Fifty Lectures for
American Mathematics Competitions
Volume 1
Copyright © 2012
http://www.mymathcounts.com/index.php
Table of Contents
Chapter 1 Algebra Manipulations
1
Chapter 2 Radicals
18
Chapter 3 Solving Radical Equations
40
Chapter 4 Absolute Values
64
Chapter 5 Solving Absolute Value Equations
85
Chapter 6 Vieta Theorem and Applications
109
Chapter 7 Square Numbers
138
Chapter 8 Divisibility
156
Chapter 9 Geometry Congruent Triangles
182
Chapter 10 Geometry Area and Area Method
202
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
232
Chapter 12 Trigonometry Six Functions
257
Index
280
50 AMC Lectures
Chapter 1 Algebraic Manipulation
BASIC KNOWLEDGE
Below is a list of useful equations to be aware of and know. They can all be derived
through expanding or factoring.
Perfect square trinomial
( x y)2 ( x y)2 4 xy
( x y)2 x 2 2 xy y 2
( x y)2 ( x y)2 4 xy
( x y)2 x 2 2 xy y 2
( x y)2 ( x y)2 2( x 2 y 2 )
( x y)2 ( x y)2 4 xy
( x y z)2 x 2 y 2 z 2 2 xy 2 xz 2 yz
x2 + y2 + z2 – xy – yz –zx =
1
[( x – y )2 + ( y –z )2 + ( z – x )2]
2
(x y z w)2 x 2 y 2 z 2 w 2 2xy 2xz 2xw 2yz 2yw 2zw
Difference and sum of two squares
x 2 y 2 ( x y)2 2 xy
x 2 y 2 ( x y)2 2 xy
x 2 y 2 x y x y
Difference and sum of two cubes
x 3 y 3 ( x y)( x 2 xy y 2 )
x 3 y 3 ( x y)( x 2 xy y 2 )
x y 3 x3 3x2 y 3xy 2 y3 x3 y3 3xyx y .
x y 3 x3 3x2 y 3xy 2 y3 x3 y3 3xyx y .
x n y n x y x n 1 x n 2 y y n 1 for all n.
x n y n ( x y)( x n 1 x n 2 y ... y n 1 ) for all even n.
x n y n ( x y)( x n 1 x n 2 y ... y n 1 ) for all odd n.
1
50 AMC Lectures
Chapter 1 Algebraic Manipulation
EXAMPLES
Example 1: (1987 AMC) If (x, y) is a solution to the system xy = 6 and x2 y+ xy2 + x + y
= 63, find x2 + y2.
Solution: 69.
We first factor: x2 y+ xy2 + x + y = xy ( x + y) + (x + y) = (6 + 1)(x + y).
We are given that the expression above equals 63, so
(6 + 1)(x + y) = 63
x + y = 9.
We know that x2 + y2 = ( x + y)2 – 2xy, and substituting in 6 for xy and 42 for (x + y)2
yields x2 + y2 = 81 – 12 = 69.
Example 2: Find m3
1
1
if m 2 .
3
m
m
Solution: 2.
Method 1:
1
2 by m:
m
m2 + 1 = 2m
m2 – 2m + 1 = 0
From this equation, we get m = 1.
1
1
Therefore m3 3 13 3 1 1 2 .
m
1
We multiply both sides of m
(m – 1)2 = 0.
Method 2:
3
We know that x y x3 3x 2 y 3xy 2 y3 x3 y3 3xy x y .
So x3 y 3 x y – 3xy ( x y)
1
Letting x = m and y
in (1) gives us:
m
1
1
1
1
m3 3 (m )3 3m (m ) 23 3 2 2 .
m
m
m
m
3
Method 3:
2
(1)
50 AMC Lectures
Chapter 1 Algebraic Manipulation
We know that x 3 y 3 ( x y)( x 2 xy y 2 ) .
1
Letting x = m and y
in (1), the above equation becomes:
m
1
1
1
1
1
m3 3 (m )(m2 m 2 ) 2[(m )2 3] 2(22 3) 2
m
m
m m
m
Method 4:
1
1
2 by (m2 2 1) we get:
m
m
1
1
1
1
1
(m2 2 1) (m ) 2 (m2 2 1)
m3 3 = 2[(m ) 2 2 1] 2 .
m
m
m
m
m
Multiplying both sides of m
Example 3: Find m 4
1
1
if m 4 .
4
m
m
Solution: 194.
1
2
2
2
2 as y into x y ( x y ) 2 xy gives us:
m
2
2
2
1 2 1
1
2
4
m 4 m 2 2 = m 2 2 16 2 2 194 .
m
m
m
Substituting m2 as x and
Example 4: Find m8
1
1
if m 1 .
8
m
m
Solution:
Substituting m4 as x and
1
2
2
2
4 as y into x y ( x y ) 2 xy gives us:
m
2
2
2
1
1
1
m 4 m2 2 2 = m 2 2 9 2 7 .
m
m m
1
Substituting m4 as x and 4 as y into x 2 y 2 ( x y)2 2 xy yields:
m
4
2
1
1
m 8 m4 4 2 49 2 47 .
m m
8
3
50 AMC Lectures
Example 5: Find m3
Chapter 1 Algebraic Manipulation
1
if y m2 m 1 m 1 m2 , where both m and y are real
3
m
numbers.
Solution: 4.
Since y is a real number, the expressions under the square roots must be greater than or
equal to 0, so we know that
m2 m 1 0
m 1 m2 0 m2 m 1 0
Since m2 m 1 must be greater than or equal to 0 and less than or equal to 0,
1
m 1
m2 m 1 0
( m ≠ 0 ).
m
2
1
1
1
1
1
m3 3 m m2 2 1 m m 3 4 .
m
m
m
m
m
Example 6: Find
Solution:
2 x 3xy 2 y 1 1
if 3 .
x 2 xy y
x y
3
.
5
Method 1:
Since the denominator of a fraction cannot be 0, we know that x 0, y 0, and x – 2xy –
y 0.
2 x 3xy 2 y
Dividing each term of
by xy yields:
x 2 xy y
1 1
2 2
3 2 3
23 3 3
y x
x y
.
1 1
3
2
5
1
1
2 2
x y
y x
Method 2:
1 1
3 by xy gives us y x 3xy .
x y
2 x 3xy 2 y 2( y x) 3xy 6 xy 3xy 3
.
Hence
x 2 xy y
( y x) 2 xy 3xy 2 xy 5
Multiplying both sides of
4
50 AMC Lectures
Chapter 1 Algebraic Manipulation
Method 3:
Solving for x in the equation
y
1 1
.
3 gives us x
3y 1
x y
2y
3y2
6 y2 2 y
2 x 3xy 2 y 3 y 1 3 y 1
2 y 3y2 6 y2 2 y 3y2 3
3y 1
.
Hence
y
2 y2
3y2 y
x 2 xy y
y 2 y2 3y2 y
5 y2 5
3y 1 3y 1 3y 1
Method 4:
2 x 3xy 2 y
Let
k.
x 2 xy y
2x 3xy 2y k(x 2xy y)
2x 3xy 2y kx 2kxy ky
2y 2x ky kx 3xy 2kxy
(1)
(2 k )( y x) (3 2k ) xy
Since 1 1 3 , multiplying both sides by xy yields y x 3xy . (2)
x y
Dividing equation (2) by equation (1) gives us
3 2k
3
k .
(1) (2): 2 k
3
5
2 x 3xy 2 y 3
k= 2
.
x 2 xy y 2 5
Example 7: Find
2 x xy 3 y
if
x xy y
x y 3 y
x 5 y . Expanding both sides gives us
x
x 5 y . x and y are positive
numbers.
Solution: 2.
We are given that
x
x xy 3 xy 15y
Factoring gives us
x y 3 y
x 5 y
x 2 xy 15y 0.
x 3 y 0.
x 0, y 0 ,
Since x 3 y 0 because
and only if
x 5 y 0 x 25 y .
5
x 5 y
x 3 y can only equal 0 if
50 AMC Lectures
Substituting in 25y as x into
Chapter 1 Algebraic Manipulation
2 x xy 3 y
gives us
x xy y
2 x xy 3 y 2 25 y 5 y 3 y 58
= 2.
25 y 5 y y
29
x xy y
Example 8. Find the value of
3 2
x x 1 if the value of 3x 2 2 x 6 is 8.
2
Solution: 2.
Method 1:
Since 3x 2 2 x 6 = 8, 3x 2 2 x 2 .
3 2
1
1
x x 1 (3x 2 2 x) 1 2 1 2 .
2
2
2
Method 2:
We are given that 3x 2 2 x 6 = 8
or 3x2 2 x 2 4 .
(1)
Dividing both sides of (1) by 2 gives us the answer:
3 2
x x 1 2 .
2
Example 9. Calculate (2 1)(22 1)(24 1) (232 1) 1 .
Solution: 264 .
Notice that (2 – 1)(2 + 1) = 22 – 1.
We multiply the given expression by 1 = 2 – 1:
[(2 1)(2 1)](2 2 1)(2 4 1) (2 32 1) 1 [(2 2 1)(2 2 1)](2 4 1) (2 32 1) 1
[(2 8 1)(2 8 1)(216 1)...(2 32 1)] 1
(2 32 1)(2 32 1) 1 2 64 11 2 64.
Example 10. (1963 AMC) The expression x 2 y 2 z 2 2 yz x y z has:
(A) no linear factor with integer coefficients and integer exponents
(B) the factor x y z
(C) the factor x y z 1
(D) the factor x y z 1
(E) the factor x y z 1.
6
50 AMC Lectures
Chapter 1 Algebraic Manipulation
Solution: (E).
Method 1 (Official Solution):
x 2 y 2 z 2 2 yz x y z
= x 2 y 2 2 yz z 2 x y z x 2 y z x y z
= x y z x y z x y z = x y z x y z 1 .
2
Method 2:
Let x 2 y 2 z 2 2 yz x y z = 0
Rearrange the terms to give us the quadratic: y 2 (2 z 1) y z z 2 x 2 x 0 .
Use the quadratic formula to get the solutions:
(2 z 1) [(2 z 1)]2 4 1 ( z z 2 x 2 x) (2 z 1) (2 x 1) 2
2
2
2z 1 2x 1
2
2z 1 2x 1
2z 1 2x 1
y1
z x 1 and y2
z x.
2
2
Therefore, we have
x 2 y 2 z 2 2 yz x y z = (y (z x 1))(y (z x)) x y z x y z 1 .
y1, 2
2
2
1
1
Example 11: Both a and b are real numbers with m3 3 a m b 0 .
m
m
2
Prove: b b 3 a .
Solution:
Since the square of a number is always greater than or equal to 0, in order for the sum of
1
1
two squares to be 0, m 3 3 a 0 and m b 0 .
m
m
1
1
m3 3 a and m b
m
m
2
1 1
1
1
1
1
3 m ( m 2 m 2 ) m m 3 b(b 2 3).
So a m 3
m m
m
m
m
m
7
50 AMC Lectures
Chapter 1 Algebraic Manipulation
Example 12. Find the greatest positive integer not exceeding
6
7 3 .
Solution: 7039.
Let x 7 3 and y 7 3 .
x y ( 7 3) ( 7 3) 2 7 .
xy ( 7 3)( 7 3) 7 3 4 .
x 2 y 2 ( x y)2 2 xy = 20
x6 y 6 ( x 2 y 2 )3 3( x 2 y 2 ) x 2 y 2 7040 .
( 7 3)6 ( 7 3)6 7040 .
( 7 3)6 7040 ( 7 3)6 .
7 3 1.
The greatest positive integer not exceeding 7 3 is 7040 – 1 = 7039.
6
We know that 0 7 3 1 , so 0
6
Olympiad 1978) Determine the largest real number z such
Example 13. (Canadian Math
that
x yz 5
xy yz xz 3
and x, y are also real.
Solution:
Since x y z 5 , we have x y 5 z .
Squaring both sides gives us ( x y)2 (5 z )2 .
Since xy yz xz 3 , we have xy 3 xz yz 3 z( x y) 3 z(5 z) .
( x y )2 ( x y )2 4 xy (5 z ) 2 4[3 z (5 z )]
3z 2 10 z 13 (13 3z )(1 z )
Since (x y)2 0, this means that
(13 3z)(1 z) 0 .
Solving the inequality gives us
8
50 AMC Lectures
1 z
Chapter 1 Algebraic Manipulation
13
.
3
The largest real number z is
13
1
when x y .
3
3
Example 14: (1988 AIME) Find a if a and b are integers such that x2 x 1 is a factor
of ax17 + bx16 + 1.
Solution:
Using the quadratic formula, we see that the quadratic equation x2 x 1 =0 has the
1 5
1 5
solutions: p
and q
.
2
2
We can observe that p + q = 1 and pq = 1.
Since x2 x 1 is a factor of ax17 + bx16 + 1, p and q are also the roots of ax17 + bx16 + 1
= 0.
Therefore we have ap17 + bp16 = 1
(1)
and aq17 + bq16 = 1
(2)
(1) × q16 ap17q16 + bp16q16 = –q16 ap( 1)16 + b( 1)16 =–q16
ap + b = q16
(3)
(2) × p16: aq + b = p16
(4)
(3) – (4): a(p q) = p16 q16 .
So
p16 q16 ( p 8 q 8 )( p 8 q 8 ) ( p 8 q 8 )( p 4 q 4 )( p 4 q 4 )
a
p q
p q
pq
( p 8 q 8 )( p 4 q 4 )( p 2 q 2 )( p 2 q 2 ) ( p 8 q 8 )( p 4 q 4 )( p 2 q 2 )( p q)( p q)
.
p q
p q
( p 8 q 8 )( p 4 q 4 )( p 2 q 2 )( p q)
Since (p + q)2 = p2 + q2 +2pq, we can calculate that:
9
50 AMC Lectures
Chapter 1 Algebraic Manipulation
p2 q2 (p q)2 2pq 1 2 3.
Similarly,
p 4 q 4 ( p 2 q 2 )2 2( pq)2 9 2 7
and
p8 q8 ( p 4 q 4 )2 2( pq)4 49 2 47 .
Therefore a 47 7 3 1 987 .
10
50 AMC Lectures
Chapter 1 Algebraic Manipulation
PROBLEMS
Problem 1: Find m 2
1
1
if m 4 .
2
m
m
2
1
1
Problem 2: (1954 AMC) If a 3 , then a 3 3 equals:
a
a
10 3
(A)
(B) 3 3
(C) 0
(D) 7 7
(E) 6 3 .
3
Problem 3: Find m6
1
1
if m 4 .
6
m
m
Problem 4: Find a quadratic equation that has two roots of m 2
if m
1
1
and m3 3
2
m
m
1
2.
m
Problem 5: If x
1
3 13
, find the value of x .
x
2
Problem 6: Find the value of xy + yz + zx if x + y + z = 0 and x2 + y2 + z2 = 1.
Problem 7: Find x 6 y 6 if x 5 5 and y 5 5 .
2
1
1
Problem 8: (1952 AMC) If r 3 , then r 3 3 equals
r
r
(A) 1
(B) 2
(C) 0
(D) 3
Problem 9: If x
1
3 13
, find the value of x 2 2 .
2
x
11
(E) 6
50 AMC Lectures
Problem 10: Find
Chapter 1 Algebraic Manipulation
4 x3 3x 2 y 4 y 3 x
if 3 .
2 x3 2 xy 2 2 y 3 y
Problem 11: (1975 AMC) If a b , a3 b3 19x3 and a b x , which of the following
conclusions is correct?
(A) a 3x
(E) a 2 x
(B) a 3x or a 2 x
(C) a 3x or a 2 x (D) a 3x or a 2 x
Problem 12: (2012 AMC 10 A #24) Let a, b, and c be positive integers with a b c
such that a 2 b2 c2 ab 2011 and a 2 3b2 3c2 3ab 2ac 2bc 1997 . What is
a?
(A) 249
(B) 250
(C) 251
(D) 252
(E) 253
Problem 13: (1990 AIME) Find the value of (52 6 43 )3 / 2 (52 6 43)3 / 2 .
Problem 14: Find the greatest positive integer not exceeding
12
7 5 .
6
50 AMC Lectures
Chapter 1 Algebraic Manipulation
SOLUTIONS TO PROBLEMS
Problem 1: Solution: 14.
Method 1:
Since ( x + y)2 = x2 + 2xy +y2, we have:
x2 + y2 = ( x + y)2 – 2xy
(1)
2
Let x= m and y
1
1
1
1
in (1): m2 2 m 2m 42 2 14 .
m
m
m
m
Method 2:
1
1
1
1
m2 2 m2 2 2 2 m2 2 m 2 2
m
m
m
m
2
[m 2 2 m
1
1
1
( ) 2 ] 2 m 2 42 2 14.
m m
m
Problem 2. Solution: 0.
3
a3
1
1
1
a 3 a 3 3 3 3 0 .
3
a
a
a
Problem 3:
Solution: 2702.
2
We know that m2
m6
1
1
m 2 42 2 14.
2
m
m
2
1 2 1 4 1
2 1 2 1
m
m
1
=
m
m
3 14 142 3 2702
6
2
4
2
2
m
m
m
m
m
Problem 4:
Solution: x 2 4 x 4 0 .
2
1
1
m 2 m 2 4 2 2
m
m
2
13
50 AMC Lectures
m3
Chapter 1 Algebraic Manipulation
2
1
1
1
m
m
3 2
3
m
m
m
The two roots of the quadratic equation are 2 and 2. Therefore, the equation is
(x 2)(x 2) x 2 4x 4 0.
Or, we can obtain the solution this way:
2 1 3 1
2 1 3 1
m 2 m 3 4 , and m 2 m 3 4 .
m
m
m
m
The sum of the roots of the equation is 4, and the products of the roots is 4, so the
equation is: x 2 4 x 4 0 .
Problem 5. Solution: 3.
3 13
Since x
2
4 x 2 12 x 4 0
2 x 3 13
Dividing both sides of (1) by x:
x 2 3x 1 0
1
x 3 0
x
(1)
x
1
3.
x
1
.
2
2
We know that x y z x 2 y 2 z 2 2 xy 2 xz 2 yz .
We are given that x + y + z = 0 and x2 + y2 + z2 =1, so
1
0 = 1 + 2(xy + yz + zx)
xy + yz + zx = .
2
Problem 6. Solution:
Problem 7. Solution: 400.
Since x 5 5 and y 5 5 , x 2 y 2 (5 5 )(5 5 ) 25 5 20 .
x 2 y 2 (5 5 ) (5 5 ) 10
x 6 y 6 (x 2 y 2 )3 3(x 2 y 2 ) x 2 y 2 103 3 10 20 400
(If you are confused about the equation above, expand out
(x 2 y 2 )3 3(x 2 y 2 ) x 2 y 2 )
Problem 8: Solution: 0.
2
We know that r 3
1
1
1
1
1
r r 2 1 2 , and r r 2 2 2 3 .
3
r
r
r
r
r
14
50 AMC Lectures
Chapter 1 Algebraic Manipulation
The second equation gives us r 2
Therefore, r 3
1
1
3 (r )(0) 0.
r
r
11.
Problem 9: Solution:
1
1
2
2 1 r 1 2 0 .
r
r
1
1 1
1
x 2 2 x 2 2 x 2 2 ( x ) 2 2 32 2 11
x x
x
x
131
Problem 10: Solution:
.
46
x
Since we are given that 3, x = 3y. Substituting this into the given expression gives
y
us:
4 x3 3x 2 y 4 y 3 4 (3 y)3 3 (3 y)2 y 4 y 3 108 y 3 27 y 3 4 y 3 131
.
2 x3 2 xy 2 2y 3 2 (3 y)3 2 (3 y ) y 2 2 y 3
54 y 3 6 y 3 2 y 3
46
Problem 11. Solution: B.
We are given that a b , a3 b3 19x3 and a b x , therefore,
a3 b3 a b a 2 ab b2 x a 2 ab b2 19x3
Dividing the last equality above by x and substituting b a x , we obtain
a 2 a(a x) (a x) 2 19x 2
a 2 a 2 ax a 2 x 2 2ax 19x 2
3a 2 x 2 3ax 19x 2
18x 2 3ax 3a 2 0 .
Factoring this quadratic yields
3a 3x a 2 x 0 , so a 3x or a 2 x , and (B) is correct.
Problem 12: Solution: (E).
Adding the two equations gives
2a 2 2b2 2c2 2ab 2bc 2ac 14 .
This equation can be rearranged to form:
15
50 AMC Lectures
Chapter 1 Algebraic Manipulation
(a 2 2ab b2 ) (b2 2bc c 2 ) (a 2 2ac c 2 ) 14 .
Or
a b2 b c2 a c2 14 .
The only way to express 14 as the sum of three squares of positive integers is
14 32 22 12 .
Because a b c , it follows that a – c = 3 and either a – b = 2 and b – c = 1, or a – b = 1
and b – c = 2. Therefore either (a, b, c) = (c + 3, c +1, c) or (a, b, c,) = (c + 3, c + 2, c).
Substituting the relationships of the variables in the first case into the given equation
a2 b2 c 2 ab 2011 yields
2011 a 2 c 2 ab b2 (a c)(a c) (a b)b 3(2c 3) 2(c 1).
Solving gives us (a, b, c) (253,251,250) . The second case will not yield an integer
solution. Therefore a = 253.
Problem 13. Solution: 828.
Method 1:
Let a (52 6 43)1 / 2 [( 43)2 2 43 3 32 ]1 / 2 43 3 and
b (52 6 43)1 / 2 43 3 .
We would like to find the value of
(52 6 43) 3 / 2 (52 6 43) 3 / 2 .
Note that a 3 b 3 (a b)(a 2 ab b 2 )
(a b)(a2 2ab b2 2ab ab) (a b)[(a b)2 3ab] .
(52 6 43)3 / 2 (52 6 43)3 / 2 6[62 3( 43 3)(( 43 3) 6[36 3(43 9)] 828.
Method 2:
Let a (52 6 43 )1 / 2 and b (52 6 43 )1 / 2 .
We wish to find a3 b3 (a b)(a2 ab b2 ) .
We can easily calculate the values of the expressions of a2 + b2 and ab:
a 2 b2 (52 6 43 ) (52 6 43) 104 .
ab [(52 6 43)]1 / 2[(52 6 43)]1 / 2 (522 62 43)1 / 2 11561 / 2 34 .
16
50 AMC Lectures
Chapter 1 Algebraic Manipulation
Now we wish to find the value of a – b.
We know that (a b)2 a 2 2ab b2 (a 2 b2 ) 2ab 104 2 34 36 a b 6 .
Therefore a3 b3 (a b)(a2 ab b2 ) 6 (104 34) 828 .
Problem 14. Solution: 13535.
Let x 7 5 and y 7 5 .
x y ( 7 5) ( 7 5) 2 7 .
xy ( 7 5 )( 7 5 ) 7 5 2 .
x 2 y 2 ( x y)2 2 xy = 24
x6 y 6 ( x2 y 2 )3 3( x2 y 2 ) x2 y 2 13536 .
( 7 5 )6 ( 7 5 )6 13536 .
( 7 5 )6 13536 ( 7 5 )6 .
7 5 1.
The greatest positive integer not exceeding 7 5 is 13536 – 1 = 13535.
6
We know that 0 7 5 1. So 0
6
17
50 AMC Lectures
Chapter 2 Radicals
BASIC KNOWLEDGE
1. Definition
n
a is a radical. The number a
is called the radicand and n is a positive integer, called the index of the radical n a .
The symbol
When n = 2,
instead of 2
n
2
is called a radical sign. The expression
is called the square root. It is customary to use the notation
for the square root.
Here is an geometric example that employs the use of a radical.
The length of a diagonal of the square whose side is 1 is 2 .
2
1. 4142135623 7309504880 1688724209 6980785696 7187537695
2. Radical In Simplest Form
A radical expression is an expression that contains radical signs.
The expression is in simplest form when the following three conditions have been met:
• No radicands have perfect square factors other than 1.
• No fractions are inside the radical.
• No radicals appear in a denominator.
Example 1: Simplify (1) 3 1458 , (2)
512 , and (3)
Solution:
(1) 3 1458 3 2 36 93 2 .
18
27
.
8
50 AMC Lectures
Chapter 2 Radicals
(2)
512 2 28 16 2 .
(3)
27
8
3 32 3 3 3 3 2 3
6.
2 22 2 2 2 2 2 4
3. Radical Notation
The exponent
For example,
1
and the radical sign
n
n
are both used to indicate the
1
th root.
n
1
n
xx .
n
4. Radical Operations
a , (a 0)
(1) m a + n a = (m + n)
(2) ( a +
b )( a –
b ) = a – b, (a 0, b 0)
(3) x n a y n a = (x y)
n
a , (a 0 if n is even)
(4) The Product Property: ab = a ∙
even).
(5) The Quotient Property
(6) (n a ) m =
(7)
m n
n
a
=
b
a
and
b
b and
n
a m , (a 0).
a mn a , (a 0).
m and n are positive integers, m , n 2.
