C and D
lie on QR, and points E and F lie on P R, with P A = QB = QC = RD = RE = P F = 5.
Find the area of hexagon ABCDEF .
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Problem 44.4.14 (AMC 12)
In △P AT, ∠P = 36◦ , ∠A = 56◦ , and P A = 10. Points U and G lie on sides T P
and T A, respectively, so that P U = AG = 1. Let M and N be the midpoints of segments
P A and U G, respectively. What is the degree measure of the acute angle formed by lines
M N and P A?
T
U
1
N
56◦
36◦
P
G
M
1
A
10
Problem 44.4.15 (AMC 12)
In △ABC with integer side lengths, cos A = 11
, cos B = 78 , and cos C = − 14 . What is
16
the least possible perimeter for △ABC?
Problem 44.4.16 (AMC 12)
Line l in the coordinate plane has equation 3x − 5y + 40 = 0. This line is rotated
45◦ counterclockwise about the point (20, 20) to obtain line k. What is the x-coordinate
of the x-intercept of line k?
Problem 44.4.17 (AIME)
Let △ABC have side lengths AB = 30, BC = 32, and AC = 34. Point X lies in
the interior of BC, and points I1 and I2 are the incenters of △ABX and △ACX,
respectively. Find the minimum possible area of △AI1 I2 as X varies along BC.
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Answers
44.1 120
44.2 033
44.3
44.4.1 37 : 1
√
44.4.2
2
2
44.4.3 120
44.4.4 30
44.4.5 20
√
44.4.6 12 3
44.4.7 011
44.4.8 1+sin1 heta
44.4.9 089
44.4.10 5.5
44.4.11 110
44.4.12 090
44.4.13 120
44.4.14 80
44.4.15 9
44.4.16 15
44.4.17 126
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Complex Numbers
Note: This topic is mainly relevant for AMC 12.
Video Lecture
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Chapter 45. Complex Numbers
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Chapter 45. Complex Numbers
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45.1
Chapter 45. Complex Numbers
Basics
Definition 45.1.1. A complex number is a number that can be expressed in the form a + bi,
where a and b are real numbers, and i represents the “imaginary unit”. a is the real part of
our number, and bi is the imaginary part. Complex numbers are often represented by the
variable z.
√
Definition 45.1.2. i = −1, i2 = −1, i3 = −i, i4 = 1
Remark 45.1.3
√
Powers of i cycle every 4 terms, so i4n = i4 = 1, i4n+1 = i = −1, i4n+2 = i2 =
−1, i4n+3 = i3 = −i
Theorem 45.1.4 (Adding Complex Numbers)
(a + bi) + (c + di) = (a + c) + (b + d)i
Theorem 45.1.5 (Subtracting Complex Numbers)
(a + bi) − (c + di) = (a − c) + (b − d)i
Theorem 45.1.6 (Multiplying Complex Numbers)
(a + bi) · (c + di) = (ac − bd) + (bc + ad)i
Definition 45.1.7 (Real and Imaginary Parts). The imaginary part of a complex number
a + bi is b and the real part is a.
Remark 45.1.8
The imaginary part does not include a factor of i.
Definition 45.1.9. A complex conjugate is found by flipping the sign of the imaginary part
of complex number, and is represented as z̄.
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Theorem 45.1.10 (Finding Conjugates)
z = a + bi = a − bi
Definition 45.1.11 (Complex Plane). A complex number can also be represented geometrically by expressing a + bi as (a, b) on the Complex plane. The x-axis represents the Real axis
and the y-axis represents the Imaginary axis.
Definition 45.1.12. The magnitude of a complex number is represented by |z|, and is the
distance of a complex number (a, b) from the origin.
Theorem 45.1.13 (Magnitude of a Complex Number)
|z| = |a + bi| =
√
a2 + b 2
Theorem 45.1.14 (Multiplying Complex Numbers with their Conjugates)
(a + bi)(a − bi) = a2 + b2
Theorem 45.1.15
For a complex number z, z × z = z̄ 2
Remark 45.1.16
This is derived from the theorem above. It is super useful!
