Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
AMC 8 Geometry Workout 1 (from the 2008 AMC8 solution manual)
Pythagorean Theorem Workout 1
1. Draw the diagram. Because you walk east (30), and then west (6), the right triangle
that takes you back to the stump is a 24-45-C triangle. Do 452  24 2 to find C (51
paces) and you’re done. Also, notice that 24-45-51 is just three times as big as the
more common 8-15-17 Pythagorean Triple.
Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
2. The radius of the two semi-circles is 1 (from the origin to (0,1), or to (0,-1)). So the
combined area of the two semi-circles is 12   1 . The radius of the large circle is the
hypotenuse of a 1-1-C right triangle (from the origin to any of the labeled points).
12  12  2
1 1

2 2
2
So the area of the large circle is 2   2 . The ratio of the areas is
Pythagorean Theorem Workout 2
1. The easiest place to start is the A-12-13 triangle. Students should recognize this is as a
5-12-13 triangle, but you can also do 132 122  5 Then you may notice the MOST
important right triangle…the 3-4-5 right triangle (4 is the missing side). To find ‘x’ you
must solve 8.52  42 , which is annoying to do without a calculator. But students may
recognize that 8.5 is half of 17, and 4 is half of 8….so it’s half of our old friend, the
8-15-17 Pythagorean Triple! So ‘x’ is half of 15, or 7.5.
2. The radii of the circles is 8, and that is 1/6 of the length of QP, so QP must be 48. Thus
each right triangle has a hypotenuse of 48 and a leg of 8. Unfortunately 482  82 is not
a whole number, so we can either keep our answer in “simplest radical form,” or 8 35
, or just use a calculator (47.33). To find the area of the entire (2-part) triangle, just do
bh
/2 (notice that the base is just 16, or twice the radius). You could express your
answer as a radical ( 64 35 ), or as a decimal (378.6).
Similar Triangles Workout 1
1. The small triangle on top is just ¼ of the bigger triangle, and they’re similar. So, set up
a proportion to find AB.
5
3

Solving gives you AB=12, so AD must be 9.
20 AB
2. (from Handbook solutions)
Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
Counting Workout 1
1. It isn’t hard to make an organized list for this problem. Just start by having the most
quarters possible, then the most dimes possible, then the number of nickels.
Quarters 2
1
1
Dimes
0
2
1
Nickels 0
1
3
Continue that pattern. Have the most quarters, then the most dimes, then fill in the nickels.
Here’s the complete table.
Quarters 2
1
1
1
0
0
0
0
0
0
Dimes
0
2
1
0
5
4
3
2
1
0
Nickels 0
1
3
5
0
2
4
6
8
10
That’s 10 ways.
How many nice little patterns do you notice?
2. On these problems, often it’s a good idea to focus on just one area. So just focus on
the top half of the diagram. There are three semi-circles on the top half. There are
four identical “halves” like the top half. So 3x4 = 12. That’s 12 semi-circles
Counting Workout 2
1. You *could* make a list here. Call the friends A,B,C,D,E,F,G, and H. First A and B
could get a drink. Or A and C, or A-D, OR A-E, A-F, A-G, and A-H. That’s 7 using A.
Then we already have A-B, so using B just start with B-C, then B-D….that’s 6 using B,
then 5 including C, then 4, then 3, then 2, then 1 using G (just G-H).
7+6+5+4+3+2+1=28 That’s not TOO bad, but using combinations it might be a bit
8!
 28
quicker. The number of ways to “choose” two things from eight is 8 C 2 
2!6!
2. For this problem, a list would take a LONG time! A better way to do it is to just to say
there are 100 choices for the 1st prize, then 99 choices for the 2nd prize, 98 for the 3rd
and 97 for the 4th. 100x99x98x97= 94,109,400 ways to distribute the prizes. Crazy,
eh?? A way to do it with permutations is to do
100
p4 
100!
 94,109,400
4!96!
For the “extension” (What’s the probability that at least ONE of the 8 kids wins a
prize), find the probability that NONE of the 8 wins a prize, then subtract that from 1.
The probability that NONE of the 8 wins the first prize is 92/100, then for the second
prize it’s 91/99, then 90/98 for the 3rd and 89/97 for the 4th prize. Multiply all those
fractions and subtract from 1 and you get 28.7%.
Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
Counting Workout 3
See “MathCounts Mini #31” at http://www.youtube.com/watch?v=i7NDFKtkyrg
But here is one solution for each:
1. First choose the “lucky kid” who gets 2 gifts. There are 5 kids to choose from, and for
the gifts there are 6 gifts to choose from for the 1st gift, then 5 for the 2nd…..BUT,
that’s overcounting because order doesn’t matter (with that one kid). So we must
divide by two. Then for the next (non “lucky”) kid, there are 4 gifts to choose from.
For the 3rd kid there are 3 gifts, for the 4th kid there are 2 gifts, then there’s the last kid
and last gift. That gives:
 65 
5
  4  3  2 1  1800
2 