19
a
=
b
n
ab =
n
n
n
a ∙
n
b , (a 0, b 0 if n is
a
, (a 0, b > 0 if n is even).
b
50 AMC Lectures
Chapter 2 Radicals
3
Example 2: (1998 AMC) If N > 1, then
1
(A) N 27
N3 N3 N =
1
1
13
(B) N 9
(C) N 3
(D) N 27
(E) N
Solution: (D).
Method 1:
3
3
3
N N N =
3
3
N NN
1
3
=
3
3
N N
4
3
=
3
NN
4
9
=
3
N
13
9
13
27
= N .
Method 2:
1 1 1
3
1
1
1
1
1
1
13
N 3 N 3 N = ( N ( N ( N ) 3 ) 3 ) 3 = ( N ( N 3 N 9 )) 3 = N 3 ∙ N 9 ∙ N 27 = N 27
5. Nested Radical Simplification
A nested radical is a radical expression that contains another radical expression.
Some nested radicals can be rewritten in a form that is not nested. Rewriting a nested
radical in this way is called denesting.
a b can be denested if and only if
Radical Theorem 1:
a > 0, b > 0
and
a2 – b = k2
(5.1)
(k > 0)
(5.2)
There are three commonly used ways to denest a nested radical:
Skill 1: Completing the squares
Case I: If
a2 b =
x
y , (x > 0, y > 0, x > y), then a 2 b = x + y 2 xy .
Case II: If we can find x, y (x > y), such that x + y = a, xy = b, then
a2 b =
x
y.
20
50 AMC Lectures
Chapter 2 Radicals
Example 3: Simplify 4 2 3 .
Solution:
We rewrite the given radical as 4 2 3 = 4 12 .
Let a = 4, b = 12.
a2 – b = 42 – 12 = 4 = 22.
(5.1) and (5.2) are both satisfied, therefore the given nested radical can be denested.
42 3 =
(3 1) 2 3 1 =
( 3 1) 2 =
3 1 .
Skill 2: Underdetermined coefficients
Example 4: Simplify 11 2 18 .
Solution:
We can rewrite the given radical as 11 2 18 = 11 72 .
We see that a = 11, b = 72, and a2 – b = 112 – 72 = 49 = 72. (5.1) and (5.2) are both
satisfied. Therefore the given nested radical can be denested.
Let 11 2 18 =
x + y.
Squaring both sides: 11 + 2 18 = x + y + 2 xy .
x y 11
Solving we get
xy 18.
11 2 18 =
9 +
x 2
y 9
2 =3+
or
2
Skill 3: Use the formula:
a b =
or
2
( a a2 b
2
a b =
1
4a 4 b
=
4
2
a a2 b )
4a 2 4b
21
x 9
.
y 2
50 AMC Lectures
or
a b =
Chapter 2 Radicals
a a2 b
2
a a2 b
.
2
Example 5: Prove that a b =
a a2 b
2
a a2 b
if a > 0, b > 0, and a2
2
– b > 0.
Solution:
Let
a b =
x +
y.
Squaring both sides: a + b = x + y + 2 xy .
b
Then we get x + y = a, and xy = .
4
Solving the system of equations, we have:
a a2 b
a a2 b
x=
and y =
, where (a > 0, b > 0, a2 – b > 0 ).
2
2
a b =
a a2 b
+
2
We can prove that
a b =
a a2 b
2
a a2 b
–
2
a a2 b
in a similar fashion.
2
Example 6: Simplify 2 3 .
Solution:
Method 1:
2 3 =
1
8 2 12
=
4
2
8 2 12 =
2 22 3
–
2
2 22 3
=
2
6 2
.
2
Method 2:
2 3 =
3
–
2
22
1
=
2
6 2
.
2
50 AMC Lectures
Chapter 2 Radicals
6. Infinitely Nested Radicals
Assume that
Let L =
Then L =
a b a b a is a real number when a 0 and b 0.
a b a b a .
a bL
L2 – bL – a = 0 .
b b 2 4a
By the quadratic formula, L =
.
2
Therefore
a b a b a =
b b 2 4a
2
Setting a = 0 in (1), we have
(6.1)
b b b = b.
Setting a = b = 1 in (1), we get 1 1 1 =
ratio .
Example 7: Find the positive integer x in x =
1 5
, which is equal to the golden
2
2 2 2 2 .
Solution:
Method 1:
1 12 4(2) 1 3
2.
Setting a = 2 and b = 1 in (1), we have x
2
2
Method 2:
Squaring both sides of (1):
x 2 2 2 2 2 ...
x2 2 x
Using the quadratic formula we get
23
x2 x 2 0 .
50 AMC Lectures
Chapter 2 Radicals
1 (1) 2 4(1)(2) 1 9 1 3
x1
2 and
2
2
2
1 (1) 2 4(1)(2) 1 9 1 3
x2
1 (extraneous) .
2
2
2
Generally, if n > 0, then:
n n n n =
1
(1 1 4n ) .
2
n n n n =
1
(1 1 4n ) .
2
7. Conjugates
Sometimes, conjugates are used to simplify radical expressions.
x + y and x – y are conjugates.
Some other useful conjugates:
p q + r s and p q – r s
m
a n and
a
m
a mn
b and a b
m a n b and m a n b
3
a 3 b and
3
a 2 3 ab 3 b 2 .
Example 8: (1950 AMC) After rationalizing the numerator of
denominator in simplest form is:
(A) 3 ( 3 + 2 ) (B) 3 ( 3 – 2 )
(D) 3 + 6
(E) none of these answers
(C)3 –
24
3
2
3 2
, the
3
50 AMC Lectures
Chapter 2 Radicals
Solution: (D).
The conjugate of
3 2
∙
3
3 2 is
3 2 . Multiplying the fraction by
3 2
, or 1, gives:
3 2
3 2
1
=
.
3 6
3 2
Radical Theorem 2: If and only if x = a and y =b, x + y ∙ m = a + b ∙ m .
a, b, c, and d are rational numbers. m 0 and is not a square number.
If and only if x = a and y =b, x + y ∙
numbers and
n
n
m = a + b ∙ n m . a, b, c, d, and m are rational
m is an irrational number. m 0 if n is even.
Example 9: Find the ordered pair of positive integers (a, b), with a < b, for which
1 21 12 3 a b .
Solution:
Squaring both sides of the given equation:
1 21 12 3 a b 2 ab
21 12 3 (a b 1) 2 ab .
We can write
21 12 3 as
21 3 122 21 432 .
212 – 432 = 1. Since 1 is a square number, so
21 12 3 21 2 36 3 21 2 108
12 2 12 9 9 ( 12 9 ) 2 12 9 3 2 3
Therefore (a b 1) 2 ab = 3 2 3 .
We have a b 1 3 and ab 3 .
Since a and b are positive integers and a < b, the only possible values are a = 1 and b = 3.
25
50 AMC Lectures
Chapter 2 Radicals
8. Comparing Two Radicals
Radical Theorem 3:
For a > 0, b > 0,
n
n
a =
a <
n
n
n
a >
b a > b;
n
b a = b;
b a < b.
Radical Theorem 4:
For a > 0, b > 0, c > 0, a > b > c >
n
a >
n
b >
n
c.
Example 10: Which one is larger 15 + 2 2 or 14 + 3?
Solution:
Let x = 15 + 2 2 and y = 14 + 3.
x2 = 23 + 4 30 = 23 + 480 .
y2 = 23 + 6 14 = 23 +
504 .
We see that x2 < y2.
Since x > 0 and y > 0, we can conclude that x < y.
Therefore 15 + 2 2 < 14 + 3.
Radical Theorem 5:
am +
amn >
Proof:
We know that a 0, m > 0, n 0.
In order to prove a m + a m n >
( am +
2
amn ) > ( a +
a +
a 2m n if a 0, m > 0, n 0.
a + a 2m n , we only need to prove:
a 2m n )2
Expanding and simplifying, we get
m2 + mn > 0.
This is true under the given conditions.
26
50 AMC Lectures
Apparently,
a2 +
amn >
am +
Therefore
If we let
Chapter 2 Radicals
a <
a1 <
a3 >
amn <
am
a2
a1 +
a 2m n .
a +
a3 <
a 2m n .
a4 , the above theorem becomes
a4 .
If we define a2 a1 = m to be the distance between a1 and a2, the theorem can be
described as follows:
Theorem 5.1: For nonnegative numbers a1, a2, a3, a4 with a1 < a2 a3 < a4, if the
distance of a1 and a2 is the same as the distance of a3 and a4, then a2 + a3 > a1 +
a4 .
Theorem 5.2: If a > 0, n = m > 0 and a1, a2, a3, a4 form an arithmetic sequence, then
a2 + a3 > a1 + a4 .
Example 11: Prove
2 +
3 >1+
4.
Solution:
Let a1 = 1, a2 = 2, a3 = 3, a4 = 4 in theorem 5.2, and we have our conclusion immediately.
Theorem 5.3: For a > 0, m > 0, n = 0, a2 = a3, 2 a2 >
Example 12: Prove
Solution:
2 +
a4 .
3 < 10 .
The original question can be written as 2
Let a1 = 2, a2 = a3 =
a1 +
10
>
4
2 +
3.
10
, a4 = 3. From theorem 5.3, we get our conclusion.
4
Theorem 5.4: When a = 0, m > 0, n > 0 that is, when a1 = 0,
Example 13: Prove that
2 +
6 >
8.
27
a2 +
a3 >
a4 .
50 AMC Lectures
Chapter 2 Radicals
Solution:
Let a1 = 0, a2 = 2, a3 = 6, a4 = 8, we get the result from Theorem 5.4.
9. Finding The Values Involving Radicals
3
2
3
2
Example 14: (1990 AIME) Find the value of (52 6 43 ) – (52 6 43 ) .
Solution:
Method 1:
We split 52 into two parts to obtain squares in each set of parentheses:
3
3
3
3
(52 6 43 ) 2 – (52 6 43 ) 2 = (43 6 43 9) 2 – (43 6 43 9) 2
3
2 2
3
2 2
= [( 43 3) ] – [( 43 3) ] = ( 43 + 3)3 – ( 43 – 3)3
= (43 43 + 3 ∙ 3 ∙ 43 + 3 ∙ 32 43 + 33) – (43 43 – 3 ∙ 3 ∙ 43 + 3 ∙ 32 43 – 33)
= 828.
Method 2:
1
2
1
2
Let = (52 6 43 ) and = (52 6 43 ) . We wish to find
3 – 3 = (– )(2 + + 2).
1
1
Now 2 + 2 = 104 and = (522 36 43) 2 = (1156) 2 = 34.
Thus ( – )2 = 2 – 2 + 2 = 104 – 68 = 36, so – = 6 and 3 – 3 = 6(104 + 34) =
828.
10. Six More Skills For Radical Simplifications
Skill 4: Using squaring and square rooting
Example 15: (1983 ARML) Compute the numerical value of
Solution:
Let y 4 7 4 7 .
28
4 7 4 7 .
50 AMC Lectures
Chapter 2 Radicals
Squaring both sides:
y 2 (4 7 ) (4 7 ) 2( 4 7 )( 4 7 ) 8 2 9 8 6 2 .
Therefore
4 7 4 7 2 .
Skill 5: Solving an equation
Example 16: Find x6 2 2 x5 3x 4 x3 2 5x 2 4 x 5 if x =
5 +
2.
Solution:
x2 – 2 2 x – 3 = 0
x2 – 2 5 x + 3 = 0
x4(x2 – 2 2 x – 3) – x(x2 – 2 5 x + 3) – x +
Example 17: Simplify
3
20 14 2 +
3
20 14 2 .
3
5 =0–0 – ( 5 +
2)+
5 = – 2.
20 14 2 .
Solution:
Let x =
3
20 14 2 +
Cubing both sides:
3
2
2
3 20 14 2 +3 3 20 14 2 × (3 20 14 2 ) + 3 (3 20 14 2 ) × 3 20 14 2 +
3
3 20 14 2 = x3.
Or 20 14 2 +3 (3 202 (14 2 ) 2 × (3 20 14 2 +
Or 40 + 3×
3
8 × (x) = x3
3
x3 – 6x – 40 = 0
20 14 2 ) + 20 14 2 = x3
(x – 4)( x2 + 4x + 10) = 0.
Since x2 + 4x + 10 = (x + 2)2 + 8 > 0, x = 4.
Skill 6: Determining the range of variables
Example 18: Simplify
x2 6x 9 +
x2 2x 1 –
Solution:
29
x2 4x 4 .
50 AMC Lectures
x2 6x 9 +
Chapter 2 Radicals
x2 2x 1 –
x2 4x 4 =
( x 3) 2 +
( x 1) 2 –
( x 2) 2 =
x 3 + x 1 – x 2 .
Let x + 3 = 0, x – 1 = 0, x – 2 = 0, x = – 3, x = 1, x = 2.
When x < – 3, we have x 3 + x 1 – x 2 = –( x + 3) – (x – 1 ) + (x – 2 ) = – x – 4
When – 3 x < 1, we have x 3 + x 1 – x 2 = ( x + 3) – (x – 1 ) + (x – 2 ) = x + 2
When 1 x 2, we have x 3 + x 1 – x 2 = ( x + 3) + (x – 1 ) + (x – 2 ) = 3x .
When x > 2, we have x 3 + x 1 – x 2 = ( x + 3) + (x – 1 ) – (x – 2 ) = x + 4.
Skill 7: Using the proportion formula
Example 19: Show
a c
if
b d
a b
=
a b
c d
.
c d
Solution:
a b
=
a b
a
b
c d
( a b) ( a b)
( c d)( c d)
=
=
( c d)( c d)
( a b) ( a b)
c d
c
.
d
Squaring both sides:
a c
.
b d
Skill 8: Rationalizing the denominator or numerator
Example 20: (1989 AMC) The set of all real numbers x for which
1
x + x2 1
x x2 1
is a rational number is the set of all
(A) integers x
(B) rational x
(D) x for which
x 1 is rational
2
(C) real x
(E) x for which x +
30
x 2 1 is rational
50 AMC Lectures
Chapter 2 Radicals
Solution: (B).
Rationalize the denominator of the third term in the given expression:
1
x x2 1
∙
x x2 1
x x2 1
=
x2 1 x
.
1
x 2 1 and x 2 1 – x are reciprocals. Hence the given expression
1
x2 1
equals
x x2 1
Thus x +
x+
x + x 2 1 – ( x 2 1 – x) = 2x
which is rational if and only if x is rational. In order to determine that this is the only
answer, we must show that each of the other sets is different from the set of rational
numbers:
(A) and (C) are not in the set of rational numbers.
The rational number x = 1 is not in the set described by (D) or (E) since neither 2 nor 1
+ 2 is rational.
Skill 9: Introducing new parameters
Example 21: Show that
1
1
1
+
+ = 1.
x
z
y
3
ax 2 by 2 cz 2 =
3
a +
3
b +
3
c if ax3 = by3 = cz3 and
Solution:
Let ax3 = by3 = cz3 = t.
t
t
t
Then ax2 = , by2 = , cz2 = .
x
z
y
t t t
1 1 1
= 3 ( )t .
x y z
x y z
1
1
1
We know that +
+ = 1.
x
z
y
Therefore
1
1
1
1
1
1
3
ax 2 by 2 cz 2 = ( +
+ ) 3 t = 3 ax 3 + 3 by 3 + 3 cz 3
x
z
x
z
y
y
3
=
ax 2 by 2 cz 2 =
3
a +
3
b +
3
3
c.
31
50 AMC Lectures
Chapter 2 Radicals
PROBLEMS
Problem 1: (1988 AMC)
(A)
26
(B) 2( 2 +
8 + 18 =
3)
(C) 7
Problem 2: (1979 AMC) The product of
(A) 7 12
(B)2 7 12
(C)
7
(E) 2 13
(D) 5 2
3
4 and
4
(D)
12
32
8 equals
32
(E)2 12 32
Problem 3: Simplify 9 2 14 .
Problem 4: Simplify 9 2 23 6 10 4 3 2 2 .
1
,
5
Problem 5: Compare
8
Problem 6: Compare
Problem 7: Prove:
Problem 9: Prove that
1
, and
11
8! with
5 +
Problem 8: Compare
3
9
8 >
5 +
5 +
6
1
.
123
9! .
2 + 11 .
6 and 3 +
20 >
8.
23 .
2 +
Problem 10: Find x3 x 2 y xy 2 y 3 if x =
Problem 11: Find (a – b +
5 3
and y =
5 3
4ab
1
5
4ab
)(a + b –
) if a =
and b = – .
ab
2
2
a b
Problem 12: Find the integer part of (1 + 2 )4.
3 2 2 – 3 2 2 is equal to
(C)4 2
(D) 6
(E) 2 2
Problem 13: (1970 AMC)
(A) 2
(B) 2 3
5 3
.
5 3
32
50 AMC Lectures
Chapter 2 Radicals
Problem 14: (1973 AMC) The fraction
(A)
2 2
3
(B) 1
(C)
Problem 15: Simplify S =
3
2 3
3
2 5 +
2( 2 6 )
3 2 3
(D)
3
4
3
is equal to
(E)
16
9
2 5 .
Problem 16: Simplify:
x 2 6 x 9 + x 2 4 x 4. if – 3 < x < 2.
Problem 17: Simplify
x 2 6 x 9 + 16 8 x x 2 .
Problem 18: Find x 2 y 2 2 xy 4 x 4 y 4 + 1 2 x x 2 –
if 0 < x < 1 < y < 2.
Problem 19: Find 3x 2 5xy 3 y 2 if x =
Problem 20: Show that
3
a
1
and y =
3 2
a 1 8a 1
+
3
3
3
a
Problem 21: (1986 AIME) Evaluate the product
( 5 + 6 + 7 )( 5 + 6 – 7 )( 5 – 6 +
y2 4 y 4
1
.
3 2
1
a 1 8a 1
= 1 if a .
8
3
3
7 )(–
5 +
6 +
7 ).
Problem 22: (2006 AIME 1) The number 104 6 468 10 144 15 2006 can be
written as a 2 + b 3 + c 5 , where a, b, and c are positive integers. Find abc.
33
50 AMC Lectures
Chapter 2 Radicals
SOLUTIONS
Problem 1: Solution: (D).
8 + 18 = 4 2 + 9 2 = 2 2 + 3 2 = 5 2 .
Problem 2: Solution: (E).
3
4
4
1
2 3
2
3
1
3 4
8 = (2 ) (2 )
3
4
= 2 2 =2
17
12
1(
= 2
5
)
12
=2∙ 2
5
12
= 2 12 32 .
7 2.
Problem 3: Solution:
We rewrite the given radical as 9 2 14 = 9 56 .
We see that a = 9, b = 72, and a2 – b = 92 – 56 = 25 = 52. Therefore the given nested
radical can be denested.
9 2 14 =
(7 2) 2 7 2 =
Problem 4: Solution:
( 7 2 )2 =
7 2.
2 + 1.
9 2 23 6 10 4 3 2 2 =
=
9 2 23 6 6 4 2 =
=
9 2 11 6 2 =
Problem 5: Solution:
3
1
=
5
6
1
.
125
1
=
11
6
1
.
121
9 2 23 6 10 4( 2 1)
9 2 23 6(2 2 )
9 2(3 2 ) =
3
1
>
11
6
1
>
123
3 2 2 =
1
.
5
34
2 +1
50 AMC Lectures
6
1
>
121
6
3
1
>
11
1
>
123
6
1
>
123
Chapter 2 Radicals
6
1
, so
125
1
.
5
Problem 6: Solution:
Let x =
8
8! and y =
9
8
8! <
9
9! .
9! .
x 72 (8!)9 = (8!)8 ∙ 8! < (8!)8 ∙ 98.
y 72 (9!)8 = (8!)8 ∙ 98.
So x 72 < y 72 .
Since x > 0, and y > 0, we conclude that x < y
8
8! <
9
9! .
Problem 7: Solution:
Let a1 = 2, a2 = 5, a3 = 8, a4 = 11. Substitute these values into Theorem 5.1, which states
that for nonnegative numbers a1, a2, a3, a4 with a1 < a2 a3 < a4, if the distance of a1
and a2 is the same as the distance of a3 and a4, then a2 + a3 > a1 + a4 , and we
get our answer.
Problem 8: Solution:
5 +
6 =
3 +
8 =
( 5 6 ) 2 = 11 2 30 .
( 3 8 ) 2 = 11 2 24 .
Since (11 + 2 30 ) – (11 + 2 24 ) = 2( 30 –
5 + 6 > 3 + 8.
24 ) > 0,
Problem 9: Solution:
Let a1 = 2, a2 = 5, a3 = 20, a4 = 23 in theorem 5.1, and we have our conclusion
immediately.
35
50 AMC Lectures
Chapter 2 Radicals
Problem 10: Solution: 496.
x3 x 2 y xy 2 y 3 = (x + y)(x2 + y2) = (x + y)[(x + y)2 – 2xy]
5 3
5 3
= 4 + 15 and y =
= 4 – 15
5 3
5 3
Therefore x + y = 8 and xy = 1.
x3 x 2 y xy 2 y 3 = 8 ∙ (82 – 2) = 496.
x=
Problem 11: Solution: 1.
4ab
4ab
)(a + b –
)
ab
a b
( a b) 2 ( a b) 2
=
∙
= (a + b)(a – b) = a2 – b2.
a b
ab
1
5
Substituting a =
, b = – into the expression above, we get:
2
2
4ab
5
1
4ab
(a – b +
)(a + b –
) = – = 1.
4
ab
4
a b
(a – b +
Problem 12: Solution: 33.
Method 1:
(1 + 2 )4 = [(1 + 2 )2]2 = (3 + 2 2 )2 = 17 + 12 2 .
We know that 2 1.4142
1.41 < 2 < 1.415.
So 16.92 < 12 2 < 16.98
33.92 < 17 + 12 2 < 33.98.
Therefore the integer part of (1 + 2 )4 is 33.
Method 2:
(1 + 2 )4 = [(1 + 2 )2]2 = (3 + 2 2 )2.
We also know that
(3 + 2 2 )2 + (3 – 2 2 )2
= 9 + 12 2 + 8 + 9 – 12 2 + 8 = 34.
Since 0 < 3 – 2 2 < 1 and 0 < (3 – 2 2 )2 < 1, we can conclude that the integer part of
(1 + 2 )4 is 33.
36
50 AMC Lectures
Chapter 2 Radicals
Problem 13: Solution: (A).
Denote the difference by d, where d is positive, and
d 2 = ( 3 2 2 )2 –2 3 2 2
3 2 2 + ( 3 2 2 )2
= 3 + 2 2 – 2 32 (2 2 ) 2 + 3 – 2 2
= 6 – 2 9 8 = 6 – 2 = 4.
Therefore d =
d2 =
4 = 2.
Problem 14: Solution: (D).
The square of the given fraction is
4(2 2 12 6)
4(8 4 3 ) 16(2 3 ) 16
=
=
=
.
9
9(2 3 )
9(2 3 )
9(2 3 )
16 4
.
9
3
(See another solution in “Chapter 16 Completing The Square”).
Hence the fraction is equal to
Problem 15: Solution: 1.
Let 3 2 5 = a, 3 2 5 = b.
We get S = a + b.
Cubing both sides: S3 = (a + b)3 = a3 + b3 + 3ab(a + b) = 4 + 3 ( – 1 )S = 4 – 3S.
S3 + 3S – 4 = 0, (S – 1)( S2 + S + 4) = 0.
Since the quadratic S2 + S + 4 = 0 has no real solutions, the only answer is S = 1.
Problem 16: Solution: 5.
x2 6x 9 + x2 4x 4 = x 3 + x 2 .
Since – 3 < x < 2, then x + 2 > 0 and x – 2 < 0.
Therefore, x 3 can be written as x + 3 and x 2 can be written as – (x – 2).
x + 3 – (x – 2) = x + 3 – x + 2 = 5.
Problem 17: Solution:
When x < 3, the original expression equals 7 – 2x.
37
50 AMC Lectures
Chapter 2 Radicals
When 3 x < 4, the original expression equals 1.
When x 4, the original expression equals 2x – 7.
Problem 18: Solution: 1.
x 2 y 2 2 xy 4 x 4 y 4 + 1 2 x x 2 –
=
( x y 2) 2 + (1 x) 2 –
y2 4x 4
( y 2) 2 = x y 2 + 1 x – y 2 .
Since 0 < x < 1 < y < 2, we know that x – y + 2 > 0, 1 – x > 0, and y – 2 < 0.
So, x y 2 becomes x – y + 2,1 x becomes 1 – x, and y 2 becomes 2 – y.
The answer is then (x – y + 2) + (1 – x) – (2 – y) = 1.
Problem 19: Solution: 25.
x=
3 2
1
=
=
( 3 2 )( 3 2 )
3 2
3 +
2
y=
3 2
1
=
=
( 3 2 )( 3 2 )
3 2
3 –
2
x + y = 2 3 and xy = ( 3 )2 – ( 2 )2 = 1.