Example 45.1 (AIME)
Let N be the number of complex numbers z with the properties that |z| = 1 and z 6! − z 5!
is a real number. Find the remainder when N is divided by 1000.
Video Solution
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Chapter 45. Complex Numbers
Example 45.2 (AIME)
The complex numbers z and w satisfy the system
z+
20i
=5+i
w
12i
= −4 + 10i
z
Find the smallest possible value of |zw|2 .
w+
Video Solution
Example 45.3 (AIME)
There is a complex number z with imaginary part 164 and a positive integer n such that
z
= 4i.
z+n
Find n.
Video Solution
45.1.1
Polar Form
Definition 45.1.17. The angle that the positive real axis makes with the ray that connects
the origin with a complex number is called the argument of that complex number and is
represented by θ.
Theorem 45.1.18 (Argument of a Complex Number)
The argument θ of a complex number a + bi is
tan θ =
b
a
.
Definition 45.1.19. The distance between 0 and a complex number is sometimes called the
modulus of that complex number and is represented by r.
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Chapter 45. Complex Numbers
Theorem 45.1.20 (Modulus of a Complex Number)
The modulus r of a complex number a + bi is
r = |a + bi| =
√
a2 + b 2
.
Definition 45.1.21. Polar form is another way to represent a complex number based on its
modulus r and argument θ.
Theorem 45.1.22 (Polar Form)
z = a + bi = r(cos θ + i sin θ) = r cis θ
Remark 45.1.23
cis θ is just short for cos θ + i sin θ
Remark 45.1.24
Trigonometric ratios tell us that cos θ = ar and sin θ = rb , which we can rearrange to see
that r cos θ = a and r sin θ = b. Plugging in these values gives us the polar form formula.
Remark 45.1.25
cos θ + i sin θ can also be written as cis θ.
Theorem 45.1.26 (Euler’s Formula)
Euler’s Formula tells us that
, which tells us that
cos θ + i sin θ = eiθ
z = a + bi = r(cos θ + i sin θ) = reiθ
.
Remark 45.1.27
Euler’s Identity is a special case of Euler’s Formula and tells us that
eπi = −1
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.
Definition 45.1.28. Roots of unity are the complex solutions to an equation xn = 1, for
some positive integer n. There will always be n solutions to xn = 1.
Theorem 45.1.29 (Roots of Unity)
The set of the nth roots of unity is
for
e2kπi/n
k ∈ {1, 2, . . . n}
.
Example 45.4 (AMC 12)
How many integers n ≥ 2 are there such that whenever z1 , z2 , ..., zn are complex numbers
such that
|z1 | = |z2 | = ... = |zn | = 1 and z1 + z2 + ... + zn = 0,
then the numbers z1 , z2 , ..., zn are equally spaced on the unit circle in the complex plane?
Video Solution
Theorem 45.1.30 (Rotating a Point)
To rotate a point θ radians counterclockwise, covert a coordinate to its corresponding
complex number and multiply it by eiθ . Converting this back to ordered pairs gives us
our answer.
Theorem 45.1.31 (De Moivre’s Theorem)
For a complex number z = reiθ and a real number n,
z n = (reiθ )n = rn [cos(nθ) + i sin(nθ)]
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Chapter 45. Complex Numbers
Remark 45.1.32
We can use this to evaluate expressions like
√
( 3 + i)8
much easier because we just convert to polar form and apply De Moivre’s Theorem.
Remark 45.1.33
DeMoivre’s Theorem is very useful when dealing with complex numbers and exponents.
Example 45.5 (AIME)
The complex numbers z and w satisfy z 13 = w, w11 = z, and the imaginary part of z is
sin mπ
, for relatively prime positive integers m and n with m < n. Find n.
n
Video Solution
Example 45.6 (AMC 12)
Let
1+i
z= √ .