2. Let’s do this one like the last one. There are six choices for “the number that gets
double-used.” Then there are 4C2 places (or
43
) places to put those two “double2
used” numbers. Then there are five choices for the 3rd die, then 4 choices for the 4th
43 
  5  4  720 But don’t forget this is probability! The total
 2 
die. That gives: 6  
number of ways the dice can turn up is 6  6  6  6  1296 and 720/1296 reduces to 5/9
Using Algebra Workout 1
1. Let’s call the numbers n, n+2, and n+4 The equation will be
5(n+4) = (n+n+4) – 6
5n+10 = 2n + 4 – 6
5n+10 = 2n – 2
3n = -12
n = -4
so if that’s the smallest number, then the middle # must be -2.
2. We could call the number of dimes D and the number of nickels N, but let’s call the
number of nickels D – 5 because we have 5 less nickels than dimes.
Then because dimes are worth 10 cents and nickels are worth 5, the equation is
10d + 5(d – 5) = 245
10d + 5d – 25 = 245
15d – 25 = 245
15d = 270
d = 18
so that means we have 13 nickels (5 less than 18).
3. Kind of like the last problem, we could call the number of cows C and the number of
ducks D, but let’s say we have C cows and (50 – c) ducks. (do you see why?...we have
Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
50 total animals…..). So, since cows have 4 legs and ducks have 2, our “leg equation”
is
2d + 4(50 – d) = 140
2d + 200 – 4d = 140
-2d + 200 = 140
-2d = -60
d = 30 and that’s our answer.
4. The formula for perimeter is P = 2L + 2W But since it tells us that L = 2W – 6, we can
substitute 2W – 6 in for L. Then the equation is
504 = 2(2w – 6) + 2w
504 = 4w – 12 + 2w
504 = 6w – 12
516 = 6w
86 = w
So the width is 86, which makes the length 166, and the AREA is 14,276
5. So Peter is 37 years old now (if six years from now he’ll be 43). In 13 years Peter will
be 50. So if we call Andrew’s age (now) A, then our equation is
50 + (A + 13) = 87
63 + A = 87
A = 24 So Andrew is 24 now.
6. In that mixture John has, has 4 ounces of pure salt (20% of 20 ounces). If we add more
pure salt to it, we’ll add more salt, but also more “overall mixture.” So if ‘x’ is the
extra pure salt we’re adding, then our equation is
4 x
 0.25 Use cross products to get
20  x
4 + x = 0.25(10 + x)
4 + x = 5 + 0.25x
0.75x = 1
x = 11/3 oz
Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
Using Algebra: Interesting Applications
1. I know it seems silly, but write the equation
x = 1.23
then multiply both sides by 100, and you get
100x  123.23
Then subtract the first equation from the second, and you get
99x = 122
So
X = 122/99, or 123/99
2. You might not see it at first, but the bottom of the fraction is the same as the whole
expression…..so the equation is
x  1
1
x
If you multiply both sides by x, you get
x2 = x + 1 Rewrite to get
x2 – x – 1 = 0 Then use the quadratic formula to find
x
1 5
2
The negative value doesn’t make sense, but the positive solution is 1.618
Have you seen that before?.....it’s the GOLDEN RATIO!! How cool!
3. Again, it seems silly, but write
x  3 2
4 8
  ... Multiply both sides by the RATE that it’s decreasing at, which is 2/3. You
3 9
get
2
4 8
x  2    ... WO….that’s the same as the top, but “slid over” a bit. Let’s subtract the
3
3 9
top equation from the bottom one, then you get:
/3 x = 3 And THAT’S easy.
x = 9 HOW COOL!
1
Number Theory Workout 1
Try this one: The prime factorization of 320 is 26x51
Because the exponents are 6 and 1, there are (6+1)(1+1) ways to make factors.
7x2 is 14. So 320 has 14 factors.
1. To be a perfect square, a number has to have squares of primes in its prime
factorization (like 22 or 52). But 56 is also a perfect square, because 53x53 is 56
So we have to be careful to count them all….in fact 20 is also a perfect square!
Session 1 Solutions
July 2013
Matt Bixby
ColoradoMathCamp.com
Okay, so from 23x34 you can get 2 perfect squares from the 2s (20 and 22), and 3
perfect squares from the 3s (30, 32 and 34). So that’s 2x3, which is 6. So 23x34 has 6
perfect square factors.
2. The “biggest negatives” are -10, -5 and -3….but we can get a smaller (more negative)
number if we use -10, 10 and 6. That product is -600.
3. First we rewrite all the numbers as primes.
20 2009  10 2000  40 9  2 n
2 4018 5 2009  2 2000  5 2000  2 27  59  2 n
2 4018  2 2027  2 n
21991  2 n
So n = 1991
Primes/Number Theory Workout 2
1. You need BOTH x and y to be integers, so we need something in the bottom of
the fraction that is a factor of 6….otherwise ‘y’ won’t be an integer. So the factors of 6 are
1,2,3 and 6….but don’t forget the negatives!
So, set 2x – 1 equal to each of the factors and solve for x.
For 6, -6, 2 and -2, the values for ‘x’ don’t come out to be integers, but look at 2x – 1 = 3
2x – 1 = 3
Try 2x – 1 = -3
And 2x – 1 = -1
and 2x – 1 = 1
2x = 4
2x = -2
2x = 0
2x = 2
X = 2 that worked!
x = -1 hooray!
X=0
x=1
Add all those x values together and we’re done: 2 + (-1) + 0 + 1 = 2
2. For the three sets, the first one blinks every 6 seconds (60 divided by 10), the
second set blinks every 5 seconds, and the third one blinks every 3 seconds. So
what we need is the LCM of 6,5 and 3, which is 30. So after 30 seconds they’ll all
blink (once) at the same time.
3. “Units Digits” problems are very common, and almost always are easy to solve
(because of patterns that form). Let’s look at the units digit of the powers of 19:
The units digit of 191 is 9, then for 192 it’s 1, then for 193 it’s 9, then for 194 it’s
1….see the pattern? For every odd numbered power (1, 3, 5, etc) the units digit is
9, and for every even power (2,4, 6, etc) the units digit is 1. It’s actually exactly the
same for the powers of 99. So that means the units digits of 1919 and 9999 are 9 and
9. 9+9 is 18, but we just care about the units digit, which is 8