3x 2 5xy 3 y 2 = 3(x2 + y2) – 5xy = 3(x + y)2 – 11xy .
Substituting in the values that we calculated for x + y and xy, we get
3(x + y)2 – 11xy = 3(2 3 )2 – 11 = 25.
Problem 20: Solution: 1.
Let
8a 1
= t.
3
We know that a
1
. Hence t is real.
8
Solving for a in terms of t, we get a =
3t 2 1
.
8
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50 AMC Lectures
Chapter 2 Radicals
8a 1 3 3t 2 1 (t 2 3)t
=
+
8
8
3
1 t
= 1. Therefore, t is 1.
2
3
3t 2 1 (t 2 3)t
=
8
8
3
(1 t )3
+
8
3
(1 t )3
1 t
=
+
8
2
Problem 21: Solution: 104.
( 5 +
6 + 7 )( 5 + 6 – 7 ) = (( 5 + 6 ) +
= ( 5 + 6 )2 – ( 7 )2 = 11 + 2 30 – 7 = 4 + 2 30 ,
( 5 +
6 + 7 )( 5 +
= – 4 + 2 30 ,
6 –
7 ) = ( 7 )2 – ( 5 +
7 )(( 5 +
6)–
7)
6 )2 = 7 – (11 – 2 30 )
The product then equals (4 + 2 30 )(– 4 + 2 30 ) = (2 30 )2 – 42 = 120 – 16 = 104.
Problem 22: Solution: 936.
Expand ( a 2 + b 3 + c 5 )2 to obtain 2a2 + 3b2 + 5c2 + 2ab 6 + 2ac 10 + 2bc 15 .
Setting this equal to 104 6 468 10 144 15 2006 , we can conclude that
2a2 + 3b2 + 5c2 = 2006,
2ab =104,
2ac = 468,
and 2bc = 144.
Therefore
ab = 52 = 22 ∙ 13,
ac = 234 = 2 ∙ 32 ∙ 13,
and bc = 72 = 23 ∙ 32.
Multiplying these three equations together, we get
a2b2c2 = ab ∙ ac ∙ bc = 26 ∙ 34 ∙ 132.
Taking the square root of both sides yields abc = 23 ∙ 32 ∙ 13 = 936.
abc 23 32 13
a can be written as a
= 13 and similarly b = 4 and c = 18.
bc
23 32
Previously we concluded that 2a2 + 3b2 + 5c2 = 2006.
It’s important to check this by substituting in the determined values of a, b, and c into the
equation 2a2 + 3b2 + 5c2. We do in fact get 2006.
abc = 13 4 18 = 936.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
BASIC KNOWLEDGE
In this lecture we are going to show you nine commonly used methods to solve radical
equations.
1. Solving the radical equations directly by observing the domain of x.
Example 1: Solve
x 1 + 2 = 0.
Solution:
Since x 1 ≥ 0, then x 1 + 2 ≥ 2.
No matter what value we choose for x, x 1 + 2 will not equal 0.
Therefore there is no solution to the given equation.
2. Solving the radical equations by substitution.
2.1 Introducing one variable
Example 2: Solve
x
+
1 x
1 x
13
=
.
x
6
Solution:
Let
x
= y. We have
1 x
Substituting
1
for
y
1 x
1
= .
x
y
1 x
and y for
x
x
into the given equation, we have
1 x
13
1
=
.
6
y
This can be simplified into the quadratic:
6y2 – 13y + 6 = 0.
y+
Solving for y: y1 =
2
3
; y2 = .
3
2
We have
40
50 AMC Lectures
x
2
=
1 x
3
Chapter 3 Solving Radical Equations
x1 =
4
.
13
x2 =
9
.
13
Or
x
3
=
1 x
2
4
9
and x2 =
back into the original equation, we can confidently
13
13
say that these are indeed the solutions.
Substituting in x1 =
Example 3: (AIME) What is the product of the real roots of the equation
x2 + 18x + 30 = 2 x 2 18x 45 ?
Solution:
Let y = x2 + 18x + 30.
The original equation can be written as y 2 y 15 .
Squaring both sides: y 2 4( y 15)
y 2 4 y 60 0 .
Solve for y: y1= 10 or y2 = 6 (extraneous root).
Therefore x2 + 18x + 30 = 10
x 2 18x 20 0 .
18 (18) 2 4 20 18 2 61
9 61 .
2
2
The product of the roots is x1 x2 20 .
x1, 2
Note that we will give two more solutions to this problem in Example 9.
2.2 Introducing two variables
Example 4: Solve
x2 5x 2 –
x 2 5x 5 = 1.
Solution:
41
50 AMC Lectures
Chapter 3 Solving Radical Equations
x2 5x 2 , v =
u–v=1
u2 – v2 = 3
(2) ÷ (1):
u+v=3
Let u =
x2 5x 5 .
(1)
(2)
(3)
From (1) and (3), we can solve for u and v to get u = 2 and v = 1, satisfying u ≥ 0, v ≥ 0.
x2 5x 2 = 2
x1 = 1, x2 = – 6.
We can check that both x1 = 1 and x2 = – 6 are indeed the solutions by plugging these
values back into the original equation.
Note (a): If we had used the value of v to find x, the results would have been the same.
(b): IMPORTANT! We needed to make sure that both u ≥ 0 and v ≥ 0.
Example 5: Solve
3
2 x +
Solution:
Let u = 3 2 x , and v =
We have:
u+v=1
u3 + v2 = 1
x 1 = 1.
x 1 .
(1)
(2)
From (1) and (2), we get u3 + u2 – 2u = 0.
The three solutions to this cubic equation are u1 = 0, u2 = 1, u3 = –2.
Plugging in these values into (1) to solve for v, we get
v1 = 1, v2 = 0 and v3 = 3. These three values satisfy v ≥ 0.
Since x = 2 – u3,
x1 = 2 – 0 = 2,
x2 = 2 – 1 = 1, and
x3 = 2 + 8 = 10.
We can check that these three values are indeed the solutions by plugging these values
back into the original equation.
Note: Here, we only needed to ensure that v ≥ 0.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
3. Solving radical equations by introducing a parameter by taking the average
Example 6: Solve
Solution:
The average value of
2 x 1 + 2 x 1 = 4.
2 x 1 and 2 x 1 is 2.
Therefore we can let
2 x 1 = 2 + t and 2 x 1 = 2 – t where (– 2 ≤ t ≤ 2).
Squaring these two equations gives us the following two equations:
2x + 1 = 4 + 4t + t2
4(x – 1) = 4 – 4t + t2
(1)
(2)
2 × (1) – (2):
6 = 4 + 12t + t2
t 6 38 .
Since – 2 ≤ t ≤ 2, the only solution for t is t 6 38 .
53
4 38 .
So 2 x 1 = 4 38
x
2
53
4 38 is indeed the solution by plugging it back into the
We can check that x
2
original equation.
Example 7: Solve
x
5
–
x
5
5
= x.
x
Solution:
The average value of
We can let
1
5
x = x +t
2
x
x
5
and –
x
5
5
1
is x .
x
2
(1)
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50 AMC Lectures
5
–
Chapter 3 Solving Radical Equations
5
1
= x –t
x
2
(2)
(1)2 – (2)2: x – 5 = 2xt
So
x
t=
1
5
(1 ) .
2
x
5
5
5
1
1
5
5
= x + (1 ) , or ( x ) – 2 x + 1 = 0 ( x – 1)2 = 0.
x
x
x
2
2
x
x
5
= 1 x2 – x – 5 = 0.
x
1
Solving for x: x (1 21) .
2
1
We can check that x (1 21) is the only solution by plugging it into our original
2
equation.
Thus, x
4. Solving the radical equations by factorization.
Example 8: Solve
x2 5x 6 +
3x 2 8x 5 = 3x – 3.
Solution:
We factor each radical:
( x 1)( x 6) +
( x 1)(3x 5) = 3(x – 1).
Since x2 + 5x – 6 ≥ 0, 3x2 – 8x + 5 ≥ 0, and 3x – 3 ≥ 0, we know that the domain of x is
5
x≥ .
3
Therefore
( x 1)( x 6) +
( x 1)(3x 5) = 3 ( x 1) 2 .
Taking out the like terms: we get
x 1 ( x 6 +
3x 5 3 x 1 ) = 0.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
We have:
x 1 = 0
or ( x 6 +
3x 5 3 x 1 ) = 0 x = 10 or x
x = 1.
22
5
(extraneous since x ).
13
3
We can check that both x = 1 and x = 10 are the solutions by plugging it into our original
equation.
Example 9: (AIME) What is the product of the real roots of the equation x2 + 18x +30 =
2 x 2 18x 45 ?
Solution:
Method 1:
Rewrite the original equation as
x2 + 18x + 45 – 2 x 2 18x 45 – 15 = 0.
Or ( x 2 18 x 45 + 3)( x 2 18 x 45 – 5) = 0.
We have
x 2 18 x 45 = 3 (extraneous) or
Squaring both sides of
x 2 18 x 45 = 5.
x 2 18 x 45 = 5:
x2 + 18x + 45 = 25 x2 + 18x + 20 = 0
Since the discriminant of the quadratic = 182 – 4 × 20 > 0, the two roots are real and by
Vieta’s Theorem, the product of two roots is then 20.
Method 2:
Rewrite the original equation as
x2 + 18x + 45 – 2 x 2 18x 45 + 1 = 16.
Or ( x 2 18x 45 1)2 = 16
Take the square root of both sides:
When
x 2 18x 45 1 = – 4,
x 2 18x 45 1 = ± 4.
x 2 18 x 45 = – 3 (extraneous).
When x 2 18x 45 1 = 4, we have
x 2 18 x 45 = 5.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
Squaring both sides of x 2 18 x 45 = 5:
x2 + 18x +45 = 25 x2 + 18x + 20 = 0.
Since = 182 – 4 × 20 > 0, by Vieta’s Theorem, the product of two real roots is then 20.
5. Solving the radical equations by the Property of Proportions
Property: If
a c
ab c d
, then
(a – b ≠ 0, and c – d ≠ 0).
b d
a b c d
Proof:
a c
We
know that , or ad = bc.
b d
Multiply both sides of (1) by 2: 2ad = 2bc
Re-write (2) as bc – ad = – bc + ad
(1)
(2)
(3)
Add ac – bd to both sides of (3):
ac + bc – ad – bd = ac – bc + ad – bd
(4)
Factor both sides of (4) into the following:
(a + b)(c – d) = (a – b)(c +d)
(5)
Since a – b ≠ 0 and c – d ≠ 0, we can divide both sides of the equation by (a – b)(c – d) to
get:
ab c d
. QED.
a b c d
Example 10: Solve
5 x x3
=
5 x x3
9 2x x 3
.
9 2x x 3
Solution:
By the property of proportion, we know that
5 x x3 5 x x3
=
5 x x3 5 x x3
9 2x x 3 9 2x x 3
9 2x x 3 9 2x x 3
Therefore
46
50 AMC Lectures
5 x
=
x3
5 x
We have
9 2x
x3
x 3
Chapter 3 Solving Radical Equations
x 3
x 3 0
9 2x
5 x
9 2x
x 3
5 x 9 2x
x 3 0.
5 x 9 2x 0 or x 3 0 .
Solving 5 x 9 2x 0 , we get x = 4.
Solving x 3 0 , we
get x = 3.
We substitute these two values into the original equation to find that only x = 3 is the
solution.
5 x x3
(When x = 4, the denominator of
is 0, which cannot be true.)
5 x x3
Note: When you get
5 x
=
x3
9 2x
, it is dangerous to simplify it into 5 x =
x3
9 2 x by multiplying both sides by x 3 , because we have to be aware of the
possibility of the denominator being 0.
6. Solving radical equations by using conjugates (rationalization)
Example 11: Solve
3x 1 –
x 4 = 7.
Solution:
Method 1:
We see that
3x 1 – x 4 = 7
and
(3x + 1) – (x + 4) = 2x – 3
(1)
2x 3
, or 7 3x 1 + 7 x 4 = 2x – 3
7
Simplifying (1) × 7 + (3), we get
7 3x 1 = x + 23.
(2) ÷ (1):
3x 1 +
x4 =
Squaring both sides of the above equation and factoring yields:
47
(2)
(3)
50 AMC Lectures
Chapter 3 Solving Radical Equations
(x – 5)(x – 96) = 0
(4)
Solving for x: x = 5, x = 96.
We substitute these two values into the original equation to find that only x = 96 works.
Method 2:
Rationalize the left hand side of the given equation by multiplying it by
3x 1 x 4
1=
.
3x 1 x 4
We get:
( 3x 1 x 4 )( 3x 1 x 4 )
7
3x 1 x 4
2x 3
7
3x 1 x 4
3x 1 +
x4 =
2x 3
.
7
The rest follows Method 1.
3
3
6.
Example 12: Find all values of a such that
1 a 2 1 a 2
Solution:
First we divide both sides by 3:
1
1
2
1 a 2 1 a 2
(1)
We see that 1 a 2 and 1 a 2 are conjugates and
( 1 a 2 )( 1 a 2 ) = 3 – a.
Therefore from (1) we have
2
1 a 2 1 a 2
2
2
3a
(1 a 2 )(1 a 2 )
1
1
3 a 1 a 2.
3 a
We can check that a 2 is the solution by plugging it into our original equation.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
7. Solving the radical equations by integer properties
Example 13: How many pairs of integers x and y satisfy the following equation
x y = 99 ?
Solution:
Method 1:
It is clear that 0 ≤ x ∙ y ≤ 99.
We can rewrite the given equation as
x =
99 –
y.
Squaring both sides: x = 99 + y – 2 99 y x – 99 – y = – 6 11y .
Or
6 11y = y + 99 – x
(1)
Since both x and y are integers, 6 11y must be integer and 11y is a square number.
Then we have 0 ≤ y = 11t2 ≤ 99 0 ≤ t2 ≤ 9.
Therefore the values of t are t = 0, 1, 2, 3.
When t = 0, 1, 2, and 3, the corresponding values of y are 0, 11, 44, and 99 and
subsequently the corresponding values of x are 99, 44, 11, and 0. All satisfy 0 ≤ x ≤ 99.
Therefore there are 4 pairs of solutions.
Method 2:
The original equation can be written as x + y = 3 11 .
Both x and y are in the form of 11t2.
Let x 11t12 , y 11t22 , where both t1 and t2 are nonnegative integers.
Therefore we have t1 + t2 = 3.
3 2 1
4 .
The number of nonnegative solutions to the above equation is
2 1
Thus there are 4 pairs of solutions.
Method 3:
We can simply write out all the solutions:
49
50 AMC Lectures
99 = 3 11 =
99 = 3 11 =
99 = 3 11 = 2
99 = 3 11 = 3
Chapter 3 Solving Radical Equations
0+
99
11 + 2 11 = 11 +
11 + 11 =
11 +
0 =
44
44 + 11
99 +
0
There are 4 pairs of solutions.
8. Solving the radical equations by solving an equivalent equation
Theorem: The equation f ( x) h( x) g 2 ( x) + g ( x) h( x) f 2 ( x) = h(x) has the same
solutions as the equation f 2 ( x) g 2 ( x) h( x) , f(x) ≥ 0, g(x) ≥ 0.
Proof:
For convenience, we can ignore the letter x in our following expressions.
The equation f h g 2 + g h f 2 = h is equivalent to
[ f 2 2 f h g 2 (h g 2 ) ] + [ g 2 2 g h f 2 (h f 2 ) ] = 0
f h g2
( f h g 2 )2 + ( g h f 2 )2 = 0
Also g h f 2
f 2 + g2 = h with f ≥ 0, g ≥ 0.
Example 14: Solve 2 x 2 1 +
x2 4 x2 .
Solution:
Let f(x) = 2, g(x) = 1, h(x) = x2.
From our theorem above, we know that the original equation has the same solutions as
22 + 1 = x2.
Solving for x: x1 5 , x2 5 .
9. Solving radical equations by using an inequality
Example 15: Solve
x5 +
3
= 2 3.
x5
50
50 AMC Lectures
Chapter 3 Solving Radical Equations
Solution:
We observe that
x 5 > 0 and
3
> 0.
x5
We know that a + b ≥ 2 ab (a > 0, b > 0).
In our case,
3
= 2 3.
x5
3
Equality holds if and only if x 5 =
.
x5
So the original equation is equivalent to x + 5 = 3
x5 +
3
≥ 2
x5
x5
x = – 2.
We can check that x 2 is the solution by plugging it into our original equation.
51
50 AMC Lectures
Chapter 3 Solving Radical Equations
PROBLEMS
5x 3 +
3x 1 +
Problem 1: Solve
Problem 2: Solve x2 –
x 1 = 2 2 .
x 2 9 = 21.
Problem 3: Solve 2(x+1) – 2 x( x 8) =
Problem 4: Solve
3
x 45 –
Problem 5: Solve x =
Problem 6: Solve x +
x
3
x x8.
x 16 = 1.
1
1
+ 1 .
x
x
x
x 1
2
=2 2.
Problem 7: Solve 2x2 + x 3x 2 1 = 4.
Problem 8: Solve
x x3
= 2x – 5.
x x3
Problem 9: Solve
x2 5x 2 –
4x 1
–
x3
Problem 10: Solve
x 2 5x 5 = 1.
x2
= 1.
x3
Problem 11: Solve 3 x 4 2 x 2 1 + 2 x x 4 2 x 2 8 = (x2 + 1)2.
2 x 3 3 1 x = 0.
Problem 12: Solve
Problem 13: Solve
Problem 14: Solve
4
x +
4
97 x = 5.
x5 +
x 3 = 4.
52
50 AMC Lectures
Chapter 3 Solving Radical Equations
Problem 15: Solve 2 1 2 x 2 +
2 8x2
1
.
x
Problem 16: (AMC 12) The equation x x 2 4 has:
(A) 2 real roots
(B) 1 real and 1 imaginary root
(C) 2 imaginary roots
(D) no roots
(E) 1 real root
Problem 17: (AMC 12) The equation x 4 –
(A) no root
(B) one real root
(C) one real root and one imaginary roots
(D) two imaginary roots
(E) two real roots
x 3 + 1 = 0 has:
Problem 18: (AIME) What is the sum of the solutions of the equation
12
4
x =
74 x
x 1 +
Problem 19: Solve
Problem 20: Solve
3
x 1 x 2 1 = x.
45 x + 3 16 x = 1.
Problem 21: (IMO) For which real numbers x do the following equations hold:
(a)
x 2x 1 +
x 2x 1 =
(b)
x 2x 1 +
x 2 x 1 = 1,
(c)
x 2x 1 +
x 2 x 1 = 2?
2,
53
50 AMC Lectures
Chapter 3 Solving Radical Equations
SOLUTIONS
Problem 1: Solution:
1
Since 3x – 1 ≥ 0, x ≥ .
3
3
Since 5x – 3 ≥ 0, x ≥ .
5
Since x – 1 ≥ 0, x ≥ 1.
Therefore the domain is x ≥ 1 as shown in the figure below.
When x ≥ 1, we have
3x 1 ≥ 2 , 5 x 3 ≥
Therefore
3x 1 +
2 , and
5x 3 +
x 1 ≥ 0.
x 1 ≥ 2 2 .
The equality holds if and only if x = 1.
We can check this by plugging it into the original equation.
Problem 2: Solution: 5, –5.
Let
x 2 9 = y. We have x2 = y2 + 9.
Substituting into the given equation we have y2 – y – 12 = 0.
Solving the quadratic equation for y, we get: y1 = 4, y2 = – 3 .
We have
Solving
x 2 9 = 4 or
x 2 9 = – 3 (extraneous).
x 2 9 = 4 for x: x1 = 5, x2 = – 5.
We can check that both x1 = 5 and x2 = – 5 are the solutions by plugging it into the
original equation.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
Problem 3: Solution: 1.
Let
x x 8 = y.
Squaring both sides we get:
y2 = x + x + 8 – 2 x( x 8) .
Therefore 2x + 2 – 2 x( x 8) = y2 – 6.
The original equation can be written as y2 – 6 = y.
Solving for y: y1 = 3, y2 = – 2.
We have
x x 8 = 3 or
x x 8 = – 2.
x x 8 , x x 8 = 3 cannot be true, so we only solve
x x 8 = – 2. We get x = 1.
Since
We can check that that x = 1 is the solution by plugging it into the original equation.
Problem 4: Solution: 80, –109.
x 45 , and v = 3 x 16 .
u–v=1
u3 – v3 = 61
(2) ÷ (1): u2 + uv + v2 = 61
Let u =
3
(1)
(2)
(3)
From (1) and (3), we get v1 = 4, v2 = – 5.
Since v3 = x – 16, x = v3 + 16.
x1 = 64 + 16 = 80, x2 = –125 + 16 = –109.
We can check that both x1 = 80 and x2 = –109 are the solutions by plugging it into the
original equation.
Note: We do not need to check to make sure that u ≥ 0 and v ≥ 0, since x + 45 or x – 16
can be negative.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
1 5
2
1
1
x = u and 1 = v.
x
x
Problem 5: Solution: x
Let
We have u + v = x.
1
1
x 1
u 2 v2
x
x = 1 1 .
Since u – v =
,u–v=
x
x
uv
1
Then we get 2u = x + 1
2u = u2 + 1.
x
1
Therefore u = 1 (u = –1 is extraneous) and x = 1.
x
1 5
1 5
1
Solving x = 1 we get x
(x
is extraneous since x must be positive).
x
2
2
1 5
We can check that x
is a solution by plugging it into the original equation.
2
Problem 6: Solution:
2.
The average value of x and
x
x 1
2
is
2.
Therefore we let
x=
2 + t and
x
x 1
2
=
2 – t.
x2 = 2 + 2 2 t + t2
x2
= 2 – 2 2 t + t2
x2 1
From (1), we have: x2 – 1 = 1 + 2 2 t + t2
1
From (2), we have: 2
= 1 – 2 2 t + t2
x 1
(3) × (4): 1 = (1 + 2 2 t + t2)(1 – 2 2 t + t2).
Simplifying: t4 – 6t2 = 0.
Solving for t: t = 0 or t = 6 .
56
(1)
(2)
(3)
(4)
50 AMC Lectures
So x =
2 or x =
We see that only x =
equation.
Chapter 3 Solving Radical Equations
2 6.
2 is a solution after plugging these two values into the original
Problem 7: Solution: x = 1 and x = – 4.
Multiply both sides of the given equation by 2:
4x2 + 2 x 3x 2 1 = 8.
Then, add 1 to both sides:
x2 + 2 x 3x 2 1 + 3x2 + 1 = 9.
The original equation can then be written as
( x 3x 2 1)2 = 9.
We have x 3x 2 1 = 3
We also have x 3x 2 1 = – 3
x = 1 or x = – 4.
x = – 1 or x = 4.
We can check that both x = 1 and x = – 4 are the solutions by plugging it into the original
equation.
Problem 8: Solution: x = 3 and x = 4.
By the property of proportion, we know that
2x 5 1
x x3 x x3
=
.
2x 5 1
x x3 x x3
We have
x
x2
x3 x3
x x 3 x 2 x 3
x x 3 x 2 x 3 0 [ x x 3 x 2] x 3 0.
So,
x x 3 x 2 0
or
x3 0.
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50 AMC Lectures
Solving
Solving
Chapter 3 Solving Radical Equations
x x 3 x 2 0, we get x = 4.
x 3 0 , we get x = 3.
We can check that both x = 3 and x = 4 are the solutions by plugging them into the
original equation.
Problem 9: Solution: 1, – 6.
We are given the equation
x2 5x 2 –
x2 5x 5 = 1
Rationalizing the left hand side by multiplying it by
get:
3
x 5x 2 x 2 5x 5
2
x 2 5x 2 x 2 5x 5
x 2 5x 2 x 2 5x 5
(1)
= 1, we
1
x 2 5x 2 x 2 5x 5 3.
(2)
[(1) + (2)]/2:
x2 5x 2 = 2
(3)
[(2) (1)]/2:
x2 5x 5 = 1
(4)
Solving (3) for x: x = 1 or x = – 6.
Solving (4) for x: x = 1 or x = – 6.
We can check that both x = 1 and x = – 6 are the solutions by plugging them into the
original equation.
Problem 10: Solution: 6.
We are given that
4x 1
x2
–
=1
x3
x3
(1)
We can see that
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50 AMC Lectures
Chapter 3 Solving Radical Equations
4 x 1 x 2 3x 3
x3 x3 x3
(2) ÷ (1):
4x 1
+
x3
(2)
x2
3x 3
=
x3
x3
(3)
(1) + (3):
2
4x 1 4x 6
=
.
x3
x3
Solving the equation above, we get x = 6.
We can check x = 6 is the solution by plugging it into the original equation.
Problem 11: Solution: 2.
The original equation can be rewritten as
3 ( x 2 1)2 4 x 2 + 2 x ( x 2 1)2 9 = (x2 + 1)2.