2
What is
z
12
+z
22
+z
32
+ ··· + z
122
1
1
1
· 12 + 22 + 32 + · · · + 122 ?
z
z
z
z
1
Video Solution
Concept 45.1.34
Complex numbers and their relations to circles makes them easy to work with for many
geometry problems, especially when dealing with polygons such as equilateral triangles
or squares.
How to solve geometry problems using complex numbers:
1. Assign a complex number to 1 or more of the coordinates
2. To find the complex numbers for other points, multiple/divide by eiθ
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3. Use the information you have to solve for what you are asked in the problem
Remark 45.1.35
We can also view algebraic complex number problems geometrically.
Example 45.7 (AIME)
The points (0, 0) , (a, 11) , and (b, 37) are the vertices of an equilateral triangle. Find
the value of ab .
Video Solution
Example 45.8 (AIME)
Given f (z) = z 2 − 19z, there are complex numbers z with the property that z, f (z), and
f (f (z)) are the vertices of a right triangle in the complex plane with a right√angle at
f (z). There are positive integers m and n such that one such value of z is m + n + 11i.
Find m + n.
Video Solution
45.2
Complex Numbers in Polynomials
Definition 45.2.1. A polynomial of degree n has n roots, and these roots may be complex.
For binomials, if our discriminant is negative we have complex roots.
Example 45.9 (AIME)
2018
2017
The polynomial
+ cz 2016 has real coefficients not exceeding 2019,
√ f (z) = az √+ bz
1+ 3i
and f
= 2015 + 2019 3i. Find the remainder when f (1) is divided by 1000.
2
Video Solution
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Chapter 45. Complex Numbers
Example 45.10 (AMC 12)
The equation
x10 + (13x − 1)10 = 0 has 10 complex roots r1 , r1 , r2 , r2 , r3 , r3 , r4 , r4 , r5 , r5 , where the
bar denotes complex conjugation. Find the value of
1
+ r21r2 + r31r3 + r41r4 + r51r5 .
r1 r1
Video Solution
45.3
Complex Numbers in Trigonometry
Theorem 45.3.1 (Sin and Cos values in terms of complex numbers)
sin θ =
eiθ + ei(180−θ)
2i
eiθ + ei(−θ)
cos θ =
2
tan θ =
eiθ + ei(180−θ)
i
eiθ + ei(−θ)
Remark 45.3.2
By using these substitutions, we can bash out the value of trigonometric expressions
easily without clever manipulation of trigonometric identities that would be needed to
solve the problem otherwise.
Example 45.11 (AMC 12)
Suppose that the roots of the polynomial P (x) = x3 + ax2 + bx + c are cos 2π
, cos 4π
,
7
7
and cos 6π
,
where
angles
are
in
radians.
What
is
abc?
7
Video Solution
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Chapter 45. Complex Numbers
Example 45.12 (PUMAC)
√
The value of sin(2°)sin(4°) . . . sin(90°) is p2505 . Find p.
Video Solution
45.4
Practice Problems
Problem 45.4.1 (AMC 12)
Of the following complex numbers z, which one has the property that z 5 has the greatest
real part?
√
√
√
√
(A) − 2
(B) − 3 + i
(C) − 2 + 2i
(D) − 1 + 3i
(E) 2i
Video Solution
Problem 45.4.2 (AMC 12)
Let Q(z) and R(z) be the unique polynomials such that
z 2021 + 1 = (z 2 + z + 1)Q(z) + R(z)
and the degree of R is less than 2. What is R(z)?
Video Solution
Problem 45.4.3 (AMC 12)
Let Z be a complex number satisfying 12|z|2 = 2|z + 2|2 + |z 2 + 1|2 + 31. What is
the value if z + z6 ?
Video Solution
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Additional Problems
Problem 45.4.4 (AMC 12)
There are 24 different complex numbers z such that z 24 = 1. For how many of these is z 6
a real number?
Problem 45.4.5
[AMC 12] What is the sum of the roots of z 12 = 64 that have a positive real part?