Let f(x) = 3, g(x) = 2x, h(x) = (x2 + 1)2.
The original equation has the same solutions as 9 + 4x2 = (x2 + 1)2, (x ≥ 0).
( x 2 1) 9 0
x4 2x2 1 9 4x2
x4 2 x2 8 0
( x 2 1 3)( x 2 1 3) 0
There is one solution x = 2.
Problem 12: Solution: No solutions.
3
Since 2x – 3 ≥ 0, x .
2
Since 1 – x ≥ 0, x ≤ 1.
There exists no such value for x satisfying both the inequalities listed above.
Therefore there is no solution to the given equation.
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50 AMC Lectures
Chapter 3 Solving Radical Equations
Problem 13: Solution: 16, 81.
Let u =
4
x,v=
4
97 x .
u+v=5
u4 + v4 = 97
(1)
(2)
We know that
u4 + v4 = (u + v)4 – 4uv(u + v)2 + 2 u2v2 = 625 – 100uv + 2u2v2 = 97.
Therefore u2v2 – 50uv + 264 = 0
(3)
From (1) and (3), we get u1 2 , u2 3 and v1 = 3, v2 = 2, which satisfies u ≥ 0, v ≥ 0 .
In other words , 4 x = 2 or 4 x = 3.
Solving for x: x1 1 6, x2 81 .
We can check that both x1 = 16 and x2 = 81 are the solutions by plugging them into the
original equation.
Problem 14: Solution: 4.
We are given the equation x 5 +
x3 = 4
Rationalizing the left hand side by multiplying it by
8
=4
x5 x3
x5 – x3 = 2
(1)
x 5 x 3
, we get:
x 5 x 3
(2)
x5 = 3
x3 = 1
(1) (2):
Solving (3) for x: x = 4.
Solving (4) for x: x = 4.
(1) + (2):
(3)
(4)
We can check that 4 is the solution by plugging it into the original equation.
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50 AMC Lectures
Problem 15: Solution: x
Chapter 3 Solving Radical Equations
6
.
6
We will employ the following theorem:
The equation f ( x) h( x) g 2 ( x) + g ( x) h( x) f 2 ( x) = h(x) has the same solutions as
the equation f 2 ( x) g 2 ( x) h( x) , f(x) ≥ 0, g(x) ≥ 0.
The original equation can be rewritten as 2 x 1 ( 2 x)2 +
2 x 1 (2 x)2 = 1.
Let f(x) = 2x, g(x) = 2 x , h(x) = 1.
The original equation has the same solution as (2x)2 + ( 2 x )2 = 1, (x ≥ 0).
6
Solving: x
.
6
Problem 16: Solution: (E).
x 2 4 x , x – 2 = 16 – 8x + x2, 0 = x2 – 9x + 18 = (x – 6)(x – 3); x = 6, x = 3.
We can check that only x = 3 is a solution by plugging the obtained values into the
original equation.
Problem 17: Solution: (A).
x4 =
x 3 – 1;
x + 4 = x – 3 – 2 x 3 + 1;
3 = – x 3.
This is impossible since the left side is positive while the right side is negative.
Problem 18: Solution: 337.
Let 4 x = y.
The given equation can be written as y2 – 7y + 12 = 0.
Solving: y = 3 or y = 4.
Therefore x = 34 or 44.
The sum is 34 + 44 = 337.
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50 AMC Lectures
Problem 19: Solution: x
Chapter 3 Solving Radical Equations
5
.
4
The domain is x ≥ 1, so the provided equation is equivalent to the equation below:
x 1 +
x 1 = x + x 2 1 ≥ 0 2( x 1 +
x 1 +
x 1 = 2 x 1 = 2 –
x 1 ) = ( x 1 +
x 1 )2
x 1 4 4 x 1 x 1
x 1
2 x 1 0
1
5
x 1
2 x .
4
x 1 2
Therefore the solution to the original equation is x
5
.
4
Problem 20: Solution: x1 = –109 and x2 = 80.
Method 1:
Let
3
45 x = u, 3 16 x = v
u v 1,
3 3
u v 61
From (1) and (2), we can get
uv = – 20.
(1)
(2)
(3)
u 4, u 5,
From (1) and (3), we get
or
v 5;
v 4.
3 16 x = 5 or 3 16 x = – 4.
Solving we get x1 109, x2 80.
We can check that both x1 = –109 and x2 = 80 are the solutions by plugging it into the
original equation.
Method 2:
The given equation can be written as
3
45 x + 3 16 x +
62
3
1 = 0.
50 AMC Lectures
Let a =
3
Chapter 3 Solving Radical Equations
45 x , b = 3 16 x , c =
3
1 .
We observe that a + b + c = 0.
Therefore we have a3 + b3 + c3 = 3abc.
Or (45 + x) + (16 – x) + ( – 1) = 3 3 45 x ∙ 3 16 x ∙
3
1
3
(45 x)( x 16) = 20.
Solving for x we get x1 109, x2 80 .
We can check that both x1 = –109 and x2 = 80 are the solutions by plugging it into the
original equation.
Problem 21: Solution:
x 2x 1 +
1
( 2x 2 2x 1 + 2x 2 2x 1 )
2
1
=
( 2x 1 1 + 2x 1 1 )
2
2 2x 1
2 2 x 1, if 2 x 1 1, or x 1;
2
=
2 2 , if 2 x 1 1, or 1 x 1.
2
2
x 2x 1 =
(a) The equation holds for 1/2 ≤ x ≤ 1.
(b) The equation has no solution.
(c) The equation holds for 4x −2 = 4 ⇒ x = 3/2.
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50 AMC Lectures
Chapter 4 Absolute Values
BASIC KNOWLEDGE
1. Definition
The absolute value of a number x is denoted as x . It is the distance from the point x to
zero on the number line.
As shown in the figure below, 3 indicates that the distance from 3 to 0 is 3 and 3
indicates that the distance from 3 to 0 is 3.
2. Properties Of Absolute Value:
x
(1). x 0
x
if x 0
x
if x 0 or x
x
if x 0
if x 0
x
or x
if x 0
x
(2). For all real numbers x and y:
x y if and only if x = y or x = – y.
(3). x 0 . The equality holds if and only if x = 0.
(4). x x
(5). x y y x
(6). xy x y , and
x
x
y
y
y 0.
64
if x 0
if x 0
50 AMC Lectures
Chapter 4 Absolute Values
3. Graphing Absolute Values
(1). To get the graph of y = f (x) , we first plot y = f (x). Next, we flip the part that is
below x-axes up about the x-axes. Note that the part below the x-axes should not be there
anymore. The other parts of the graph do not change.
(2). Graph of y = kx b : Draw y1 = kx +b (Figure a). Flip the part of the graph below the
x-axes up about the x-axes (Figure b). The solid lines comprise of the graph of y = kx b
(Figure c).
Figure (a)
Figure (b)
Figure (c)
(3). To get the graph of y = f ( x ) , we first plot y = f (x) (Figure d). Then we flip the part
that is on the right side of the y-axes to the left of the y- axes (Figure e). The solid lines
comprise of the graph of y = f ( x ) (Figure f). Note that the part originally on the left of
the y-axes is erased while the part on the right of the y-axes is still kept.
Figure (d)
Figure (e)
65
Figure (f)
50 AMC Lectures
Chapter 4 Absolute Values
(4). Some graphs of absolute values
y=x
y= x
y = x 1
y= x –1
y = ax2 + bx + c
y=– x
y = x 1
y=1– x
y= x +1
y = x 1 + y = x 2
y = ax2 + bx + c
y = ax 2 bx c
66
y = ax 2 bx c
50 AMC Lectures
Chapter 4 Absolute Values
Example 1: Which of the figures is y = x 2 5 x 24 ?
(A)
(B)
(C)
(D)
Solution: (A).
We first disregard the absolute value sign, and graph the quadratic y = x 2 5x 24 as
follows:
We then flip part of this graph that is to the right of the y-axes about the y-axes to the left
to get the graph of y = x 2 5 x 24 . Note two things:
(1) After flipping, the part that is on the left side of the y-axes became replaced and is
thus not there anymore, and
(2) The part on the right of the y-axes is still kept.
The answer is(A).
4. Evaluating Absolute Values
Example 2: (1978 AMC) If x < 0, then x ( x 1) 2 equals
(A) 1
(B) 1 – 2x
(C) – 2x – 1
(D) 1 + 2x
(E) 2x – 1
Solution: (B).
We know that
( x 1) 2 x 1 . By property (5), we also have x 1 1 x .
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50 AMC Lectures
Chapter 4 Absolute Values
Since x is negative, 1 x is positive. So
( x 1)2 1 x
x ( x 1) 2 = x (1 x) 2 x 1 .
Similarly 2 x 1 = 1 2 x and 1 – 2x is positive.
Therefore x ( x 1) 2 = 1 – 2x.
Example 3: (1974 AMC) If x < – 2 then 1 1 x equals
(A) 2 + x
(B) – 2 – x
(C) x
(D) – x
(E) – 2
Solution: (B).
a when a 0
By definition a =
.
a when a 0
If x < – 2, then 1 + x < 0 and 1 x (1 x) and 1 1 x = 1 1 x = 2 x .
If x < – 2, then 2 + x < 0 and 2 x = – 2 – x.
Example 4: Calculate:
1
1
1
1
1
1
.
1992 1991 1993 1992 1993 1991
Solution: 0
1
1
1
1
1
1
1992 1991 1993 1992 1993 1991
1
1
1
1
1
1
(
) (
) [(
)]
1992 1991
1993 1992
1993 1991
1
1
1
1
1
1
0.
1992 1991 1993 1992 1993 1991
5. Finding The Number Of Solutions Of Equations
Example 5: (1984 AMC) The number of distinct solutions of the equation x 2 x 1 =
3 is
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Solution: (C).
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50 AMC Lectures
Chapter 4 Absolute Values
The equation x 2 x 1 = 3 is satisfied by precisely those
x for which the vertical distance between the graph of y = x
and the graph of y = 2 x 1 is 3. These two graphs are
sketched in the adjoining figure; the latter graph is V1
shaped with vertex ( , 0) and sides of slope 2. Thus
2
there are exactly two x values which satisfy the original
equation, because the vertical distance between the graphs
1
increases steadily as one moves right or left from x , and the vertical distance for
2
1
that x is .
2
Example 6: Find real number m such that x 2 4 x 5 m has four distinct real roots.
Solution: 1 < m < 5.
Graph y = f (x) = x 2 4 x 5 .
We see that when m = 1, there is one root. When m = 5, there
are three roots. When 1 < m < 5, there are four roots.
Example 7: Find real number a such that x 2 ax 4 has only three distinct real roots.
Solution: 4.
Let y1 x 2 ax and y2 = 4.
We know that the equation has three distinct real solutions. This tells us that the graphs
y1 x 2 ax and y2 = 4 have three points of intersection.
We can get the graph of y1 x 2 ax using the following method:
(1) graph y1 x 2 ax ;
(2) flip the part of the graph that is below x-axes up about x-axes.
We obtain the adjoining graph.
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50 AMC Lectures
Chapter 4 Absolute Values
We can see that the graphs y1 x 2 ax and y2 = 4 have three points of intersection if and
a2
only if
= 4.
4
Solving we get a = 4.
6. Finding The Enclosed Area
Theorem: If a and b are real numbers and c is positive integer, the shape of the region
formed by x a + y b = c is a square and the side of the square is 2c .
CASE 1: a x b y c
Basic procedure:
Let y = 0, and our original equation becomes x
two points on x-axis: x1
c
. The solutions to this equation are the
a
c
c
, x2
a
a
c
.
b
The solutions to this equation are the two points on y-axis:
c
c
y1 , y2 .
b
b
Then, we connect these four points to form a rhombus. The enclosed area will be the area
of this rhombus:
c
c
(2 ) (2 )
a
b 2 c c
(1)
A
2
a b
Let x = 0, and our original equation becomes y
CASE 2. x y x y c
Basic procedure:
Let y = 0, and our original equation becomes
c
x x c x
2
c
Similarly we also get y
2
70
50 AMC Lectures
Chapter 4 Absolute Values
c
The enclosed area is then the area of a square of side length (2 ) .
2
The area is:
c
c
A (2 ) (2 ) c 2
(2)
2
2
Example 8: Find the number of square units in the area of the region determined by
x 3 y 6 5.
Solution: 50.
We can ignore–3 and 6 in the given equation, and re-write the equation as x y 5 .
(– 3 and 6 are only distracters. They shift the figure along the Cartesian plane but do not
affect the area of the figure).
We divide both sides by 5 and then follow the steps in the solution of Problem 1.
The enclosed area will be A = 50.
Example 9: Find the number of square units in the area of the region determined by
2 x 3 3 y 6 12 .
Solution: 48.
Ignoring –3 and 6, we can re-write equation as 2 x 3 y 12 , or
2 x 3 y 12 .
Divide both sides by 12 to obtain:
x y
1
6 4
A = 2 6 4 = 48.
Example 10: What is the number of square units in the area of the
region determined by x y x y 4 ?
Solution: 16.
Let y = 0, and our original equation becomes
x x 4
x 2.
Similarly we get y 2
The enclosed area A = 4 4 = 16.
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50 AMC Lectures
Chapter 4 Absolute Values
7. Finding The Smallest / Greatest Value Of Absolute Values
Properties For any real a and b, we have
(7). a b a b . The equality holds when ab 0.
Proof:
The given inequality is equivalent to:
( a b )2 (a b)2
a 2 2 ab b 2 a 2 2ab b 2
It is clear that the equality holds when ab ab
ab ab .
ab 0 .
(8). a b a b . The equality holds when ab 0.
Proof:
The given inequality is equivalent to:
(a b) 2 ( a b ) 2 a 2 2ab b 2 a 2 2 ab b 2
always true.
It is clear that the equality holds when ab = ab
ab 0.
(9). a b = a b
(10). a b = a b
ab ab , which is
ab 0.
ab 0
(11). For real numbers a1, a2, . . . an,
a1 a2 an a1 a2 an
The equality holds when a1, a2, . . . an have the same sign, i.e. ai a j 0 (i, j = 1, 2, . . . ,
n, i j).
Example 11: Find the smallest value of y = x 2 + x 1 .
Solution: 3.
By property (5), we have x 2 + x 1 = 2 x + x 1 .
(Why do we want to do this? Because we want to cancel out x).
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50 AMC Lectures
Chapter 4 Absolute Values
By property (8), we have 2 x + x 1 ≥ 2 x x 1 3 3 .
The smallest value of 3 is achieved when (2 x)( x 1) 0 ( x 2)( x 1) 0 . The
solution to the inequality is 1 ≤ x ≤ 2.
Example 12: Find the smallest value of y x a x b x c if a < b < c.
Solution: c – a.
From property (6), we have y x b ( x a c x ) .
Since x b 0 , we have x b ( x a c x ) x a c x .
From property (8), we have x a c x ( x a) (c x)
= ca
= c – a.
The equality holds when
x b 0
( x a)(c x) 0
(a < c)
x=b
Therefore, when x = b, ymin = c – a.
Example 13: (1983 AIME) Let f (x) = x p + x 15 + x p 15 , where 0 < p < 15.
Determine the minimum value taken by f (x) for x in the interval p x 15.
Solution: 15.
Method 1:
Since 0 < p x 15, x p = x – p, x 15 = 15 – x, and
x ( p 15) = p + 15 – x.
Therefore f (x) = (x – p) + (15 – x) + (p +15 – x) = 30 – x.
When x is the greatest, f (x) is the smallest. The smallest value is 15.
Method 2 (our solution):
By property (6), we have f (x) = x p + x 15 + x p 15 = x p + 15 x +
p 15 x .
By property (11), we have
x p + 15 x + p 15 x ≥ x p 15 x p 15 x 30 x 30 x .
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50 AMC Lectures
Chapter 4 Absolute Values
When x is the greatest value, 15, f (x) will take on the smallest value. The smallest value
of f (x) is 15.
Property (12): For y = x a1 + x a2 + + x an , if a1, a2, . . . an satisfy a1 < a2
< … < an, we have:
(1) if n = 2m + 1 is odd, y has the following smallest value when x = am + 1:
a1 a2 am + am 2 + am3 + + an
(2) if n is even, n = 2m, y has the following smallest value when x = am or x = am +1:
a1 a2 am + am1 + + an.
The smallest distance is achieved when x is a point in the middle of n points in the
number line. If n is odd, x is the point in the middle. If n is even, x can be any of the two
points in the middle of the n points on the number line.
Example 14: What is the smallest value of x 1 + x 2 + x 3 + x 4 + x 5 for
any real x?
Solution: 6.
By property (12), we know that the smallest value of the expression is achieved when x =
3, the point right in between –1 and –5.
This smallest value is 3 1 + 3 2 + 3 3 + 3 4 + 3 5 = 2 + 1 + 0 + 1 +
2 = 6.
Example 15: Find the smallest value of x + x 1 + x 2 + . . . + x 2007 .
(A) 10032.
(B) 10042.
(C) 20062.
(D) 20072
Solution: (B).
Let x = y – 1. The given equation becomes y 1 + y 2 + y 3 + . . . + y 2008
By property (12), we know that the smallest value is achieved when y = 1004 (or 1005).
When y = 1004, y 1 + y 2 + y 3 + . . . + y 2008 = (1004 – 1) + (1004 – 2)
+…+ (1004 – 1003) + (1004 – 1004) + (1005 – 1004) + (1006 – 1004) +…+ (2008 –
1004) = 1003 + 1002 +…+ 1 + 0 + 1 + 2+…+ 1004 =
(1 1003) 1003 (1 1004) 1004
10042 .
2
2
74
50 AMC Lectures
Chapter 4 Absolute Values
PROBLEMS
Problem 1: y = f (x) is graphed as follows:
Which of the figures is y = f (x 2) – 1?
(A)
(B)
(C)
(D)
Problem 2: Simplify 2 2 x 2 if x < – 1.
Problem 3: Simplify 6 x x 10 if 5 x 5 x and 3 x x 3
Problem 4: Find the number of distinct solutions of the equation x 1 + x 2 = x + 3.
Problem 5: Find k if y1 x 2 1 and y = x + k intercept exactly at three points.
Problem 6: Find the number of square units in the area of the region determined
by 2 x 3 y 6 x y 16
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50 AMC Lectures
Chapter 4 Absolute Values
Problem 7: Find the smallest value of y x a x b if a < b.
Problem 8: Find the smallest value of y x a x b x c x d . a < b < c < d.
Problem 9: What is the greatest value of y 2 x 6 4 x 1 x 1 ?
Problem 10: Simplify b a a c c b . The coordinates of a, b, and c are shown on
the number line below.
Problem 11: What is the sum of all values of x for which x 3 3 x 1 3?
Problem 12: Simplify x 2 + 2 x 1 – x 1 .
Problem 13: Which of the following figures in solid line represents the graph of
y 1 x x2 ?
(A)
(B)
(C)
Problem 14: Find the maximum value of y x x 2 x 1 when x 1 6.
76
(D)
50 AMC Lectures
Chapter 4 Absolute Values
Problem 15: (1982 AMC) Let f (x) = x 2 + x 4 – 2 x 6 , for 2 x 8. The sum of
the largest and smallest values of f (x) is
(A) 1
(B) 2
(C) 4
(D) 6
(E) none of these
Problem 16: Find a b c + a b c + a b c + a b c if a, b, and c are sides
of a triangle.
(A) 0.
(B) 2a + 2b + 2c.
(C) 4a.
(D) 2b – 2c.
Problem 17: (1987 AIME) Find the area of the region enclosed by the graph of x 60 +
y =
x
.
4
Problem 18: (2002 AIME II) Given that
(1) x and y are both integers between 100 and 999, inclusive;
(2) y is the number formed by reversing the digits of x; and
(3) z = x y
How many distinct values of z are possible?
Problem 19: (1974 AMC) If f (x) = 3x + 2 for all real x, then the statement:
“ f ( x) 4 < a whenever x 2 < b and a > 0 and b > 0” is true when
(A) b
a
3
(B) b >
a
3
(C) a
b
3
(D) a >
b
3
(E) The statement is never true.
Problem 20: (2003 AMC12 B #24) Positive integers a, b and c are chosen so that a < b
< c, and the system of equations 2x + y = 2003 and y = x a + x b + x c has
exactly one solution. What is the minimum value of c?
(A) 668
(B) 669
(C) 1002
(D) 2003
(E) 2004
Problem 21: (2002 ARML) If f ( x) ( x 3)2 1 , compute the set of real numbers such
that f ( x ) f ( x) .
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50 AMC Lectures
Chapter 4 Absolute Values
SOLUTIONS
Problem 1: Solution: (C).
The graphs of y = f ( x) , y = f [( x 2)] , y = f (x 2) , and y = f (x 2) – 1
are shown as follows, respectively.
Problem 2: Solution: 2 x .
By the property (5), we have x 2 2 x .
Since x is negative, 2 x is positive.
So 2 x 2 = 2 (2 x) x , and
2 2 x2 = 2 x .
Since x is negative, by property (1), we have x x .
Therefore 2 x 2 ( x) 2 x .
Problem 3: Solution: 16 2 x .
Since 5 x 5 x , 5 – x 0
x 5.
Since 3 x x 3 , x – 3 0
x 3.
6 x x 10 = 6 x (10 x) 16 2 x .
Problem 4: Solution: 2.
Let y = x 1 + x 2 and y = x + 3.
The number of solutions is the same as the number of
intersections of the graphs
y = x 1 + x 2 and y = x + 3.
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50 AMC Lectures
Chapter 4 Absolute Values
We see from the figure above that y = x 1 + x 2 and y = x + 3 have two points of
intersection. Therefore
x 1 + x 2 = x + 3 has two solutions.
5
Problem 5: Solution: 1 or .
4
Graph y = x2 – 1 and flip the part where y < 0 up.
We know that y = x + k passes through (– 1, 0) or intercepts
with y x 2 1 , where ( – 1 x 1). Therefore k
5
= 1 or .
4
Problem 6: Solution: 32.
Let y = 0.
We have 2 x 6 x 16 or 8 x 16 , or x 2 .
Similarly, let x = 0, we get y 4
The area is A = 2 4 4 = 32.
Problem 7: Solution: b – a.
From property (6), we have y x a x b = x a b x .
From property (8), we have x a b x ( x a) (b x)
= ba
(a < b)
= b – a,
The equality holds when (x – a)(b – x) 0
a x b.
Therefore y has the smallest value b – a when a x b.
Problem 8: Solution: c + d – a – b.
From properties (6) and (8), we have
y ( x a d x) ( x b c x)
(d x) ( x a) ( x b) (c x)
(a < b < c < d)
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50 AMC Lectures
Chapter 4 Absolute Values
= d a cb
= d – a + c – b = c + d – a – b,
The equality holds when
( x a)(d x) 0
bxc
( x b)(c x) 0
Therefore, when b x c, ymin = c + d – a – b.
Problem 9: Solution: 6.
x 1,
5 x 11,
y
3 3 x,
1 x,
x 3
3 x 1
1 x 1
x 1
Case 1: When x 3 , y x 1 4 , and ymax y(3) 4 .
Case 2: When 3 x 1 , y 5x 11 6 , and ymax y(1) 6 .
Case 3: When 1 x 1, y 3 3x 6 , and ymax y(1) 6 .
Case 4: When x 1 , y 1 x 0 , and ymax y(1) 0 .
Therefore, ymax max 4,6,6,0 6 .
Problem 10: Solution: 2c
From the number line we know that a > 0, b > 0, c < 0, and c a b 0 .
b a a c c b = (a b) (a c) (b c) 2c .
Problem 11: Solution: 6.
Method 1:
x 3 3 x 1 3
x 3 3 x 4
(x – 3 + 3 – x + 4)( x – 3 + 3 – x – 4)[x – 3 – (3 – x) + 4][ x – 3 – (3 – x) – 4)] = 0
4 (– 4)(2x – 2)(2x – 10) = 0 (2x – 2)(2x – 10) = 0.
x = 1 and x = 5
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We can check both values by plugging them into the original equation to see that they are
the solutions. Their sum is 1 + 5 = 6.
Method 2:
x 3 3 x 1 3
x 3 x 3 4
2x3 4
x3 2
x = 1 and x = 5
The sum is 1 + 5 = 6.
Problem 12: Solution:
When x < – 2, the original expression can be simplified to – 2x – 2.
1
When – 2 x < , the original expression can be simplified to 2.
2
1
When x < 1, the original expression can be simplified to 4x.
2
When x 1, the original expression can be simplified to 2x + 2.
Problem 13: Solution: (B).
Method 1:
The roots of x 2 x = 0 are 0 and 1.
Therefore
y 1 x x2 1 x2 x .
2
0 x 1;
1 x x,
=
2
x 0, or x 1.
1 x x
Hence the answer is (B).
Method 2:
Since y = 1 x x 2 1, the graph should be below y = 1.
Therefore answers (A), (C), (D) can be eliminated, giving the answer (B).
Problem 14: Solution:16.
Solving x 1 6, we get – 7 x 5.
Case I: 0 < x 5.
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Chapter 4 Absolute Values
y x 2 2 x 1 ( x 1) 2
The greatest value of y is (5 – 1)2 = 16..
Case II: – 7 x 0.
y x 2 2 x 1 2 ( x 1) 2
The greatest value of y is 2.
Therefore when – 7 x 5, the greatest value of y is 16.
Problem 15: Solution: (B).
When 2 x 3, f (x) = (x – 2) – (x – 4) + (2x – 6)
= – 4 + 2x.
Similar algebra shows that when 3 x 4, f (x) = 8 – 2x;
and when 4 x 8, f (x) = 0. The graph of f (x) in the
adjoining figure shows that the maximum and minimum
of f (x) are 2 and 0, respectively. The sum of these values equals 2, so the answer is (B).
Note: Since a linear function reaches its extreme values at the endpoints of an interval,
and since the given function is linear in each subinterval, it suffices to calculate f (2) = 0,
f (3) = 2, f (4) = f (8) = 0.
Problem 16: Solution: (B).
Since a, b, and c are the sides of a triangle,
a + b > c, b + c > a, c + a > b.
The original expression can be written as
(a + b + c) + (b + c – a) + (a + c – b) + (a + b – c)
= 2(a + b + c).
Problem 17: Solution: 480.
Method 1:
Since the graph is symmetrical to the x-axes, we just need to calculate two times the area
of region formed by the graphs:
x
y x 60,
4
y 0
In order to figure out what this region is, we need to draw
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Chapter 4 Absolute Values
x
and y2 x 60 .
4
The region formed by the difference of the two graphs that is above the x axes since
y 0, is DAB, shown in the figure to the right.
The coordinates of the vertices of the triangles can be
calculated as D(60, 15), A(48, 12), B(80, 20).
AB CD
240 .
The triangle’s area is S DAB
2
The total area of the original equation is two times the
area of triangle DAB, or 240 2 = 480.
y1
Method 2:
The equation is symmetrical about the x-axes.
The absolute equation can be written as the union of two equations:
x
5
y = – (60 – x) = x – 60, x 60
4
4
and
x
3
y = – (x – 60) = 60 – x , x > 60
4
4
In (1), when y = 0, x = 48, we have the point on the graph A(48, 0).
(1)
(2)
In (2), when y = 0, x = 80, we have the point on the graph B(80, 0).
D: (60, 15).
The region enclosed by the given graph consists of two congruent triangles, with the
coordinates of the three vertices of one of triangles being A, B, and C.
1
The area of this region equals 2S ADB 2 (80 48) 15 480 .
2
Problem 18: Solution: 9.
Let the hundreds, tens, and units digits of x be h, t, and u, respectively. Then
x = 100h + 10t + u, y = 100u + 10t + h, and z = 99(h u) = h u. Since h
and u are between 1 and 9, inclusive, h u must be between 0 and 8, inclusive.
Thus there are 9 possible values for z.
Problem 19: Solution: (A).
Consider f ( x) 4 = 3x 2 4 = 3 x 2 .
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50 AMC Lectures
Now whenever x 2 <
Chapter 4 Absolute Values
a
, then f ( x) 4 < a. Consequently whenever x 2 < b and b
3
a
, we have f ( x) 4 < a.
3
Problem 20: Solution: (C).
Since the system has exactly one solution, the graphs of the two equations must intersect
at exactly one point.
If x < a, the equation y = x a + x b + x c is equivalent to y = – 3x + (a + b + c).
Through similar calculations, we obtain
3x (a b c),
x (a b c),
y=
x (a b c),
3x (a b c),
if
if
if
if
xa
a xb
bxc
c x.
Thus the graph consists of four lines with slopes – 3, – 1, 1, and 3, and it has corners at
(a, b + c – 2a), (b, c – a), and (c, 2c – a – b). On the other hand, the graph of 2x + y =
2003 is line whose slope is – 2. If the graphs intersect at exactly one point, that point
must be (a, b + c – 2a). Therefore 2003 = 2a + (b + c – 2a) = b + c.
Since b < c, the minimum value of c is 1002.
Problem 21: Solution: 0 x 2 or x 4.
The leftmost diagram is the graph of f ( x) ( x 3)2 1 , the middle is the graph of
f ( x) ( x 3)2 1 , and the rightmost is the graph of f ( x ) ( x 3)2 1 .
The solution set consists of x-values for which the middle and right hand graphs overlap,
namely, 0 x 2 or x 4. Alternately, [0, 2] [4, ).
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BASIC KNOWLEDGE
1. Solving Absolute Value Equations
(1). Solve x a x b c
(1)
x a is the distance from x to point a, and x b is the distance from x to point b.
We call points a and b “critical points”. a, b, and c are real numbers.
If b a = c, (we denote b a as the critical distance) equation (1) has infinite many
solutions. (No matter where we move x along the x-axis between a and b, the distance
represented by x a x b will always equal the value of c).
If b a < c, equation (1) has two solutions. (One is to the left side of a, and the other
solution is to the right side of b).
If b a > c, equation (1) has no solutions. (No matter where we put x, the equation will
not be satisfied).
Example 1: Find all integer solutions to x 2 x 3 1 .
Solution: 2 and 3.
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Method 1:
The critical points are 2 and 3. The critical distance is then 3 – 2 = 1, which is the same
value as the value of the right hand side of the equation. This means that the equation has
infinitely many solutions. However, the integer solutions can only be achieved at the two
ends, where x = 2 and x = 3.
Method 2:
We use the basic formula.
We can write the equation as: x 2 1 x 3 . According to the basic formula, we can
write:
x 2 1 x 3
(1)
.
x 2 (1 x 3 )
(2)
Re-write equation (1) as x 3 3 x . Thus, x – 3 = 3 – x or x – 3 = – (3 – x) x = 3.
Re-write equation (2) as x 3 x 1 . Thus, x – 3= x – 1 or x – 3 = – (x – 1) x = 2.
The solutions will be x = 2 and x = 3.
Method 3:
We have the following three intervals:
{– ; 2], [2, 3], [3, }.
On {-, 2], we have x – 2 = 2 – x and x – 3 = 3 – x.
Our original equation thus becomes
2–x+3–x=1
Solve for x, x = 2.
On [2, 3], we have x – 2 = x – 2 and x – 3 = 3 – x.
Our original equation thus becomes
x – 2 + 3 – x = 1 from where we obtain the identity 1 = 1.
This means that all the numbers on this interval are solutions to this equation.
On [3, }, we have x – 2 = x – 2 and x – 3 = x – 3.
Our original equation thus becomes
x – 2 + x – 3 = 1 from where x = 3.
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Upon assembling all this, the integer solutions will be x = 2 and x = 3.
Example 2: How many integer solutions are there to the equation x 1 x 3 4 ?
(A) 3
(B) 4
(C) 5
(D) Infinitely many
Solution: (D).
We know that x 1 x (1) .
The critical distance is 4. Therefore any point between 1 and 3 will
satisfy the equation. The integers that are solutions to the equation
are 1, 0, 1, 2, and 3.
(2). Solve f1 ( x) f 2 ( x) g ( x)
(2)
Theorem 1: The solutions to the equation
f1 ( x) f 2 ( x) g ( x)
(2-1)
are among the solutions to the system of equations:
f1 ( x) f 2 ( x) g ( x) 0
f1 ( x) f 2 ( x) g ( x) 0
f1 ( x) f 2 ( x) g ( x) 0
f1 ( x) f 2 ( x) g ( x) 0
(2-2)
Proof:
To solve the equation f1(x) f 2 (x) g(x) , we need to deal with four cases:
Case 1: f1 ( x) 0 and f 2 ( x) 0
f1 ( x) f 2 (
x) g ( x)
f1 ( x) + f 2 ( x) + g (x) =0
f1 ( x) f 2 ( x) g (x)
(2-3)
Case 2: f1 ( x) 0 and f 2 ( x) 0
f1 ( x) f 2 ( x) g ( x)
f1 ( x) + f 2 ( x) g (x)
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f1 ( x) f 2 ( x) + g (x) =0
Case 3: f1 ( x) 0 and f 2 ( x) 0
f1 ( x) f 2 ( x) g ( x)
(2-4)
f1 ( x) f 2 ( x) = g (x)
f1 ( x) f 2 ( x) g (x) =0
(2-5)
Case 4: f1 ( x) 0 and f 2 ( x) 0
f1 ( x) f 2 ( x) g ( x)
f1 ( x) + f 2 ( x) = g (x)
f1 ( x) + f 2 ( x) g (x) =0
(2-6)
Similarly, for f1 ( x) f 2 ( x) g ( x) , we also get the solution set (2-2).
Note that after we get the solutions to the set of equations in (2-2), we need to plug them
in to equation (2-1) to check to see if they are valid solutions..
Example 3: Find the integer solutions of x 2 x 5 2 x 1 .
Solution: 2 or – 3.
x2 x 5 2x 1
2
x – x + 5 – 2x + 1 = 0
x2 – 3x + 6 = 0
(no integer solutions)
x2 – x + 5 – 2x – 1 = 0
x2 – 3x + 4 = 0
(no integer solutions)
2
2
x – x – (5 – 2x) – 1 = 0
x +x –4=0
(no integer solutions)
2
2
x – x – (5 – 2x) + 1 = 0
x +x –6=0
x = 2 or x = – 3 (checked)
When we substitute in x = 2 or x = – 3 back into the original equation, we see that these
values do satisfy the equation.
Example 4: Solve the equation x 1
Solution: x =
x +1+x –
1
2 x 3 5x 2 .
2
1
10
3
+ 5x + 2 = 0
2
When we substitute in x = –
7x +
3
=0
2
x=–
3
(Checked, no)
14
3
back into the original equation, we see that the value
14
does not satisfy the equation.
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3
5
5
– (5x + 2) = 0 – 3x – = 0
x = – (Checked, no)
2
2
6
5
When we substitute in x = – back into the original equation, we see that this value
6
does not satisfy the equation.
x +1+x –
3
9
9
) + 5x + 2 = 0 5x + = 0
x=–
(Checked, no)
2
2
10
9
When we substitute in x = –
back into the original equation, we see that this value
10
does not satisfy the equation.
x +1–(x –
3
1
1
) – (5x + 2) = 0
–5x + = 0 x =
(Checked, yes)
2
2
10
1
When we substitute in x =
back into the original equation, we see that this value does
10
satisfy the equation.
1
The only solution is x =
.
10
x +1–(x –
(3). Solve f ( x) g ( x) ( x)
(3)
Theorem: The equation f ( x) g ( x) ( x) with 0, can be solved as follows:
(We use f, g, to represent f (x), g (x), (x), respectively).
Case I:
If (f + g)2 > 2, (3) and (f – g)2 = 2 have the same solution.
Case II:
If (f + g)2 = 2, (3) and fg 0 have the same solution.
Case III:
If (f + g)2 < 2, (3) has no solution.
Case I Proof:
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2
2
( f g )
f g
f 2 2 fg g 2 2
2
f 2 fg g 2 2
2 fg 2 fg 0
2
2
( f g )
2
2
2
f 2 fg g
We can prove Cases II and III similarly.
( f g ) 2 2
.
( f g ) 2 2
Example 5: Solve x 3 2 x 1 = 3x.
(1)
Solution:
When x 0, (1) has no solution.
When x > 0, [(x + 3) + (2x – 1)]2 = (3x + 2)2 > (3x)2.
Therefore (1) has the same solutions as
[(x + 3) – (2x – 1)]2 = (3x)2.
(2)
Solving (2) we get x1 = 1, and x2 = – 2.
– 2 is an extraneous solution, so the solution to (1) is x = 1.
Example 6: Solve x 2 2 x 1 2 x 5 x 2 2 .
(1)
Solution:
We have [( x 2 2 x 1) (2 x 5)]2 = ( x 2 4) 2 > ( x 2 2) 2 .
Therefore (1) has the same solutions as
[( x 2 2 x 1) (2 x 5)]2 = ( x 2 2) 2
Solving (2) we get: x = – 2, x = 1
(2)
3.
Therefore (1) has the solutions: x = – 2 , x = 1
90
3.
50 AMC Lectures
Example 7: Solve 7 x
Chapter 5 Solving Absolute Value Equations
1 1
2 x = 5x.
x x
Solution:
Only if x > 0 will the equation have solutions, since the right hand side must equal an
absolute, non-negative value.
1
1
Therefore [(7 x ) ( 2 x)]2 (5 x) 2 has the same solutions as
x
x
1 1
(7x – )( – 2x) 0.
(1)
x x
We know that x > 0 (since 5x equals an absolute value).
2
7
Solve for x in (1): x
or 0 < x
.
2
7
7
2
Therefore the solutions of the original equation are (0,
] or [
, + ).
7
2
4. Solve equations f1 ( x) + f 2 ( x) = f1 ( x) f 2 ( x) and f1 ( x) + f 2 ( x) = f1 ( x) f 2 ( x)
Properties
f1 ( x) f 2 ( x)
0
(1) If f1 ( x) + f 2 ( x) =
, then f1 ( x) f 2 ( x)
f1 ( x) f 2 ( x)
0
f1 ( x) f 2 ( x)
0
(2) If f1 ( x) f 2 ( x)
, then f1 ( x) + f 2 ( x) =
.
f1 ( x) f 2 ( x)
0
Proof:
We know that f1 ( x) + f 2 ( x) = f1 ( x) f 2 ( x) .
Therefore [ f1 ( x) + f 2 ( x) ]2 = f1 ( x) f 2 ( x) 2 = [f1 (x) + f2 (x)]2
f1 ( x) 2 + 2 f1 ( x) ∙ f 2 ( x) + f 2 ( x) 2 = f12 ( x) + 2 f1 ( x) f 2 ( x) + f 22 ( x)
Or f1 ( x) ∙ f 2 ( x) = f1 ( x) f 2 ( x) .
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Hence f1 ( x) f 2 ( x) = f1 ( x) f 2 ( x) .
From the definition of the absolute values, we have f1 ( x) f 2 ( x) 0.
When f1 ( x) f 2 ( x) 0, f1 ( x) + f 2 ( x) = f1 ( x) f 2 ( x) .
Similarly we can prove that
if f1 ( x) + f 2 ( x) = f1 ( x) f 2 ( x) , f1 ( x) f 2 ( x) 0, and
if f1 ( x) f 2 ( x) 0, f1 ( x) + f 2 ( x) = f1 ( x) f 2 ( x) .
Example 8: How many real solutions are there to the equation 2 x 1 x 2 = x 1 ?
(A) 1 (B) 2 (C) 2
(D)3
(E) Infinite many.
Solution: (E).
The original equation can be rewritten as 2 x 1 x 2 = (2 x 1) ( x 2) .
By property 1, we have (2x – 1)(x – 2) 0.
1
Solving we get:
x 2.
2
Therefore, the equation has infinitely many solutions.
Example 9: How many real solutions are there to the equation x 2 x 3 = 1?
(A) 0
(B) 1
(C) 2
(D) 3
(E) more than 3.
Solution: (E).
The original equation can be rewritten as x 2 x 3 = ( x 2) ( x 3) .
By property 1, we have
(x – 2)(x – 3) 0
Solving we get: 2 x 3.
Therefore, the equation has more than 3 solutions.
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Example 10: Solve 2 x 1 x 3 4 .
Solution: x = 8 or x = – 2.
The equation can be written as
4 + x 3 2x 1
(1)
From property 2, we have
Case I: If 4(x + 3) 0 or x – 3, equation (1) can be written as 4 ( x 3) 2 x 1 , or
x + 7 = (2x – 1).
Solving we have x = 8 or x = – 2. After substituting in these values into the original
equation, we see that they are indeed the solutions.
Case II: If 4(x + 3) 0, or x – 3 , equation (1) can be written as
4 ( x 3) 2 x 1
1 – x = (2x – 1).
2
. After substituting in these values into the original
3
equation, we see that they are not the solutions. Therefore, the only solutions are x = 8 or
x = – 2.
Solving we have x = 0 or x =
2. Solving System Of Absolute Value Equations
Example 11: (1988 AMC) If x + x + y = 10 and x + y – y = 12, find x + y.
(A) – 2
(B) 2
(C)
18
5
(D)
22
3
(E) 22
Solution: (C).
We label the equations:
x + x + y = 10
(1)
x + y – y = 12
(2)
Case I: x 0.
When x 0, x + x =0. So (1) becomes y = 10. Substituting this y value into (2):
x + 10 – 10 = 12
solutions in this case.
x = 12 and this contradicts x 0. Therefore, there are no
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Case II: x > 0.
From (1), we have 2x + y = 10
(3)
If y 0, from (2) we get x = 12.
Substituting this x value into (3): 2 12 + y =10.
y = – 14 < 0 which contradicts y 0.
Therefore, y < 0.
2 x y 10
The original system of equations becomes
.
x 2 y 12
32
x 5 ,
Solving we get:
.
y 14
5
18
x+y=
.
5
Example 12: (1997 AMC 28) How many ordered triples of integers (a, b, c) satisfy
a b + c = 19 and ab + c = 97?
(A) 0
(B) 4
(C) 6
(D) 10
(E) 12
Solution: (E).
If c 0, then ab – a b = 78, so
(a – 1)(b – 1) = 79 or (a + 1)(b + 1) = 79.
Since 79 is prime, its only two factors are 1 and 79. Thus,
{a, b} can be {2, 80}, { –78, 0}, {0,78}, or{ – 80, – 2}.
Hence, a b = 78 or a b = 82, and from the first equation our initial hypothesis, it
follows that c < 0, which is a contradiction.
On the other hand, if c < 0, then ab + a b = 116, so
(a + 1)(b + 1) = 117 or (a – 1)(b – 1) =117.
Since 117 = 32 ∙ 13, we have the following cases:
{a, b} = {0, 116} yields c = – 97;
{a, b} = {2, 38} yields c = – 21;
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{a, b} = {8, 12} yields c = – 1;
{a, b} = { – 116, 0} yields c = – 97;
{a, b} = { – 38, – 2 } yields c = – 21;
{a, b} = { – 12, – 8} yields c = – 1.
Since a and b are interchangeable, each of these cases leads to two solutions, for a total of
12 solutions.
x y 1,
Example 13: Solve
x 2 y 3.
Solution:
(1)
(2)
From (1) we get x – y = 1 or x – y = – 1.
Therefore the original system of equations is equivalent to the following two systems of
equations:
x y 1,
(I)
x 2 y 3.
x y 1,
(II)
x 2 y 3.
Eliminating x in (I) by substituting in x = y + 1 into the second equation, we get
y 1 2 y = 3.
Solving we get y =
2
4
or y = .
3
3
Substituting these values into x = y + 1, we can find the values of x:
1 4
5 2
(x, y) = ( , ), ( , ).
3 3
3 3
5 2
1 4
Similarly for (II), we have (x, y) = ( , ), ( , ).
3 3
3 3
1 4
5 2
5 2
1 4
The solutions for the given problem are (x, y) = ( , ), ( , ), ( , ), ( , ).
3 3
3 3
3 3
3 3
Example 14: (Eighth IMO Problem 5) Solve the following system of equations:
a1 a2 x2 + a1 a3 x3 + a1 a4 x4 = 1,
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a2 a1 x1 + a2 a3 x3 + a2 a4 x4 = 1,
a3 a1 x1 + a3 a2 x2 + a3 a4 x4 = 1,
a4 a1 x1 + a4 a2 x2 + a4 a3 x3 = 1,
where a1, a2, a3, and a4 are mutually distinct real numbers.
1
Solution: x1 x 4
, x2 = x3 = 0.
a1 a4
Without loss of generality, assume that a1 > a2 > a3 > a4. Then we have
(1)
(a1 a2 ) x2 + (a1 a3 ) x3 + (a1 a4 ) x4 = 1,
(a1 a2 ) x1 + (a2 a3 ) x3 + (a2 a4 ) x4 = 1,
(a1 a3 ) x2 + (a2 a3 ) x2 + (a3 a4 ) x4 = 1,
(2)
(a1 a4 ) x1 + (a1 a4 ) x2 + (a3 a4 ) x3 = 1,
(4)
(3)
(1) – (2), (2) – (3), (3) – (4):
(a1 a2 )( x2 x3 x4 x1 ) = 0,
(a1 a3 )( x3 x4 x1 x2 ) = 0,
(a3 a4 )( x4 x1 x2 x3 ) = 0.
x2 x3 x4 x1 = 0,
x3 x4 x1 x2 = 0,
x4 x1 x2 x3 = 0.
(5)
(6)
(7)
(5) + (7): x1 = x4.
Substituting in x4 as x1 into (6) we get x2 = x3.
Substituting in x4 as x1 and x2 as x3 into (5) we have
2x2 = 0
x2 = 0.
x2 = x3 = 0.
1
.
a1 a4
1
Therefore when a1 > a2 > a3 > a4, the solutions are x4 x1
, x2 = x3 = 0.
a1 a4
Substituting these values into (1) and (4), we get x4 x1
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PROBLEMS
Problem 1: (1968 AMC) The sum of the real values of x satisfying the equality
x 2 2 x 2 is:
(A)
1
3
(B)
2
3
(C) 6
(D) 6
1
3
(E) 6
2
3
Problem 2: (1958 AMC) The symbol x means x if x is not negative and – x if x is not
positive. We may then say concerning the solution of
x 2+ x –6=0
That:
(A) there is only one root (B) the sum of the roots is +1
(C) the sum of the roots is 0 (D) the product of the roots is +4
(E) the product of the roots is – 6
Problem 3: Solve x 2 3x 2 – 2x = 2.
Problem 4: Solve: x 2 7 x 12 = x2 – 7x +12.
Problem 5: Find the integer solution(s) of xy – 2 x + y = 4.
Problem 6: What is the sum of all values of x for which x 3 3 x 1 3?
Problem 7: Solve: 1 x = 1 + x .
Problem 8: Find the real solutions of x: x 2 11x 10 = 2 x 2 x 45 .
Problem 9: Solve x2 – x – 1 =0.
(A)
1 5
.
2
(B)
1 5
2
(C)
1 5
1 5
or
2
2
Problem 10: Solve 2 x 1 + 3 = 4x.
97
(D)
1 5
2
50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
Problem 11: Solve x 3 2 x 4 + 1= 0.
Problem 12: (1990 AMC) The number of real solutions of the equation x 2 + x 3 =
1 is
(A) 0
(B) 1
(C) 3
(D) 3
(E) more than 3
Problem 13: What is the sum of roots of x2 – 2x – 5 x 1 + 7 = 0?
(A) – 2
(B) 0
(C) 2
(D) 4
Problem 14: Find the number of solutions to the equation x ∙
(A) 1
(B) 2
(C) 3
(D) 4
( x 1) 2 + 1 = 0.
Problem 15: Find m if x2 – 2 x + 2 = m has 3 real roots.
x 1 y 2 5
Problem 16: Solve the following system of equations
.
x 1 y 3
x2 2 y 2 6
Problem 17: Solve the following system of equations
.
2
x 2 2 y 4
Problem 18: How many sets of negative real solutions are there to the system of
x yz 2
equations:
?
x 1 y 1 z 1
(A) 5
(B) 4
(C) 2
(D) 1.
Problem 19: Solve for real x: x 1 + x 2 – x 3 = 4.
Problem 20: Solve for real x: x 4 x 2 6 = x 4 4 – x 2 2 .
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50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
SOLUTIONS
Problem 1: Solution: (E).
Method 1:
If x 2 or x – 2, then x + 2 and x – 2 are both non-negative or both non-positive. So the
given equation is equivalent to x + 2 = 2(x – 2) or, equally, – (x + 2) = – 2(x – 2).
Hence in this case, x = 6.
If – 2 < x < 2, then x + 2 is positive and 2(x – 2) is negative, so the given equation is
2
equivalent to x + 2 = – 2(x – 2), x = .
3
After substituting in these values into the original equation, we see that they are indeed
the solutions.
2
2
The required sum of all values of x satisfying the given equation is 6 +
=6 .
3
3
Method 2:
Since the absolute values of two real numbers are equal if and only if their squares are
equal, the given equation yields
x2 + 4x + 4 = 4(x2 – 4x + 4)
which, when simplified, gives the quadratic equation
3x2 – 20x + 12 = 0
20
2
with real roots. The sum of its roots is – (–
), or 6 .
3
3
Problem 2: Solution: (C).
Method 1:
If x > 0, x2 + x – 6 = 0; (x – 2)(x + 3) = 0;
x + 3 0; x = 2.
If x < 0, x2 + x – 6 = 0; (x – 3)(x + 2) = 0;
x – 3 0; x = – 2;
Method 2:
( x + 3)( x – 2 ) = 0; x + 3 0;
x = 2, i. e, x = 2 or – 2.
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Chapter 5 Solving Absolute Value Equations
Problem 3: Solution: 0, 5.
We know that x 2 3x 2 = 2 + 2x.