Problem 45.4.6 (AMC 12)
The polynomial f (x) = x4 +ax3 +bx2 +cx+d has real coefficients, and f (2i) = f (2+i) = 0.
What is a + b + c + d?
Problem 45.4.7 (AMC 12)
√
A function f is defined by f (z) = iz, where i = −1 and z is the complex conjugate of z. How many values of z satisfy both |z| = 5 and f (z) = z?
Problem 45.4.8 (AIME)
Find c if a, b, and c are positive integers which satisfy c = (a + bi)3 − 107i, where
i2 = −1.
Problem 45.4.9 (AMC 12)
For what value √
of n is i + 2i2 + 3i3 + · · · + nin = 48 + 49i?
Note: here i = −1.
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Problem 45.4.10 (AIME)
There are real numbers a, b, c, and d such that −20 is a root of x3 + ax +√b and −21 is a
root of x3 + cx2 + d. These two polynomials
share a complex root m + n · i, where m
√
and n are positive integers and i = −1. Find m + n.
Problem 45.4.11 (AMC 12)
What is the sum of the roots of z 12 = 64 that have a positive real part?
Problem 45.4.12 (AMC 12)
The complex numbers z and w satisfy z 13 = w, w11 = z, and the imaginary part
, for relatively prime positive integers m and n with m < n. Find n.
of z is sin mπ
n
Problem 45.4.13 (AMC 12)
A function f is defined by f (z) = (4 + i)z 2 + αz + γ for all complex numbers z,
where α and γ are complex numbers and i2 = −1. Suppose that f (1) and f (i) are both
real. What is the smallest possible value of |α| + |γ| ?
Problem 45.4.14 (AIME)
The complex number z is equal to 9 + bi, where b is a positive real number and i2 = −1.
Given that the imaginary parts of z 2 and z 3 are the same, what is b equal to?
Problem 45.4.15 (AMC 12)
z+i
for all complex numbers z =
̸ i, and let zn = F (zn−1 ) for all posiz−i
1
tive integers n. Given that z0 =
+ i and z2002 = a + bi, where a and b are real
137
numbers, find a + b.
Let F (z) =
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Problem 45.4.16 (AIME)
Given that z is a complex number such that z + z1 = 2 cos 3◦ , find the least integer
1
.
that is greater than z 2000 + z2000
Problem 45.4.17 (AMC 12)
A sequence of complex numbers z0 , z1 , z2 , ... is defined by the rule
zn+1 =
izn
,
zn
where zn is the complex conjugate of zn and i2 = −1. Suppose that |z0 | = 1 and z2005 = 1.
How many possible values are there for z0 ?
Problem 45.4.18 (AIME)
For how many positive integers n less than or equal to 1000 is (sin t + i cos t)n =
sin nt + i cos nt true for all real t?
Problem 45.4.19 (AMC 12)
A sequence (a1 , b1 ), (a2 , b2 ), (a3 , b3 ), . . . of points in the coordinate plane satisfies
√
√
(an+1 , bn+1 ) = ( 3an − bn , 3bn + an ) for n = 1, 2, 3, . . ..
Suppose that (a100 , b100 ) = (2, 4). What is a1 + b1 ?
Problem 45.4.20 (AIME)
Let N be the number of complex numbers z with the properties that |z| = 1 and
z 6! − z 5! is a real number. Find the remainder when N is divided by 1000.
Problem 45.4.21 (AIME)
Let w and z be complex numbers such that |w| = 1 and |z| = 10. Let θ = arg
w−z
z
.
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The maximum possible value of tan2 θ can be written as pq , where p and q are relatively
prime positive integers. Find p + q. (Note that arg(w), for w =
̸ 0, denotes the measure
of the angle that the ray from 0 to w makes with the positive real axis in the complex
plane)
Problem 45.4.22 (AMC 12)
The solutions to the equation (z + 6)8 = 81 are connected in the complex plane to
form a convex regular polygon, three of whose vertices are labeled A, B, and C. What is
the least possible area of △ABC?