Case I:
x2 – 3x + 2 = 2 + 2x
x = 0 or x = 5.
Case II: x2 – 3x + 2 = – (2 + 2x)
No solution.
After substituting in x = 0 and x = 5 into the original equation, we see that they are indeed
the solutions.
Problem 4: Solve: x 2 7 x 12 = x2 – 7x +12.
Solution: x 3 or x 4.
We are given that x 2 7 x 12 = x2 – 7x +12.
We have x2 – 7x +12 0.
Solving this inequality gives us x 3 or x 4.
The solutions are all real numbers with x 3 or x 4.
Problem 5: Solution: (0, 4), (0, – 4), (1, 3), (1, – 3), (– 1, – 3), (–1, 3).
The original equation can be written as ( x + 1)( y – 2) = 2.
We know that x + 1 > 0.
Case I:
x 1 1,
y 2 2,
Case II:
x 1 2,
y 2 1,
(1)
(2)
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Chapter 5 Solving Absolute Value Equations
x 0,
x 0
Solving (1) we get:
or
.
y 4.
y 4
x 1
x 1,
Solving (2) we get
or
, or
y 3
y 3
x 1
, or
y 3
x 1
y 3.
The integer solutions are (x, y) = (0, 4), (0, – 4), (1, 3), (1, – 3), (– 1, – 3), (–1, 3).
Problem 6: Solution: 6.
Method 1: x 3 3 x 1 3
x 3 3 x 4.
(x – 3 + 3 – x + 4)( x – 3 + 3 – x – 4)[x – 3 – (3 – x) + 4][ x – 3 – (3 – x) – 4)] = 0
4 (– 4)(2x – 2)(2x – 10) = 0
(2x – 2)(2x – 10) = 0
x = 1 and x = 5.
After substituting in x = 1 and x = 5 into the original equation, we see that they are indeed
the solutions. Their sum is 1 + 5 = 6.
Method 2: x 3 3 x 1 3
x 3 x 3 4
2x3 4
x3 2.
Solving, we get x = 1 and x = 5
After substituting in x = 1 and x = 5 into the original equation, we see that they are indeed
the solutions. Their sum is 1 + 5 = 6.
Problem 7: Solution: x 0.
Squaring both sides of the given equation:
1 – 2x + x2 = 1 + 2 x + x2.
x = – x 0
x 0.
Problem 8: Solution: x1, 2 6 91 , and x3, 4
Case I:
x2 – 11x + 10 = 2 x2 + x – 45
5 130
.
3
x2 + 12x – 55 = 0
Case II:
101
x 6 91
50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
x2 – 11x + 10 = – (2 x2 + x – 45)
3 x2 – 10x – 35 = 0
The solutions are x1, 2 6 91 , and x3, 4
Problem 9: Solution:
x
5 130
3
5 130
.
3
1 5
.
2
This problem was solved in Example 16. Here we have two more ways to solve the
problem.
Method 1:
Case I: When x 0, x = x.
Thus, the original equation becomes
x2 – x – 1 = 0.
Solving this quadratic, we get:
1 5
x1 = –
2
1 5
x2 = –
(extraneous since x 0)
2
Case II:
When, x < 0, x = – x.
Thus, the original equation becomes
x2 + x – 1 = 0.
Solving this quadratic, we get:
1 5
x1 =
2
1 5
x2 =
(extraneous since x < 0)
2
Combining case I and case II, we see that the solutions are x =
Method 2:
102
1 5
.
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50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
The original equation can be written as x2 – 1 = x .
Since x 0, x2 – 1 0.
So x 1 or x – 1.
This means that the roots are not between – 1 and 1. So (A), (B), (C) can be eliminated,
giving us the answer (D).
Problem 10: Solution: x
5
.
6
Method 1:
Case I: x 1.
The original equation becomes: 2(x – 1) + 3 = 4x.
1
Solving for x: x (extraneous solution because x 1).
2
Case II: x < 1.
The original equation becomes: 2(1 – x) + 3 = 4x.
Solving for x: x
5
.
6
The solution is then x
5
.
6
Method 2:
The original equation can be written as 2 x 1 = 4x – 3.
From the definition of absolute values, we get:
1
x .
Case I: 2(x – 1) = 4x – 3
2
5
x .
Case II: 2(x – 1) = – (4x – 3)
6
1
5
5
After substituting in x and x into the original equation, we see that only x
2
6
6
is the solution.
Method 3:
The original equation can be written as 2 x 1 = 4x – 3.
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Chapter 5 Solving Absolute Value Equations
Squaring both sides: 4(x – 1)2 = (4x – 3)2 12x2 – 16x + 5 = 0.
1
5
Solving we get x or x .
2
6
1
5
5
After substituting in x and x into the original equation, we see that only x
2
6
6
is the solution.
Problem 11: Solution: x = 4 or x = 5 –
5.
Since 2x – 4 0, x 2.
Case I: 2 x < 3.
The original equation becomes: 3 – x –
Solving we get: x = 5 +
2 x 4 + 1 = 0.
5 (extraneous solution), or x = 5 –
Case II: 3 x
The original equation becomes: x – 3 –
5
2 x 4 + 1 = 0.
Solving we get: x = 2(extraneous solution), or x = 4
The solutions are x = 4 or x = 5 –
5.
Problem 12: Solution: (E).
The critical points of the absolute value equation x 2 + x 3 = 1 are 2 and 3. The
critical distance is then 3 – 2 = 1, which equals the right hand side of the equation. This
means that no matter what x we chose, x 2 + x 3 will always equal 1, so the
equation has infinitely many solutions.
Problem 13: Solution: (D).
The equation can be written as (x2 – 2x + 1) – 5 x 1 + 6 = 0.
Or x 1 2 – 5 x 1 + 6 = 0.
Hence x 1 = 2 or x 1 = 3.
Therefore x1 = – 1, x2 = 3, x3 = – 2, x4 = 4.
x1 + x2 + x3 + x4 = 4.
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50 AMC Lectures
Problem 14: Solution:
Chapter 5 Solving Absolute Value Equations
1 5
.
2
The equation can be written as x ∙ x 1 + 1 = 0
x ∙ x 1 = – 1,
(1)
x ∙ x 1 < 0,
x < 0,
x – 1 < 0.
Equation (1) becomes – x (x – 1) + 1 = 0.
1 5
x2 – x – 1 = 0.
x=
.
2
1 5
x=
> 0 (extraneous).
2
1 5
Therefore the solution is x =
.
2
Problem 15: Solution: 2.
The nonzero roots of x2 – 2 x + 2 = m come in pairs, so the number of nonzero roots of
x2 – 2 x + 2 = m must be even.
Since the equation has 3 roots, one of the roots must be 0.
Let x = 0 and substituting this value into x2 – 2 x + 2 = m, we get m = 2.
x 1
x 3
Problem 16: Solution:
and
.
y 5
y 5
We know that y – 3 = x – 1 0, so y 3. Then y – 2 = 0 and y – 2 = y – 2.
Therefore we have two cases:
Case I: When x 1, the original equations can be written as:
x 1 y 2 5
x 3
.
x 1 y 3
y 5
Case II: When x < 1, the original equations can be written as:
( x 1) y 2 5
x 1
.
( x 1) y 3
y 5
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50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
x 3
The solutions of the equations above as well as the original equation are
and
y 5
x 1
.
y 5
Note that we can double check to make sure that they are also the solutions to the original
equation by substituting in these values into the original equation.
x 6
x 6
Problem 17: Solution:
and
.
y 4
y 4
We know that 2y – 4 = x2 – 2 0, so y 2. Then y – 2 = 0 and y – 2 = y – 2.
Therefore we have two cases:
Case I: when x2 2, the original equations can be written as:
2
x 6
x 2 y 2 6
.
2
y
4
x
2
2
y
4
Case II: when x2 < 2, the original equations can be written as:
( x 2 2) y 2 6
No real solutions.
2
(
x
2
)
2
y
4
x 6
The solutions of the equations above as well as the original equation are
and
y
4
x 6
.
y
4
Note that we can double check to make sure that they are also the solutions to the original
equation by substituting in these values into the original equation.
Problem 18: Solution: (D).
Since we are looking for negative solutions, x < 0, y < 0, and z < 0.
Therefore x 1 < 0, y 1 < 0, and z 1 < 0.
The original equations can be written as:
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50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
x yz 2
x yz 2
1 x 1 y 1 z
x y z
Solving for x, we get: x = 2 or x = 1 (extraneous).
Therefore the solutions are x = y = z = 2.
x x2 2 .
Problem 19: Solution:
When x – 2, – (x – 1) – (x + 2) + (x – 3) = 4
–x – 4 = 4, x = – 8.
When–2 < x 1,
– (x – 1) + (x + 2) + (x – 3) = 4, x = 4. ( extraneous)
When 1 < x 3,
(x – 1) + (x + 2) + (x – 3) = 4, x = 2.
When x > 3,
(x – 1) + (x + 2) – (x – 3) = 4, x = 0 (extraneous).
The solutions are x1 = – 8, x2 = 2.
Problem 20: Solution: x
3 and x –
3.
Method 1:
( x 2 2)( x 2 3) = ( x 2 2)( x 2 2) – x 2 2 .
Since x2 + 2 >0, x 2 3 = x 2 2 – 1.
When 0 x2 2, 3 – x2 = 2 – x2 – 1 3 = 1. This is not true, so we can disregard this
case.
When 2 < x2 < 3, 3 – x2 = x2 – 2 – 1, x2 = 3. Since 3 is not in the given range, we can
also disregard this case.
When x2 3, x2 – 3 = x2 – 2 – 1 x2 – 3 = x2 – 3. This is true for all values in the
given range.
Therefore the solutions are x 3 and x – 3 .
Method 2:
We know that the equality of a b = a b holds if and only if both a and b are
nonnegative or nonpositive.
We are given that x 4 x 2 6 + x 2 2 = x 4 4 .
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50 AMC Lectures
Chapter 5 Solving Absolute Value Equations
Since x 2 2 is always nonnegative, then x 4 x 2 6 and x4 – 4 must both be
nonnegative.
4
2
x2 – x – 6 0
So x 3. This gives us
x 3 or x – 3 .
2
(x
+ 2)( x2 – 3) 0
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
BASIC KNOWLEDGE
1. Vieta’s Theorem
If x1 and x2 are two roots of a quadratic equation ax 2 bx c 0 , (a ≠ 0) ,
then
b
x1 x2
(1.1)
a
c
x1 x2
(1.2)
a
Proof:
Let x1 and x2 be the two roots of a quadratic equation
ax 2 bx c 0 , (a ≠ 0).
b b 2 4ac
b b 2 4ac
, x2
.
2a
2a
The sum of x1 and x2 is obtained by adding the two equations together:
x1
b b 2 4ac b b 2 4ac
2a
2a
2b
b
.
2a
a
x1 x2
The product of x1 and x2 is obtained by multiplying the two equations together:
x1 x2
b b 2 4ac b b 2 4ac
(b)2 ( b 2 4ac )2 4ac c
2 .
2a
2a
4a 2
4a
a
Note: When we derived Vieta’s Theorem, the roots could have been real or not.
However, if there is a problem stating that the roots are real (or positive), you must
consider 0 and a 0. If it is necessary, you need to check if 0 or not.
2. Useful Forms Of Vieta’s Theorem
x12 x22 ( x1 x2 )2 2 x1x2
(2.1)
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
b( x1 x2 ) 2c
a
2
b 2ac
x12 x22
a2
x12 x22
(2.2)
(2.3)
x13 x23 ( x1 x2 )[( x1 x2 )2 3x1x2 ]
(2.4)
x13 x23
b( x12 x22 ) c( x1 x2 )
a
(2.5)
x13 x23
3abc b3
a3
(2.6)
b( x1n 1 x2n 1 ) c( x1n 2 x2n 2 )
x x
a
(2.7)
( x1 x2 )2 ( x1 x2 )2 4 x1x2
(2.8)
n
1
n
2
x1 x 2 x1 x2 2 x1 x2
( x1 0, x2 0 )
1 1 x1 x2
b a
b
x1 x2
x1 x2
a c
c
1
1
2
2
x1 x2
b(
(2.10)
1 1
) 2a b( b ) 2a
b 2 2ac
x1 x2
c
c
c
c2
1
1 b3 3abc
x13 x23
c3
(2.11)
(2.12)
1
1
1 1
b[( ) 2 ( ) 2 ] a( )
1
1
x1
x2
x1 x 2
( )3 ( )3
x1
x2
c
(2.9)
110
(2.13)
50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
In all the above formulas, x1 and x2 represent the two roots of the quadratic equation
ax 2 bx c 0 .
3. Converse Of Vieta’s Theorem
If x1 and x2 satisfy the following:
b
c
x1 x2 , and x1 x2
a
a
then x1 and x2 are two roots of a quadratic equation ax 2 bx c 0 , (a ≠ 0).
This theorem can be used to construct a quadratic equation.
Proof:
ax 2 bx c a[ x 2
b
c
x ]
a
a
b
c
( x1 x2 ) , x1 x2 .
a
a
b
c
ax 2 bx c a[ x 2 x ] a[ x 2 ( x1 x2 ) x x1x2 ]
a
a
Substituting x1 for x into the left hand side of the above equation:
Since
ax12 bx1 c a[ x12 ( x1 x2 ) x1 x1x2 a[ x12 x12 x1x2 x1x2 ] 0.
So x1 is the root of ax 2 bx c 0 .
Similarly, we have ax22 bx2 c a[ x22 x22 x1x2 x1x2 ] 0 .
So x2 is also the root of ax 2 bx c 0 .
4. Generalized Vieta’s Theorem:
(1). Let x1, x2, and x3 be the roots for a 3-degree polynomial a0 x3 a1 x 2 a2 x1 a3 0 .
Then
a
x1 x 2 x 3 1
(4.1)
a0
a
x1 x 2 x2 x3 x3 x1 2
(4.2)
a0
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50 AMC Lectures
x1 x2 x3
Chapter 6 Vieta’s Theorem and Applications
a3
a0
(4.3)
(2). Let x1, x2, x3, and x4 be the roots for a 4-degree polynomial
a0 x 4 a1x3 a2 x 2 a3 x1 a4 0 .
Then
a
x1 x 2 x3 x4 1
a0
a
x1 x2 x1 x3 x1 x 4 x2 x3 x2 x4 x3 x 4 2
a0
a
x1 x2 x3 x1 x2 x4 x1 x3 x4 x2 x3 x4 3
a0
a
x1 x2 x3 x 4 4
a0
(4.4)
(4.5)
(4.6)
(4.7)
Example 1: If 2 is a root of 4 x 2 11x 6 0 , find the second root.
Solution:
Method 1:
Let x1 be the unknown, second root.
Using Vieta’s Theorem, the product of the two roots is equal to
6
3
2 x1
x1 .
4
4
Method 2:
Let x1 be the unknown root. Using Vieta’s Theorem, the sum of the two roots is equal to
x1 2
11
4
x1
Example 2: Find
1
3
3
.
4
1
3
if and are two roots of the equation 2 x 2 5x 3 0 .
Solution:
Method 1:
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
By Vieta’s Theorem, the sum and product of the two roots of the quadratic equals
5
3
, .
2
2
1
1
1 1
b[( ) 2 ( ) 2 ] a( )
1
1
x1
x2
x1 x 2
We also know by (2.13) that ( ) 3 ( ) 3
.
x1
x2
c
1
b
5
,
c 3
2
1 2
1 2 1 1
1 1
5
2 37
( ) ( ) 2 ( )2 2( )
3
3
9
Since
1
1
1
( ) ( )
3
3
5
37
5
2
9
3 215 7 26 .
3
27
27
Method 2:
We know that a = 2, b = 5 and c = 3.
1
1 b3 3abc
We also know by (2.12) that 3 3
.
x1 x2
c3
Therefore,
1
3
1
3
b3 3abc 53 3 2 5 (3) 125 90 215
26
=
7
3
3
c
(3)
27
27
27
5. A Different Form Of Vieta Theorem
x2 x1
b 2 4ac
a
(5.1)
If a > 0, then (1) can be simplified as
x2 x1
b 2 4ac
a
a
(5.2)
Proof:
According to Vieta’s Theorem, we have
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
b
c b2 4ac
.
( x2 x1 )2 ( x1 x2 )2 4 x1 x2 ( )2 4
a
a
a2
Hence:
b 2 4ac
.
a
a
x1 x2
x1 x2
b b 2 4ac b b 2 4ac
b 2 4ac
.
2a
2a
a
Example 3: (1970 AMC 14) Consider x 2 px q 0 , where p and q are positive
numbers. If the roots of this equation differ by 1, then p equals
(A) 4q 1 (B) q 1
(C) 4q 1
(D) q 1
(E) 4q 1
Solution:
Method 1 (Official solution):
Call the roots r and r + 1. Their sum is p 2r 1 and their product is q r (r 1) . Thus
p 1
p 1
r
, r 1
,
2
2
and
( p 1)( p 1) p 2 1
q r (r 1)
;
22
4
p 2 4q 1, and p 4q 1 .
Method 2 (our solution):
Let and be the two roots of the equation.
By Vieta’s Theorem:
p
q
(1)
(2)
1
(3)
Squaring both sides of (3):
2
1
2 2 2 1
Substituting (1) and (2) into (4):
( p) 4q 1
2
114
( )2 4 1
p 4q 1 .
2
(4)
50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Since p is positive, p 4q 1 .
Method 3 (our solution):
b 2 4ac
x2 x1
a
a
Since p is positive, p 4q 1 .
1=
p2 4 1 q
p 2 4q 1
1
6. Applications Of Vieta’s Theorem
(1). Problems with quadratic equations:
Example 4: If the roots of the equation 2kx2 (3k 2) x 2k 1 0 differ by 1, find the
positive value of k.
Solution:
x2 x1 1 , and we know by (5.1) that x2 x1
b 2 4ac
.
a
Thus,
[(3k 2)]2 4 2k (2k 1)
1
a
2k
11k 2 20k 4 0 . k1 2 ,
2
.
11
Since k is positive, k = 2.
k2
(2). Distance between two x-intercepts of a parabola
The quadratic function y ax 2 bx c meets the x-axes at two points A(x1, 0) and B(x2,
0). x1 and x2 satisfy ax 2 bx c 0 .
The distance between A and B is:
AB x2 x1
b 2 4ac
.
a
a
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Example 5: The quadratic function y 3 px 2 6 px 3 p 2 intersects the x-axes at two
points. Find p such that the distance between two points of intersection is
6
.
3
Solution:
(6 p) 2 4 3 p(3 p 2)
6
.
3p
3p
3
Solve for p: p = 4.
(3). Simplification of roots testing
Example 6: Show that x1
5 43
5 43
and x2
are two roots of the equation
9
9
9 x 2 10 x 2 0 .
Solution:
By Vieta’s Theorem, we have:
5 43 5 43 10
x1 x2
9
9
9
5 43 5 43 18
2
x1 x2
.
9
9
81
9
By the converse of Vieta’s Theorem, x1 and x2 are the two roots of the given equation.
(4). Determination of the signs of roots
Example 7: Determine the sign of the roots of 6 x 2 5x 1 0 without actually solving
for the roots.
Solution:
Since the quadratic’s discriminant 52 4 6 1 0 , the equation has two real roots.
c 1
Since x1 x2 0 , the product of the roots is positive, so the sign of the two roots
a 6
must be the same.
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
b
5
, the sum of the roots is negative. Since the sign of the two roots
a
6
are the same, the sign of the two roots is negative.
Since x1 x2
(5). Finding values
Example 8: Find
b a
2
2
2 2 if a b, a 3a 1,and b 3b 1.
a b
Solution:
Since a2 3a 1
can bewritten as a2 3a 1 0 and b2 3b 1 can be written as
2
b 3b 1 0, we know that a and b are two roots of the quadratic equation
x 2 3x 1 0. The discriminate of this quadratic equation (3)2 4 (1) 13 0 ,
so the two roots are also real.
Hence, by Vieta’s Theorem, a b 3 and ab 1 .
Therefore,
b a a3 b3
a3 b3 (a b)(a2 ab b2 ) (a b)[(a b) 2 3ab] 3(32 3) 36.
a2 b2
(ab) 2
(6). Solving a system of equations
x xy y 11
Example 9: Solve 2
2
x y xy 30
Solution:
The original system of equations can be factored into
( x y ) xy 11
( x y ) xy 30
By Vieta’s Theorem we know that x + y and xy are two roots of the quadratic equation
z 2 11z 30 0 .
Solve for z: z1 6 , z2 5.
So x + y = 6 and xy = 5 or x + y = 5 and xy = 6.
From Vieta’s Theorem, the values above tell us that x and y are two roots of the quadratic
equation
m2 5m 6 0 or m2 6m 5 0 .
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The solutions to the system of equations are
x3 2,
x1 5,
x2 1,
x4 3,
y1 1,
y2 5, y3 3,
y4 2.
(7). Geometry problems
Example 10: As shown in the figure, circle O of radius r is inscribed to
∆ABC at D, E, and F. C =60. The length of the side opposite to C is
c 3 . Find the range of r.
Solution:
Let the lengths of the sides opposing A and B be a and b, respectively. Connect
segments OE, OF, and OC. OC is the angle bisector of C, so
OCE = OCF = 30 and in right triangles OEC and OFC,
CE CF r 3 .
AC BC AB CE CF
Substituting in the known variables, the equation above can be written as
a b 3 2 3r .
1
1
We also know that SABC ab sin 60 r (a b c) .
2
2
Simplifying yields ab 4r (1 r ) .
With our knowledge of the sum and product of a and b, and Vieta’s Theorem, we see that
a and b are two roots of the quadratic equation x 2 ( 3 2 3r ) x 4r (1 r ) 0 .
The discriminant of this quadratic must be greater than or equal to 0 in order for the
quadratic to attain real number solutions, so
( 3 2 3r )2 16r (1 r ) 0 .
Solving, we get
3
1
r .
2
2
1
Since r > 0, the range of r is 0 r .
2
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Chapter 6 Vieta’s Theorem and Applications
(8). Proving problems
Example 11: Show that there is one and only one of real numbers x, y, and z exceeding
3
4 if x y z 0 and xyz 1 .
Solution:
Since the product of three numbers x, y, and z, is positive and their sum is equal to 0, only
one among the three is positive, while the other two are negative.
Let z be positive. We have the following system of equations:
x y z
1
xy z
1
By Vieta’s Theorem, x and y are the two roots of the quadratic equation m2 zm 0 .
z
The discriminant of this quadratic must be greater than or equal to 0 in order for the
quadratic to attain real number solutions, so
1
z2 4 0
z3 4.
z3 4 0
z
Since x and y are negative, among these three numbers, only one exceeds 3 4 . QED.
7. Applications Of The Converse Of Vieta’s Theorem
If a problem contains the sum and the product of two numbers, the converse of Vieta’s
Theorem can be used to construct a quadratic equation directly.
We have included examples demonstrating the use of the converse of the theorem above;
however, here are a few more.
Example 12: Solve for real x, y, and z:
x y 2
2
xy z 1
Solution:
Since x y 2 and xy 1 z 2 , we know that x and y are two real roots of
(1)
t 2 2t 1 z 2 0
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Chapter 6 Vieta’s Theorem and Applications
The discriminant of this quadratic must be greater than or equal to 0 in order for the
quadratic to attain real number solutions, so
4 (1 z 2 ) 0
z2 0
But we know that z 2 0 , so z = 0.
Substituting z = 0 into (1), we have t 2 2t 1 0 .
Solve for t: t1 t2 1 . Therefore, x = y =1.
The solutions:
x y 1
.
z 0
If the problem contains the sum of two numbers, we can find the product of these two
numbers, and then construct the quadratic equation.
Similarly, if the problem contains the product of two numbers, we can find the sum of
these two numbers, and then construct the quadratic equation. See the example below.
Example 13: Find the greatest real value of z such that x + y + z = 5 and xy yz zx 3 .
x and y are real numbers.
Solution:
x y 5 z
(1)
xy 3 z( x y) 3 z(5 z) z 5z 3
2
(2)
From (1) and (2), we know that x and y are two real roots of the quadratic
t 2 (5 z )t z 2 5z 3 0 .
The discriminant of this quadratic must be greater than or equal to 0 in order for the
quadratic to attain real number solutions.
Therefore (5 z )2 4( z 2 5z 3) 0 .
3z 2 10 z 13 0 .
13
Solve for z: 1 z .
3
13
The greatest value of z is
.
3
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Chapter 6 Vieta’s Theorem and Applications
Sometimes the problem does contain the sum (or product) of two numbers and we are
able to find the product (or sum) of these two numbers. However, we may still have
difficulty solving the problem. In that case, we can create new variables based off of our
given variables and find the sum or product of the two new variables and construct a
quadratic equation. See the example below.
Example 14: Show that 1 a b
4
8
and a 2 b 2 1 if a + b + c = 1, and
3
9
a 2 b 2 c 2 1.