Problem 45.4.23 (AMC 12)
Let R be the region in the complex plane consisting of all complex numbers z that
can be written as the sum of complex numbers z1 and z2 , where z1 lies on the segment
with endpoints 3 and 4i, and z2 has magnitude at most 1. What integer is closest to the
area of R?
Problem 45.4.24 (AIME)
There are 2n complex numbers that satisfy both z 28 − z 8 − 1 = 0 and | z |= 1. These
numbers have the form zm = cos θm + i sin θm , where 0 ≤ θ1 < θ2 < . . . < θ2n < 360 and
angles are measured in degrees. Find the value of θ2 + θ4 + . . . + θ2n .
Problem 45.4.25 (AMC 12)
Let c be a real number, and let z1 and z2 be the two complex numbers satisfying
the equation z 2 − cz + 10 = 0. Points z1 , z2 , z11 , and z12 are the vertices of (convex)
quadrilateral Q in the complex plane. When the area of Q obtains its maximum possible
value, c is closest to which of the following?
Problem 45.4.26 (AIME)
Let P be the product of the roots of z 6 + z 4 + z 3 + z 2 + 1 = 0 that have a positive
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imaginary part, and suppose that P = r(cos θ◦ + i sin θ◦ ), where 0 < r and 0 ≤ θ < 360.
Find θ.
Problem 45.4.27 (AMC 12)
√
√
√
The solutions to the equations z 2 = 4 + 4 15i and z 2 = 2 + 2 3i, where i = −1, form
the vertices of a parallelogram in
√ the complex plane. The area of this parallelogram can
√
be written in the form p q − r s, where p, q, r, and s are positive integers and neither
q nor s is divisible by the square of any prime number. What is p + q + r + s?
Answers
45.1 440
45.2 40
45.3 697
45.4 2
45.5 71
45.6 36
45.7 315
45.8 230
45.9 53
45.10 850
1
45.11 32
45.12 192
√
45.4.1 − 3 + i
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45.4.2 −z
45.4.3 −2
45.4.4 12
√
√
45.4.5 2 2 + 6
45.4.6 9
45.4.7 2
45.4.8 198
45.4.9 97
45.4.10 330
√
√
45.4.11 2 2 + 6
45.4.12 071
√
45.4.13 2
45.4.14 015
45.4.15 275
45.4.16 000
45.4.17 22005
45.4.18 250
45.4.19 2198
45.4.20 440
45.4.21 100
√
45.4.22 32 2 − 32
45.4.23 13
45.4.24 840
45.4.25 4.5
45.4.26 276
45.4.27 20
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Additional Techniques and Strategies
46.1
Meta-solving Techniques
Video Lectures
Meta-solving Techniques
Definition 46.1.1. Meta-solving is finding the answer to a problem without actually solving
it.
Remark 46.1.2
These techniques may not work for all problems. These techniques are especially useful
when the problem provides answer choices.
Remark 46.1.3 (Meta-Solving Warning)
Don’t get too carried away with these techniques to the point where you don’t even try
to solve the problem legitimately.
Concept 46.1.4 (Engineering Induction)
Engineering Induction is the process of trying and finding the value to small cases and
assuming it’s true for larger ones.
Steps for Engineering Induction Problems:
1. Try small cases
2. Look for a pattern amongst those small cases (there may not always be one)
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3. Assume the pattern can continue for larger cases and find the answer
Remark 46.1.5
We can try to apply engineering induction when we see the values in the problem seem
hard/impossible to compute.
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Concept 46.1.6 (Looking for unique properties of numbers)
Rather than computing the exact answer, we can compute unique properties of your
answer choices so that you can eliminate all answer choices that don’t satisfy the property
and so that you will be left with 1 answer choice (or possibly more in which case you
can just guess from the remaining ones)
Some unique properties you can look for in your answer choices and try to compute are
• Units Digit
• Last 2 digits
• Parity (Even, Odd)
• Perfect square or not one
• Prime/composite
• Modulus (remainders when divided by 3, 4, 5, etc.)