Solution:
Let a c x and b c y . Then x y 0 .
x y (a b c) 3c 1 3c
(1)
Since (a b c) (a b c ) 2(ab bc ca) 0 ,
2
2
2
2
ab c(a b) c(1 c) c 2 c .
xy (a c)(b c) ab c(a b) c 2 3c 2 2c
(2)
From (1) and (2) we know that x and y are two distinct positive roots of
t 2 (1 3c)t 3c 2 2c 0
Therefore we have:
(1 3c) 2 4(3c 2 2c) 0
1 3c 0
3c 2 2c 0
1
Solving, we have c 0 .
3
Since a b c 1, then a b 1 c . Therefore 1 a b
Since a 2 b2 c 2 1, then a 2 b2 1 c 2 and 0 c 2
Therefore
8
a 2 b2 1 .
9
121
1
9
4
.
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
If the problem contains only the sum (or only the product) of two numbers, we can find
the hidden relationship for the product (or for the sum), and then construct the quadratic
equation. See the example below.
Example 15: Find the range of the function y 2 x x 1 .
Solution:
Squaring both sides of the given function y 2 x x 1 and rearranging a few
terms, we get:
1
2 x x 1 ( y 2 1) .
2
So far, we see the product of two numbers, 2 x and
hidden relationship for the sum.
Note that ( 2 x )2 ( x 1)2
1.
x 1 ; however there exists a
1
( 2 x x 1) 2 ( 2 x ) 2 2( 2x 1)( x 1) ( x 1) 2 1 2 (y 2 1) y
2
Thus,
2 x x 1 1.
So
2 x and
x 1 are two nonnegative roots of the quadratic equation
1
t 2 yt ( y 2 1) 0 .
2
1 2
2
( y ) 4 2 ( y 1) 0
y 0
y2 1 0
Solving we get: 1 y 2 .
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If the problem contains neither the sum nor the product of two numbers, we might be able
to calculate them and then construct the quadratic equation. See the example below.
Example 16: If the sum of two squares of the real roots for the quadratic equation
1
2 x 2 ax 2a 1 0 is 7 . The value of a is
4
(A) – 11 or 3
(B) – 11
(C) 3
(D) 5
Solution: (C)
Let x1 and x2 be the roots of the quadratic equation. By Vieta’s Theorem,
2a 1
x1 x2
2
a
x1 x2
2
a
2a 1 1 2
x12 x22 ( x1 x2 )2 2 x1 x2 ( )2 2
(a 8a 4) .
2
2
4
1 2
1
(a 8a 4) 7
(a 11)(a 3) 0
4
4
a = – 11 or a = 3
a 2 8(2a 1) a 2 16a 8.
When a = – 11, (11)2 16 (11) 8 0.
Since the discriminant is negative, the equation will not have real roots when a = – 11, so
therefore a = 3.
Note that when we deal with the real roots of a quadratic equation, we must consider the
discriminant to avoid making mistakes.
Example 17: Find the greatest possible value of x12 x22 if x1 and x2 are two real roots of
x 2 (k 2) x (k 2 3k 5) 0 . k is real.
5
(A) 19
(B) 18
(C) 5
9
Solution: B.
(D) not exist
By Vieta’s Theorem:
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Chapter 6 Vieta’s Theorem and Applications
x12 x22 ( x1 x2 )2 2 x1x2 (k 2)2 2(k 2 3k 5)
(k 5)2 19
Since the equation has two real roots, the discriminant of the quadratic equation must be
greater than or equal to 0, so
(k 2)2 4(k 2 3k 5) 0
Or 3k 2 16k 16 0 .
4
Solving this inequality gives: 4 k .
3
2
In this range, the greatest value of x1 x22 can be achieved by letting k = – 4.
So the greatest possible value of x12 x22 18 (not 19!).
Example 18: (1976 AMC 30) How many distinct ordered triples (x, y, z) satisfy the
equations
x + 2y + 4z = 12
xy + 4yz + 2xz = 22
xyz = 6?
(A) none
(B) 1
(C) 2
(D) 4
(E) 6
Solution: (E).
Method 1 (official solution):
We observe that we can find a system of symmetric equations by the following change of
variables:
1
z w
x = 2u,
y = v,
(1)
2
This substitution yields the transformed system
u + v + w = 6,
uv + vw + uw = 11,
(2)
uvw = 6.
Consider the polynomial p(t ) (t u)(t v)(t w), where (u, v, w) is a solution of
equation (2).
Then
p(t ) t 3 6t 2 11t 6,
(3)
And u, v, w are the solutions of p(t) = 0.
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Conversely, if the roots of p(t) = 0 are listed as a triple in any order, this triple is a
solution to equation (2).
It is not hard to see that p(t) = 0 has three distinct solutions. In fact,
p(t ) (t 1)(t 2)(t 3).
So the triple (1, 2, 3) and each of its permutations satisfies the equation (2).
Since the change of variables (1) is one-to-one, the original system has 6 distinct
solutions (x, y, z):
3
3
1
1
(2, 3, 1), (2,2, ) , (4, 1, ) ,(4, 3, ), (6, 1, 1) or (6, 2, ).
2
2
2
2
Method 2 (our solution):
Let a = x, b = 2y, and c = 4z.
The system of equations becomes:
a + b + c = 12
ab + bc + ac = 44
abc = 48
a, b, and c are the solutions to the following equation: t 3 12t 2 44t 48 0
We observe that 2 is a solution of the cubic above.
By the long division, we can factor the cubic into:
(t 2)(t 2 10t 24) 0
Or (t 2)(t 4)(t 6) 0
So the solutions are (2, 4, 6). This can be permutated in 6 different ways, as shown
below.
a2
b4
c6
a2
x2
b
y 2
2
c 3
z
4 2
b6
c4
125
x2
b
y 3
2
c
z 1
4
50 AMC Lectures
a4
b2
c6
a6
b2
c4
Chapter 6 Vieta’s Theorem and Applications
x4
b
y 1
2
c 3
z
4 2
a4
b6
c2
x6
b
y 1
2
c
z 1
4
a6
b4
c2
x4
b
y 3
2
c 1
z
4 2
x6
b
y 2
2
c 1
z
4 2
Thus, the original system has 6 distinct solutions.
Example 19: (1974 USAMO) Determine all the roots, real or complex, of the system of
simultaneous equations
x + y + z = 3,
2
x + y2 + z2 = 3,
x3 + y3 + z3 = 3.
Solution: Let x, y, z be the roots of the cubic equation
t 3 at 2 bt c 0.
Then, a x y z 3,
2b 2( yz zx xy ) ( x y z)2 x 2 y 2 z 2 6.
and
The cubic is satisfied by x, by y and by z; if we sum these cubics we obtain
x3 y3 z 3 a( x 2 y 2 z 2 ) b( x y z) 3c 0
3 3a 3b 3c 0.
Since a = b = 3, we can solve for c: c = 1.
The cubic equation becomes (t – 1)3 = 0, and so the only solution is (x, y, z) = (1, 1, 1).
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PROBLEMS
Problem 1: Find the quadratic equation with two roots
4
7
and .
5
2
Problem 2: If and are two real roots of quadratic equation 2 x 2 7 x 2 0 , find a
1
1
quadratic equation with two roots of
and .
Problem 3: Find k if the sum of the reciprocals of two roots of the equation
8
4 x 2 8x k 0 is .
3
Problem 4: Find 3 + 3 if and are two roots of the equation 2 x 2 5x 3 0 .
Problem 5: (1955 AMC) Two numbers whose sum is 6 and the absolute value of whose
difference is 8 are roots of the equation:
(A) x 2 6 x 7 0 (B) x 2 6 x 7 0 (C) x 2 6 x 8 0
(D) x 2 6 x 8 0 (E) x 2 6 x 7 0
Problem 6: Determine the signs of the roots of 4 x 2 7 x 2 0 without actually solving
for the roots.
Problem 7: Show that
3
2 p q 2 for real positive numbers p and q if p3 q3 2 .
3
3
x y 468,
Problem 8: Solve: 2
2
x y xy 420.
(1)
(2)
Problem 9: Find the smallest positive value of x + y if x + y = x · y.
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Chapter 6 Vieta’s Theorem and Applications
Problem 10: Show that c 2 5 if
ab = 10
a 2 b2 c 2
(1)
(2)
ax
b x
2 if a > b > 0.
b x
ax
Problem 11: Solve
Problem 12: Find the range of a if a, b, and c are real numbers satisfying
2
a bc 8a 7 0
2
2
b c bc 6a 6 0
(1)
(2)
Problem 13: (1978 AMC 13) If a, b, c, and d are non-zero numbers such that c and d are
the solutions of x 2 ax b 0 and a and b are the solutions of x 2 cx d 0 , then a +
b + c + d equals
1 5
(A) 0 (B) – 2
(C) 2
(D) 4
(E)
2
Problem 14: (2004 North Carolina State Mathematics Contest) Two of the roots of the
equation 2x3– 3x2 + px + q = 0 are 3 and –2. What is the third root?
Problem 15: Solve
x y z 9,
xy yz zx 26,
xyz 24.
Problem 16: Find
1
3
1
3
if and are two roots of the equation x 2 x 3 0
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Problem 17: How many distinct ordered triples (x, y, z) satisfy the equations
(1)
x y z 1,
2
2
2
(2)
x y z 21,
x 3 y 3 z 3 55.
(3)
Problem 18: (1975 AMC #27) If p, q and r are distinct roots of x3 x 2 x 2 0, then
p3 q3 r 3 equals
(A) – 1
(B) 1
(C)3
(D) 5
(E) none of these
Problem 19: (1996 AIME) Suppose that the roots of x3 3x2 4 x 11 0 are a, b, and
c, and that the roots of x3 rx2 st t 0 are a + b, b + c, and c + a. Find t.
Problem 20: (1984 USAMO) The product of two of the four roots of the quartic equation
x 4 18x3 kx2 200 x 1984 0 is – 32. Determine the value of k.
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Chapter 6 Vieta’s Theorem and Applications
SOLUTIONS TO PROBLEMS
Problem 1: Solution: 10 x 2 27 x 28 0.
Let the equation be x 2 px q 0 .
4
7
27 27
4
7
28
By Vieta’s Theorem, p [ ( )] [ ]
and q ( ) .
5
2
10 10
5
2
10
27
28
x
0 or 10 x 2 27 x 28 0.
The quadratic equation becomes x 2
10
10
Problem 2: Solution: x 2 7 x 4 0 .
Let the equation be x 2 px q 0 .
By Vieta’s Theorem,
1
1
7 7
p [( ) ( )] [( )
] [ ] 7
2 2
1
1
1
q ( )( )
2 11 2 4.
Therefore the quadratic equation is x 2 7 x 4 0 .
Problem 3: Solution: 3.
Let the two roots be x1 and x2.
1 1 8
.
x1 x2 3
k
By Vieta’s Theorem, x1 x2 and x1 x2 2 .
4
1 1 x1 x2 2 8
Since
,
. k = 3.
k 3
x1 x2
x1 x2
4
7
Problem 4: Solution: 26 .
8
By Vieta’s Theorem,
5
3
, .
2
2
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50 AMC Lectures
3
3
Chapter 6 Vieta’s Theorem and Applications
b( 2 2 ) c( )
37
5
5
(3)( )
2 215 26 7 .
3 3 4
2
8
8
Problem 5: Solution: x 2 6 x 7 0
Let and be the two numbers.
By Vieta’s Theorem:
6
(1)
8
(2)
(1)2 – (2)2: 4 28
7
Therefore and be the two roots of the quadratic x 2 6 x 7 0 .
Problem 6: Solution: Negative, Positive.
Since 72 4 4(2) 0 , the equation has two real roots.
By Vieta’s Theorem,
c
1
x1 x2 0 , so the product is negative and thus the two roots have opposite
a
2
signs. One root is negative and the other is positive.
b
7
Since x1 x2 , the negative root has a greater absolute value than the positive
a
4
root.
Problem 7: Solution:
Let
p+q=k>0
(1)
Since
2 p3 q3 ( p q)[( p q)2 3 pq] k (k 2 3 pq) ,
k3 2
pq
(2)
3k
From (1) and (2) and Vieta’s Theorem, we know that p and q are two positive roots of
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
k3 2
0.
3k
2 4(k 3 2)
0
k
So
3k
k 3 2 0
t 2 kt
Solve for k:
3
2 k 2 or
3
2 p q 2 . QED.
x 5,
x 7,
Problem 8: Solution:
or
y 7;
y 5.
We are given the following system of equations:
3
3
x y 468,
2
2
x y xy 420.
(1)
(2)
Substituting x3 y 3 ( x y)[( x y)2 3xy ] into (1), we get
( x y)[( x y)2 3xy ] 468
(3)
Let x y u, xy v. Substituting these values into (3) and (2), we get the following
system of equations:
u 3 3uv 468,
uv 420.
Solving, we get: u 12, v 35.
Then
x y 12,
xy 35.
x 5,
x 7,
Solving we get:
or
y 7;
y 5.
Problem 9: Solution: 4.
Let x + y = x · y = k > 0. x and y are two real roots of t 2 kt k 0 .
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Therefore (k )2 4k 0 .
k 4 so the smallest positive value of x + y, or (x y)min , equals 4.
Problem 10: Solution:
We are given the following system of equations:
ab = 10
a 2 b2 c 2
From (2), we have (a b)2 c 2 2ab
Substituting (1) into (3) gives us
(a b)2 c 2 20
or
(1)
(2)
(3)
a b c 2 20
(4)
From (1) and (4), we know that a and b are two real roots of the quadratic
t 2 c 2 20 t 10 0 .
Therefore we have: ( c 2 20 )2 40 0 .
Simplifying gives us
c 2 5.
c 2 20
QED.
1
Problem 11: Solution: x (a b)
2
ax b x
=1.
b x
ax
Notice that
By Vieta’s Theorem,
t 2 2t 1 0
ax
1
b x
ax
b x
and
are two roots of the quadratic equation
b x
ax
t1 t2 1 .
1
x ( a b) .
2
1
To double check, we can substitute x (a b) into the original equation, and see that it
2
is indeed the root of the original equation.
Therefore
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Problem 12: Solution: 1 a 9.
We are given the following system of equations:
2
a bc 8a 7 0
2
2
b c bc 6a 6 0
(1)
(2)
From (1) we have
bc a 2 8a 7
(3)
From (2) – (1) we have
b c (a 1)
(4)
From (3) and (4) and Vieta’s Theorem, we know that b and c are the two roots of
t 2 (a 1)t a 2 8a 7 0 .
Since a, b, and c are real numbers, so the discriminant the quadratic must be positive, or
(a 1)2 4(a 2 8a 7) 0 .
Solving the inequality, we have 1 a 9.
Problem 13: Solution: (B).
(Official Solution)
Since the constant term of a monic quadratic equations is the product of its roots,
b = cd,
d = ab.
Since the coefficient of x in a monic quadratic equation is the negative of the sum of its
rots, – a = c + d, – c = a + b; thus a + c + d = 0 = a + b + c, and b = d. But the equations
b = cd and d = ab imply, since b = d ≠ 0, that 1 = a = c. Therefore, b = d = – 2, and a + b
+ c + d = – 2.
1
.
2
By Vieta’s Theorem, the sum of the three roots is equal to
3
x1 + x2 + x3 = .
2
We are given that two of the roots are 3 and – 2, so substituting in these values, we get
1
3
3 – 2 – x3 =
x3 = .
2
2
Problem 14: Solution:
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Problem 15: Solution:
x 4,
x 4,
x 2,
y 2,
y 3,
y 3,
z 3;
z 2;
z 4;
x 2,
y 4,
z 3;
x 3,
y 2,
z 4;
x 3,
y 4,
z 2.
By Vieta’s Theorem, x, y, and z are the three roots of the equation below
u 3 9u 2 26u 24 0
(1)
Factoring gives us:
(u 2)(u 4)(u 3) 0.
u1 = 2,
u2 = 4,
u3 = 3.
We have the six following solutions:
x 4,
x 4,
x 2,
y 2,
y 3,
y 3,
z 3;
z 2;
z 4;
x 2,
y 4,
z 3;
x 3,
y 2,
z 4;
x 3,
y 4,
z 2.
Problem 16: Solution:
1
3
1
3
10
. (Similar to Example 2. Using formula (2.12) or (2.13).
27
10
.
27
x 1,
Problem 17: Solution: y 2,
z 4;
x 1 .
y 4,
z 2;
x 2,
y 1,
z 4;
135
x 2,
y 4,
z 1;
x 4
y 1,
z 2;
x 4,
y 2,
z 1 .
50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
Subtracting equation (2) from the square of equation (1), we get
(1)2 – (2):
xy yz zx 10.
(4)
We know that
x3 y3 z 3 3xyz ( x y z )( x 2 y 2 z 2 xy zy zx)
55 3xyz (21 10)
xyz 8.
So x, y, and z are three roots of the equation u 3 u 2 10u 8 0 .
Solving, we get: u1 = – 1, u2 = – 2, u3 = 4. Since we are looking for the number of
distinct ordered triples, we can permuate these three values to obtain the following six
solutions:
x 1,
y 2,
z 4;
x 1 .
y 4,
z 2;
x 2,
y 1,
z 4;
x 2,
y 4,
z 1;
x 4
y 1,
z 2;
x 4,
y 2,
z 1 .
Problem 18: Solution: (E).
(Official Solution):
If p, q, r are roots, then the polynomial can be factored as follows:
x3 x 2 x 2 ( x p)( x q)( x r )
x3 ( p q r ) x 2 ( pq pr qr ) x pqr.
Equating coefficients of like powers of x, we find
p + q + r = 1, pq + pr + qr = 1, pqr = 2.
In looking for the sum of the cubes of the roots of a cubic equation, let us use the fact that
each root satisfies the equation:
p3 p 2 p 2 0
q3 q 2 q 2 0
r3 r 2 r 2 0 .
Adding these, we obtain
(*) p3 q3 r 3 ( p 2 q 2 r 2 ) ( p q r ) 6 0.
We saw that p + q + r = 1 and shall determine the sum of the squares of the roots by
squaring this relation:
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50 AMC Lectures
Chapter 6 Vieta’s Theorem and Applications
( p q r )2 P2 q 2 r 2 2( pq pr qr ) 1
p 2 q 2 r 2 2(1)
1
2
2
2
p q r
1.
Substituting this into (*), we obtain
p3 q3 r 3 1 1 6 4.
Problem 19: Solution: 23.
Applying Vieta’s Theorem to the x3 3x2 4 x 11 0 :
a+b+c=–3
ab +bc +ca = 4
abc = 11
So a b 3 c , b c 3 a , and c a 3 b .
Applying Vieta’s Theorem to x3 rx2 st t 0 :
t (a b)(b c)(c a)
(3 c)(3 a)(3 b)
27 9(a b c) 3(ab bc ca) abc.
t 27 27 12 11 23
Problem 20: Solution: 86.
If r1, r2, r3, and r4 are the four roots, then for some pairing of the roots, r1r2 =
–32 , and then
r r r r 1984
r3r4 1 2 3 4
62.
r1r2
32
Consequently, for some p and q,
x 4 18 x3 kx2 200 x 1984 ( x r1 )( x r2 )( x r3 )( x r4 )
( x 2 px 32)( x 2 qx 62).
Equating like coefficients on both sides of the identity, we find
p + q = 18,
– 62p + 32q = 200, and
k = 62 + pq – 32.
Solving the first two equations for p, q, we get p = 4, q = 14. Finally,
k = 62 + 4 · 14 – 32 = 86.
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BASIC KNOWLEDGE
1. Definition Of A Square Number
A square number, also called a perfect square, is a number of the form n2, where n is an
integer. The square numbers for n = 0, 1, 2, 3, 4, 5, 6, and 7 are 0, 1, 4, 9, 16, 25, 36, and
49, respectively.
The nth nonsquare number of all the positive integers an is given by the formula:
1
an n n ,
2
where x is the floor function. The first few nonsquare numbers are 2, 3, 5, 6, 7, 8, 10,
11.
2. Properties And Theorem
Property 1: Any even square number can be written as (2n)2 = 4n, which is a multiple of
4.
Any odd square number can be written as (2n + 1)2 = 4n2 + 4n + 1, which is a multiple of
4 plus 1.
Property 2: There are no square numbers between two consecutive square numbers.
Example 1: Show that if n2 < a < ( n + 1 )2, then a is not a square number.
Solution:
Suppose there does exist a square number, a, between n2 and (n +1)2. We can denote as a
as b2, where b is an integer. Then we have:
n2 < b2 < ( n + 1 )2
n<b< n+1
Since n and n + 1 are two consecutive numbers, and b must be greater than n and less
than n + 1, b cannot be an integer. This contradicts our original statement that b is integer
and therefore, there does not exist such a square number a.
Property 3: a2b is a square number only and if only b is a square number.
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Example 2: How many two-digit positive integers are there such that the sum of the twodigit positive integers and the number formed by reversing the digits of the two-digit
positive integers is a square number?
Solution:
Let the two-digit number be ab .
We are given the following equation:
10a + b +10b + a = m2, where m is a positive integer.
Simplifying, we get:
11(a + b) = m2
In order for 11(a + b) to be a square number, a + b must be equal to 11x2.
Since a and b are digits, the maximum a + b can be is 18, so the only possible value for a
+ b is 11.
Thus, we have the following possibilities:
(a, b) = (2, 9), (3, 8), (4,7), (5,6), (6,5), (7,4) ,(8,3), and (9,2).
The eight two-digit positive integers are 29, 38, 47, 56, 65, 74, 83, and 92.
Property 4: The last digit of a square number can only be 0, 1, 4, 5, 6, or 9.
The last digit of a square number can not be 2, 3, 7, or 8.
Example 3: If
x 2 y 2 is a positive integer, which of the following is possible?
(A) x = 25530, y = 29464
(C) x = 15123, y = 32477
(B) x = 37615, y = 26855
(D) x = 28326, y = 28611
Solution:
If x 2 y 2 is integer, x2 + y2 must be a square number and its last digit must be one of
the following: 0, 1, 4, 5, 6, 9.
Due to this fact, (C) and (D) can be excluded.
For option (B), x = 37615, y = 26855, the last two digits of x2 + y2 is 50. For any square
number, if the last digit is 0, the tens digit must also be 0. Therefore, x2 + y2 is not a
square number in option (B). We can eliminate choice (B), and the resulting answer is
then (D).
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Property 5: The number of factors of a square number is odd.
Property 6: Any integer, when divided by 4, will have the remainder 0, 1, 2, or 3. In
other words, any integer can be written in one of the following forms 4k, 4k + 1, 4k + 2,
4k + 3.
Any square number has one of the following forms: 4k, 4k + 1.
Any integer of either of the following forms: 4k + 2, 4k + 3 is not a square number.
Example 4: [MOSP 1998] Show that the sum of the squares of 4 consecutive integers is
not a perfect square.
Solution:
Let the 4 consecutive integers be n – 1, n, n + 1, and n + 2. The sum of the squares of
these 4 consecutive integers is
(n − 1)2 + n2 + (n + 1)2 + (n + 2)2 = 4(n2 + n + 1) + 2.
Any integer in the form of 4k + 2 is not a square number, so the sum of the squares of 4
consecutive integers is not a perfect square.
Example 5: If x2 + y2 + z2 = u2, where x, y, z, and u are integers, show that at least two of
x, y, and z are even.
Solution:
Case I: Both x and y are odd and only z is even.
Let x = 2m + 1, y = 2n + 1, z = 2k.
x2 + y2 + z2 = (2m + 1)2 + ( 2n + 1)2 + ( 2k )2 = 4(m2 + n2 + k2 + m + n) + 2
We know that any integer in the form of 4n + 2 is not a square number, so x2 + y2 + z2 is
not a square number if both x and y are odd and only z is even.
Case II: x, y, and z are all odd.
Let x = 2m + 1, y = 2n + 1, z = 2k + 1.
x2 + y2 + z2 = (2m + 1)2 + ( 2n + 1)2 + ( 2k + 1 )2
= 4m2 + 4m + 1 + 4n2 + 4n + 1 + 4k2 + 4k + 1
= 4m(m + 1) + 4n(n + 1) + 4k(k + 1) + 3.
We know that any integer in the form of 4n + 3 is not a square number, so x2 + y2 + z2 is
not a square number if x, y, and z are all odd.
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Therefore, if x2 + y2 + z2 = u2, where x, y, z, and u are integers, at least two of x, y, and z
must be even.
Property 7: Any integer, when divided by 3, will have the remainder 0, 1, or 2. In other
words, any integer can be written in one of the following forms 3k, 3k + 1, 3k + 2 .
We know that (3k)2 = 9k2 = 3 (3k2),
(3k + 1)2 = 9k2 + 6k + 1 = 3 (3k2 + 2k ) + 1,
(3k + 2)2 = 9k2 + 12k + 4 = 3 (3k2 + 4k + 1 ) + 1.
Therefore, any square number has one of the following forms: 3n or 3n + 1.