• Denominators of common fractions (or what they must divide)
• Multiples/Factors of numbers
• etc.
Remark 46.1.7
These last 2 properties are especially useful in combinatorics problems as you can easily
find numbers you have to multiply with each other to get your answer.
Concept 46.1.8
Look for the option choices that are the ”odd one out” or that are different from all
others
• Look for outliers (primes, large/small numbers, odd/even numbers, powers of 2,
etc)
Concept 46.1.9 (Trying all the Option choices)
In some problems, you can
• Try all the option choices into the conditions in the problem
• Look at the conditions in the problem and see which of the option choices could
work
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Chapter 46. Additional Techniques and Strategies
• etc.
After doing so, you will either have a better guess or exact answer.
Concept 46.1.10 (Elimination of Option Choices)
Tying closely to the previous technique, you can also usually eliminate option choices
based on
• Unique properties of Numbers (See Above)
• How large or small the number must be
Remark 46.1.11
We would recommend guessing ONLY if you can narrow it to 2 or 3 option choices.
Concept 46.1.12 (Estimation of Answer)
Often times in problems (especially geometry) you can easily find an approximate answer
and see which of the option choices most closely matches what you got.
In geometry, a common strategy to do so is to mark out areas approximately equal to
those of areas you know.
Concept 46.1.13 (Using Freedom in Problems)
Assuming facts when you have freedom in the problem statement can be very useful.
Essentially, as long as the problem is not telling you ”this fact is not true” (so, whatever
assumption you want to make will satisfy the problem’s conditions) you can assume the
fact is true to simplify your problem and make it really easy to solve.
For example, if you are asked to find some universal ratio in a triangle and you aren’t
specifically told that the triangle isn’t equilateral, you can just assume the triangle is
equilateral and solve the remaining problem from there.
Remark 46.1.14
Make sure not to assume false information! Be very careful that your assumption can be
true.
In our previous example, if we were told the triangle had 2 sides of length 7 and
8, then our assumption would be false, so it wouldn’t work then.
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46.2
Chapter 46. Additional Techniques and Strategies
Silly Mistakes
Silly mistakes are very common and can really lower your score on the AMC 10/12. Here are
some tips on how to avoid the different kinds of silly mistakes:
Concept 46.2.1 (Avoiding Arithmetic Errors)
A good way to avoid arithmetic errors is to
• Do your computation 2 ways (e.g. If you have to do 87 · 93, you can multiply them
with 87 on the top and with 93 on top)
• Be more organized, and write more steps
Concept 46.2.2 (Avoiding Mathematical Errors)
An easy way to avoid mathematical errors is to
• Do your work neatly!
• Make boxes for each problem on scratch paper per problem
• Don’t skip steps
• Check your work, following the tip above will make it easier to do so
• Do your steps methodically
• Try to substitute your answer back into the problem (if you can)
• Try an alternate solution to confirm your answer
• Estimate what the answer has to be, and see if your answer is close to what your
estimate is
Concept 46.2.3 (Avoid Reading Errors)
Reading the problem wrong is one of the most common mistakes. Often times, you might
forget about important key words like
• inclusive, except
• even, odd
• prime, composite
• integer, natural, real, complex
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Some strategies to avoid them are
• After solving the problem, reread the question part of the problem to make sure
you are answering what the question asks for!
• Underline key words while reading (that’s probably gonna be hard this year with
the test being online, but as an alternative you can take note of the important
words on your scratch paper)
Concept 46.2.4 (Avoid Missing a Final Step Errors)
Sometimes in problems, you might be so caught up in moving forward in the test that
you might forget an important step at the end.
For example, in a problem you might think ”I’ll multiply by 5 to whatever answer
I get” and then you find that answer but forget to multiply by 5. A way to avoid this is:
• Write ”Remember ...” big and bold on your scratch or the question paper
Remark 46.2.5
A very common reading mistake is getting confused between the words non-negative and
positive. Remember, non-negative includes 0 while positive doesn’t!
Concept 46.2.6 (Avoid Making False Assumptions)
Often times, you might just think something is true and assume it’s true for the rest
of the problem, when really it was false. Proving all your assumptions can be too time
consuming. However we recommend at least seeing some sort of reasoning for why your
assumption should be correct (unless of course you are using one of the meta-solving
techniques).
Concept 46.2.7 (Avoid Casework Errors)
Many times, in casework problems you might
• Undercount possibilities
• Overcount possibilities
• Miss edges/extreme cases
Some strategies to avoid them are
• Be methodical in your casework
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• Make sure all your cases work
– An easy way to do this is just to try a few examples in your case to see if they
actually work
– Especially, make sure to check if extremes work
– Make sure all your cases are disjoint and that you are not overcounting
anything that’s common between the cases
– Make sure your cases cover all possibilities that the problem asks for
– Solve the problem in multiple ways (for example, by both casework and
complementary counting)
Remark 46.2.8
I know that some of these strategies may take a lot of extra time to follow so we
recommend analyzing how you are making silly mistakes and from there see which
strategies you will want to follow.
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Chapter 46. Additional Techniques and Strategies
Other Strategies To Maximize Your Score
Concept 46.3.1 (Plan Your Time)
Make sure to plan your time.
• Questions 1-10 are generally easy, 11-20 are medium, 21-25 are hard
• Sometimes one of the early questions can be hard or bashy, or a later question can
be easy
• Don’t get stuck on a question. Move on and come back to it later.
• Star any question you are unsure about but you feel you can solve it, or any
questions that you solved but are not confident of your answer
• Budget your time properly
• Leave enough time for last 10 problems
• Leave some time to review starred problems and check your work
Concept 46.3.2 (Guessing Strategies)
If you want to guess, make sure to keep these things in mind:
• You get 1.5 points for leaving a question blank, so if you don’t know how to solve
the problem, just leave it blank
• Try using meta-solving strategies above
• If you can narrow down the choices to 2-3 options, only then make an educated
guess
Concept 46.3.3 (Test Day Strategies)
Here are some test day strategies:
• Don’t try to cram or study on the day before the test
– Just review a few formulas or strategies from this book
• Be relaxed
– Get a good night’s sleep
– Eat a healthy meal
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– Meditate
– Eat dark chocolate
– Listen to music
– Or whatever else makes you feel relaxed
Concept 46.3.4 (Problem Solving Strategies)
Try to remember these problem solving strategies:
• When you are stuck, try to use the information in the problem you haven’t used
yet
• When solving a problem, think of what technique would likely be at play
• Don’t get too stuck with one approach to a problem, move on, and come back to
the problem with a fresh perspective
• In problems that seem complex, try small cases to look for a pattern that can allow
you to figure out how to approach the problem and what patterns may exist (this
is similar to engineering induction)
Concept 46.3.5 (AIME Qualification)
To qualify for AIME, based on recent cutoff scores, you will need a score of at least
105. However you should aim for 110+ to be safe. For this, you’ll need a 17-0-8 split
(right-wrong-skipped), a 18-5-2 split, a 19-6-0 split or something better.
Concept 46.3.6 (USAJMO/USAMO Qualification)
To qualify for USAJMO/USAMO, you’ll probably need a score of at least 120 to have a
good chance (of course this also depends heavily on your AIME score).
Remark 46.3.7
Another strategy for the AMC 10/12B is that if you are expecting a AIME qualifying
score on the AMC 10/12A, then you should try to solve more problems and take a
”higher risk higher reward” approach for the AMC 10/12B, however if you didn’t do
good on the A, you should take a safe path.
Remark 46.3.8
Last, but most important, don’t stress out too much about how you will do! It’s just a
math contest, and you’ll probably have many more opportunities in the future.
Good luck to you on your math competitions. We hope you found this book useful!
If you have any feedback, find any errors, or think of any interesting problems that should be
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added here, please fill out this Feedback Form or email us at omegalearn.info@gmail.org.
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