Any integer of the form 3n + 2 is not a square number.
Example 6: Show that 2n6k 4n2k 11 is not a square number for any positive n and k.
Solution:
Let t = n2k.
2n6k 4n2k 11 = 2t 3 4t 11 2t (t 2 2) 9 2
= 2t[(t 2 1) 3] 9 2 2t (t 1)(t 1) 6t 9 2 2t (t 1)(t 1) 3(2t 3) 2
The first term is divisible by 3 and the second term is also divisible by 3,
so 2n6k 4n2k 11 has the form 3n + 2. Integers in the form of 3n + 2 are not square
numbers, so 2n6k 4n2k 11 is not a square number.
Property 8: Any integer, when divided by 5, will have the remainder 0, 1, 2, 3, or 4. In
other words, any integer can be written in one of the following forms: 5k, 5k + 1, 5k + 2,
5k + 3, and 5k + 4.
We know that
(5k )2 = 25k2 = 5 (5k2),
(5k + 1 )2 = 25k2 + 10k + 1 = 5 (5k2 + 2k ) + 1,
(5k + 2 )2 = 25k2 + 20k + 4 = 5 (5k2 + 4k ) + 4,
(5k + 3 )2 = 25k2 + 30k + 9 = 5 (5k2 + 6k + 1 ) + 4,
(5k + 4 )2 = 25k2 + 40k + 16 = 5 (5k2 + 8k + 3 ) + 1.
Therefore, any square number has one of the following forms: 5n, 5n + 1, or 5n + 4.
Any positive integer of the form of 5n + 2 or 5n + 3 is not a square number.
Property 9: Any positive number in the form 16m, 8m + 1, or 16m + 4 is a square
number.
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Any positive integer in the form of 8n + 2 , 8n + 3 , 8n + 5 , 8n + 6 , or 8n + 7 is not a
square number.
Property 10: Any positive integer in the form of 9n + 2 , 9n + 3, 9n + 5, 9n + 6, or 9n +
8 is not a square number.
Property 11: The units digit is odd and the tens’ digit is even of an odd square number.
Property 12: The product of the units digit and the tens’ digit of any square number is
even. (If the last two digits of a number are both odd, then the number is not a square
number).
Example 7: Show that when the units digit of a square number is odd, its tens digit must
be even.
Proof:
Let n be an odd square number and n = a2.
Since n is odd, we can let a equal 10q + r, where r = 1, 3, 5, 7, 9.
n a 2 (10q r )2 100q 2 20qr r 2
100q 2 20qr 10(10q 2qr ) and r2 can only be 1, 9, 25, 49, or 81.
Therefore the tens digit of
100q 2 20qr r 2 10(10q 2qr ) r 2 is the units digit of
10q2 + 2qr, 10q2 + 2qr + 2, 10q2 + 2qr + 4 , or 10q2 + 2qr + 8, which will always be
even.
Property 13: When the units digit of a square number is 6, its tens digit must be odd.
Proof:
We know that the units digit of n is 6.
Since n = a2, the units digit of a must either 4 or 6.
Case I:
When a = 10q + 4,
n a 2 (10q 4)2 100q 2 80q 16 10(10q 2 8q 1) 6 .
Case II: When a = 10q +6,
n a 2 (10q 6)2 100q 2 120q 36 10(10q 2 12q 3) 6.
The units digit of n is the tens digit of 10q 2 8q 1 or 10q 2 12q 3 , which will always
be odd.
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Property 14: When the tens digit of a square number is odd, its units digit must be 6.
Theorem 1: If the quadratic form ax 2 bx c (a 0) is a square of a linear expression
in rational x, then the discriminant b2 4ac 0 and a is a perfect square of a rational
number.
Proof:
ax 2 bx c (mx n)2 m2 x2 2mnx n2
where m and n are rational numbers and m 0.
Setting the coefficients of like variable equal to each other:
a m 2 ,
b 2mn,
c n 2 .
b2 4m2n2 4ac
a = m2.
b2 4ac 0 .
Theorem 2: If the discriminant b2 4ac 0 and a is a perfect square of a rational
number, then ax2 bx c (a 0) is a square of a linear expression in rational x.
Proof:
We are given that ∆ = 0 and a = m2, so b2 = 4ac and
a is a rational number.
2
b
b
.
4a 2 a
So c is a rational number.
We have b 2 ac .
Therefore
ax 2 bx c = ( a x)2 2 ac x ( c )2 ( a x c )2 .
Since, a 0, then c
2
ax 2 bx c is a square of a linear expression in x.
Example 8: Find m if (m 2) x 2 6mx (4m 1) is a square number for any integers x.
Solution:
We have
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∆ (6m)2 4(m 2)(4m 1) 0 or 36m2 16m2 36m 8 0
2
5m 9m 2 0.
1
Solving for m: m1 = 2, m2 .
5
1
9
Therefore a1 m1 2 2 2 4 and a2 m2 2 2
.
5
5
Since only a1 is a square number, the only answer for m is m = 2.
Theroem 3: If quadratic form ax2 bx c (a 0) is a square of a linear expression in
real x, then the discriminant b2 4ac 0 and a > 0.
Theroem 4: If the discriminant b2 4ac 0 and a > 0, then ax2 bx c (a 0) is a
square of a linear expression in real x.
Example 9: (2010 NC Math Contest) Find the sum of the values of m that make
f ( x) x 2 (m 5) x (5m 1) a perfect square trinomial.
a. 3
b. 4
c. 7
d. 8
e. 10
Solution:
The discriminant of the quadratic ∆ = (m + 5)2 – 4(5m + 1) = 0 and a = 1 is a square
number. Therefore we have m2 10m 1 0 . The sum of the values of m, or the roots,
10
) 10 .
according to Vieta’s Theorem, is (
1
Example 10: Determine if 3 y 2 12 y 12 is a square of a linear expression in (a) rational
x, and (b) real x.
Solution:
Refer to theorems 2 and 4.
(a). We know that ∆ = (– 12)2 – 4 3 12 = 0, and a = 3, which is not a square number.
Therefore 3 y 2 12 y 12 is not a square of a linear expression in rational x.
(b). We know that ∆ = ( – 12)2 – 4 3 12 = 0 and a = 3 > 0. Therefore
3 y 2 12 y 12 is a square of a linear expression in real x. In fact
3 y 2 12 y 12 =( 3 y 2 3 )2 .
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Example 11: Show that a = 2b – c or a = 3c – 2b if (b – c )x2 + (c – a)x – (c – b) is a
square of a linear expression in rational x .
Solution:
We are given that (b – c)x2 + (c – a)x – (c – b) is a square of a linear expression in
rational x .
Therefore, the discriminant equals 0.
∆ = (c a)2 4(b c)[(c b)] 0.
Simplifying:
(c a)2 4(b c)2 0 .
[( c – a ) + 2 (b – c )][( c – a ) – 2(b – c )] = 0
(2b – a – c) (3c – a – 2b) = 0.
Therefore
2b – a – c = 0 or 3c – a – 2b = 0.
In other words,
a = 2b – 3c or a = 2b – c.
3. Some Skills:
Example 12: (1989 AIME) Compute
(31)(30)(29)(28) 1.
Solution:
We know that x( x 1)( x 2)( x 3) 1 ( x 2 3x 1)2 .
Let x = 28,
(31)(30)(29)(28) 1 = 869.
Example 13: (1998 China Middle School Math Contest) How many of the integers
between 1 and 98, inclusive, can be expressed as the difference of the squares of two
nonnegative integers?
Solution:
The difference of the squares of two nonnegative integers can be written as
x n2 m2 (n m)(n m) , where (1 x 98) and m and n are integers.
We know that n + m and n – m have the same parity.
Therefore x must be a multiple of 4 or odd. There are 49 odd numbers and 24 numbers
that are multiples of 4 in between 1 and 98. The answer is then 49 + 24 = 73.
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Example 14: (1991 Beijing Middle School Math Contest) Find the sum of all positive
integers n for which n 2 19n 91 is a perfect square.
Solution:
We know that
n2 19n 91 (n 9)2 (10 n) .
When n > 10, (n 9)2 (10 n) (n 9)2 .
(n 10)2 n2 20n 100 (n2 19n 91) 9 n
(n 10)2 (n 9)2 (10 n) (n 9)2 .
Since a square number does not exist in between two consecutive square numbers, when
n > 10, n2 19n 91 can not be a perfect square.
Therefore in order for n2 19n 91 to be a square number, n can only be
1, 2, 3, 4, 5, 6, 7, 8, 9, or 10. After substituting in these values into n2 19n 91, we see
that only n = 9 or n = 10 yields perfect squares.
Example 15: (2005 AMC 12 B) Let x and y be two-digit integers such that y is obtained
by reversing the digits of x. The integers x and y satisfy x2 y2 = m2 for some positive
integer m. What is x + y + m?
(A) 88
(B) 112
(C) 116
(D) 144
(E) 154
Solution: (E).
By the given conditions, it follows that x > y.
Let x = 10a + b and y = 10b + a, where a > b. Then
m2 = x2 y2 = (10a + b)2 (10b + a)2 = 99a2 99b2 = 99(a2 b2).
Since 99(a2 b2) must be a perfect square,
a2 b2 = (a + b)(a b) = 11k2 for some positive integer k.
Because a and b are distinct digits, we have a b 9 1 = 8 and a + b 9 + 8 = 17. It
follows that a + b = 11, a b = k2, and k is either 1 or 2.
If k = 2, then (a, b) = (15/2, 7/2), which is impossible. Thus k = 1 and (a, b) = (6, 5). This
gives x = 65, y = 56, m = 33, and x + y + m = 154.
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PROBLEMS
Problem 1: Find n such that 4n2 + 5n is a square number.
Problem 2: Find the smallest value of a + b for any positive integers a and b such that
56a + 392b is a square number.
A. 6
B. 7
C. 8
D. 9
Problem 3: [MOSP 1998] Prove that the sum of the squares of 3 consecutive integers
is not a perfect square.
Problem 4: Show that the sum of 1 and the product of any four consecutive positive
integers is a square number.
Problem 5: Find m if (x – 1)(x + 3)(x – 4)(x – 8) + m is a square number.
(A) 32
(B) 24
(C) 98
(D) 196
Problem 6: (2001 China Middle School Math Contest) When 100 is added to a positive
integer, the result is a square number. When 168 is added to the same positive integer, the
result is a different square number. Find the positive integer.
Problem 7: Find the two digit number AB if ( AB ) 2 ( BA) 2 k 2 (k is a positive
integer).
Problem 8: Find all values of m such that m2 + m + 7 is a square number.
Problem 9: The sum of two positive integers is 1000 less than the product of them. One
of the positive integers is a square number. Find it.
Problem 10: What is n if n2 – 71 is divisible by 7n + 55? n is a positive integer.
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Problem 11: (AIME) Find the number of ordered pairs (m, n) of two integers with 1 m,
n 99, such that (m n)2 3m n is a perfect square number.
Problem 12: N is a 4-digit square number and its each digit is less than 7. A new square
number is formed when each of N’s digit is increased by 3. Find N.
Problem 13: Show that the product of any four consecutive positive integers is not a
square number.
Problem 14: Find the greatest square number such that when its last two non-zero digits
are removed, the new resulting number is still a square number.
Problem 15: The sum of 1990 and a 3-digit positive integer is a square number. How
many such 3-digit positive integers are there?
Problem 16: Find all three-digit positive integers such that when the three-digit number
is squared, the last three digits of the resulting number is the original three-digit positive
integer.
Problem 17: (1997 AIME) How many of the integers between 1 and 1000, inclusive, can
be expressed as the difference of the squares of two nonnegative integers?
Problem 18: Find all the integer solutions to m2 + 1986 = n2.
Problem 19: (2002 China Middle School Math Contest) N is a positive integer less than
2392. How many pairs of positive integers x and y are there such that N = 23x + 92 y is a
square number?
Problem 20: (1979 USAMO) Determine all non-negative integral solutions (n1, n2, . . . .,
n14) if any, apart from permutation , of the Diophantine equation n14 n24 n144 1,599.
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SOLUTIONS
Problem 1: Solution: 1.
When n = 1, 4n2 + 5n = 9 is a square number.
When n 2, (2n + 1 )2 < 4n2 + 5n < 4( n + 1)2 = ( 2n + 2 )2. In this case, 4n2 + 5n = 9 is
between two consecutive square numbers, so 4n2 + 5n = 9 will never be a square number.
Therefore n = 1.
Problem 2: Solution: 14.
We know that 56a + 392b = 23 × 7(a + 7b) is a square number. The smallest value that
(a + 7b) can be is 2 × 7 = 14.
a + 7b = 14
a = 7 and b = 1.
The smallest value of a + b is 14.
Problem 3: Solution:
Let the three 3 consecutive integers be n – 1, n, and n + 1. The sum of the squares of the
3 integers is then
(n − 1)2 + n2 + (n + 1)2 = 3n2 + 2.
Any positive integer in the form of 3k + 2 is not a square number, thus the sum of the
squares of 3 consecutive integers is not a perfect square.
Problem 4: Solution:
n(n 1)(n 2)(n 3) 1 (n2 3n 1)2
Problem 5: Solution: (D).
(x – 1)(x + 3)(x – 4)(x – 8) + m
= ( x 2 5x 4) ( x 2 5x 24) m = ( x 2 5x 10)2 196 m .
By setting – 196 + m equal to 0, we will get a square number. Therefore m = 196.
Problem 6: Solution: 156.
Let the number be x.
x + 100 = m2
(1)
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x + 168 = n2
Chapter 7 Square Numbers
(2)
(2) – (1): n2 – m2 = (n – m)(n + m) = 68
Since n – m and n + m have the same parity and their product is even, n – m and n + m
are both even. Note that 0 < n – m < n + m. 68 can be factored into 2 and 34, so
n m 2,
n m 2 17
Solving for n: n = 18.
Therefore x = 156.
Problem 7: Solution: 65.
2
2
AB BA (10A B)2 (10B A)2
= (11A + 11B)(9A – 9B)
= 32 11(A + B)(A – B) = k2
Therefore, 11(A + B)(A – B) is a square number, thus A + B is divisible by 11.
Because A and B are digits, 0 < A – B < 9 and 0 < A + B < 18.
A + B = 11.
k2 = 32 112(A – B).
Therefore A – B must be a square number.
Since 0 < A – B < 9, A – B = 1 or 4 .
Therefore we have
Case I:
A + B = 11
A–B=1
Solving: A = 6, B = 5.
Case II:
A + B =11
A–B=4
There is no solution for case II.
So AB = 65.
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Problem 8: Solution: – 7, –2, 1, 6.
Case I: When m 0, since
m2 m2 m 7 (m 3)2 , we have
m2 m 7 (m 1)2 or m2 m 7 (m 2)2 .
Solving for m: m = 6 or m = 1.
Case II: When m < 0, since (m 2)2 m2 m 7 (m 2)2 , we have
m2 m 7 (m 1)2 or m2 m 7 m2 or m2 m 7 (m 1)2 .
Solving for m: m = 6 (extraneous), – 7 or – 2.
The answers are m = – 7, –2, 1, 6.
Problem 9: Solution: 144.
Let the two positive integers be x and y and x be the square number.
xy – (x + y) = 1000.
xy – x – y + 1 = 1001.
(x – 1)( y – 1) = 7 11 13.
Since x is a square number, x – 1 = 11 13 = 143
Problem 10: Solution: 57.
n 2 71
k
Let
7n 55
n2 7kn (55k 71) 0
49k 2 4(55k 71) 49k 2 220k 284
Only when is a perfect square will n be an integer.
x = 144.
(1)
(7k 15)2 49k 2 220k 284 (7k 17)2
Therefore,
(7k 16)2
(7k 16)2 49k 2 220k 284
Solve for k: k = 7.
Substitute k = 7 into (1):
n = 57 or n = 8. Since n is a positive integer, n = 57 is the desired solution.
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Problem 11: Solution: 98.
(m n)2 3m n (m n)2 4(m n) 4 (m n 2)2
and
(m n)2 3m n (m n)2 .
So (m n)2 (m n)2 3m n (m n 2)2
Since (m n)2 3m n is a perfect square, it must be true that:
(m n)2 3m n (m n 1)2 .
Since (m n)2 3m n (m n 1)2 n m 1 , it follows that:
m n 1
n m 1 0
When n = 2, 3, …, 99, m = 1, 2,…, 98.
Therefore, the number of ordered pairs (m, n) is 98.
Problem 12: Solution: 1156.
Let N a 103 b 102 c 10 d , where a, b, c, and d are positive integers with 1 a
6, 0 b, c, d 6.
Since N = n2 (n is positive integer), n2 = N 6666.
Therefore n 81.
We are given that a new square is formed when each of N’s digit is increased by 3, so
(a + 3) ∙103 + (b + 3)∙102 + (c + 10) + (d + 3) = m2. m is positive integer.
m2 – n2 = 3333
(m + n)(m – n) = 3 11 101.
Since m + n > m – n and n 81, m + n = 101 and m – n = 33.
Therefore n = 34 and N = 1156.
Problem 13: Solution:
Let n be the first positive integer.
The product of the four consecutive integers is then
n(n + 1)(n + 2)(n + 3)
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= (n2 + 3n)(n2 + 3n + 2)
=(n2 + 3n + 1 – 1)(n2 + 3n + 1 + 1)
=(n2 + 3n + 1)2 – 1.
Since this product is between two consecutive square numbers n2 + 3n and n2 + 3n + 1,
therefore, it cannot be square number.
Problem 14: Solution: 1681.
Let the original square number be n2 and the new square number be k2.
0 n2 100k 2 100.
(1)
2
2
For n 100k 0 , we get n > 10k
or
n 10k + 1.
(2)
For n2 100k 2 100 , when k 5, we have
(10k 1)2 (10k )2 20k 1 100 .
This contradicts (1), so k must be less than or equal to 5.
When k = 4, we substitute in this value into (2) and consider n = 42.
We have 422 1600 164 100, and this contradicts (1).
Therefore n = 10k +1 = 41.
When k 3,
n2 100k 2 100 100 9 100 1 000 1681 412 .
We are not able to find any n2 that is greater than 412 so therefore 1681 = 412 is the
greatest such square number.
Problem 15: Solution: 9.
The smallest 3-digit positive integer is 100 and the largest is 999. The sum of this number
and 1990 is between 2093 and 2992, so we are looking for the number of square numbers
in between 2093 and 2992. Since
2093 < 2116 = 462 and 542 = 2916 < 2992, there are 9 square numbers: 462, 472, 482, 492,
502, 512, 522, 532, 542.
Problem 16: Solution: 625 and 376.
Let the three-digit positive integer be m.
We have m2 = 1000k + m, where k is positive integer.
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Therefore
m(m – 1) = 1000k = 23 53 k .
Since m and m – 1 are relatively prime and have different parity, one is the multiple of 23
and one is multiple of 53.
If m = 125, 375, 625, 875, m – 1 = 124, 374, 624, 874, respectively and only 624 is a
multiple of 23.
If m – 1 = 125, 375, 625, 875, m = 126, 376, 626, 876, respectively, and only 376 is a
multiple of 23.
So there are two such numbers: 625 and 376.
Problem 17: Solution: 750.
x = n2 – m2 = (n + m)( n – m) (1 x 1000). n and m are integers.
We know that n + m and n – m have the same parity, so their product, x, must be a
multiple of 4 or odd.
There are 500 odd numbers and 250 numbers that are multiples of 4 in between 1 and
1000.
The answer is then 500 + 250 = 750.
Problem 18: Solution: None.
Let m and n be solutions satisfying m2 + 1986 = n2.
Then n2 – m2 = 1986 (n + m) (n – m) = 1986.
(n – m) and (n + m) have the same parity.
If they are both odd then, (n – m)(n + m) is odd.
However, 1986 is even.
If they are both even, (n – m)(n + m) is a multiple of 4; however 1986 is not divisible by
4.
Therefore, there are no integer solutions to the given equation.
Problem 19: Solution: 5.
We have N = 23(x + 4y).
Since 23 is a prime number, in order for N to be a square number, x + 4y = 23m2.
Since N 2392,
N = 232 m2 2392
m2 2392 232 < 5.
m2 = 1 or 4.
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50 AMC Lectures
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23 x
.
4
(x, y) = (3, 5), (7, 4), (11, 3), (15, 2), (19, 1). (5 pairs).
92 x
y
When m2 = 4, x + 4y = 92
.
4
(x, y) = (4, 22), (8, 21), ∙ ∙ ∙ (88, 1) (22 pairs).
When m2 = 1, x + 4y = 23
y
In total, there are 5 + 22 = 27 pairs.
Problem 20: Solution: No non-negative integer solutions.
If x is even, or x = 2m, where m is integer, then x4 = 16m4, which means that when x is
even, x4 is divisible by 16, and the remainder when x4 is divided by 16 for an even x is 0.
If x is odd, or x = 2m + 1, where m is integer, then x2 has the remainder 1 when divided
by 8. Let x2 = 8n + 1, where n is integer.
Therefore x4 (8n 1)2 64n2 16n 1 16(4n2 n) 1 .
So, the remainder when x4 is divided by 16 for an odd x is 1.
In summary, the remainders can only be 0 or 1 when x4 is divided by 16.
Therefore, in any case, no matter what the integers x1, x2 ,, x1 4 are, when
1 x14 x24 x144 is divided by 16, the smallest possible remainder is 1 and the greatest
possible remainder is 15. In other words, 1 x14 x24 x144 is not divisible by 16 and
so 1 x14 x24 x144 1600.
4
Therefore, x14 x24 x14
1599.
The equation has no non-negative integer solutions.
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50 AMC Lectures
Chapter 8 Divisibility
BASIC KNOWLEDGE
Definition of a divisor
A divisor, also called a factor, of a number divides that number without leaving a
remainder.
(1) an odd number is of the form 2k + 1, for some integer k;
(2) an even number is of the form 2m, for some integer m;
Theorem 1:The product of k consecutive integers is divisible by k!.
Proof:
Let the k consecutive integers be
n, n + 1, , n + k – 1. . . .
n(n 1)(n k 1)
Cnk k 1 which is an
If they are all positive integers, we have
k!
integer.
k! n(n 1)(n k 1) .
(1)
If they are all negative integers, n(n 1)(n k 1) is divisible by k!. We still have
k! n(n 1)(n k 1) .
If they consist of both negative and positive integers, then one of the integers must zero.
Thus, n(n 1)(n k 1) 0 . The conclusion still holds.
Example 1. Show that for any positive integer n, n(n 1)(2n 1) is always divisible by 6.
Solution:
n(n 1)(2n 1) = n(n 1)[(n 1) (n 2)]
= (n 1)n(n 1) n(n 1)(n 2) .
By Theorem 1, we know that (n 1)n(n 1) is divisible by 3! = 6 and n(n 1)(n 2) is
also divisible by 6.
Therefore n(n 1)(2n 1) is always divisible by 6.
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Example 2. (1956 AMC 12) If n is any whole number, n2(n2 – 1) is always divisible by:
(A) 12
(B) 24
(C) any multiple of 12
(D) 12 – n
(E) 12 and 24
Solution:
By considering the special case n = 2, n2(n2 – 1) = 12, we can immediately rule out
choices (B), (C), (D), and (E), leaving us the answer (A). This is an immediate way to get
the answer, but we will also show that (A) holds:
n2(n2 – 1) = n{(n – 1)n(n + 1)} = nk,
where k is a product of three consecutive integers, n – 1, n, and n + 1, hence always
divisible by 3.
Since n is a factor of k, if n is even, k is also divisible by 2 and hence by 6. Thus, nk will
be divisible by 12.
If n is odd, since the even numbers n – 1 and n + 1 are factors of k, k is divisible also by 4
and hence by 12. Thus, nk will be divisible by 12.
Theorem 2: There must be one positive integer divisible by n among n consecutive
positive integers.
Example 3. Show that N and N 5 have the same last digit. N is any integer.
Proof:
N 5 N = (N 1)N(N + 1)(N 2 + 1)
= (N 1)N(N + 1)(N 2 4 + 5)
= (N 2) (N 1) N (N + 1) (N + 2) + 5(N 1)N (N + 1).
We know that (N 2), (N 1) N, (N + 1) and (N + 2) are 5 consecutive integers. Within
5 consecutive integers, there must be an even number and a multiple of 5.
Therefore (N 2) ( N 1) N (N + 1) (N + 2 ) is divisible by 10.
(N 1) , N and (N + 1) are 3 consecutive integers, so there must be an even integer
within the 3 consecutive numbers.
Therefore 5(N 1)N (N + 1) is divisible by 10.
Therefore (N 2) (N 1) N (N + 1) (N + 2) + 5(N 1)N (N + 1) = N5 N is divisible by
10.
Since N 5 N is divisible by 10, the units digit of the difference between N and N 5 equals
0, and therefore N and N 5 both have the same last digit.
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50 AMC Lectures
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Theorem 3
a
r
q , where a and b are
b
b
integers, b > 0, q is the quotient, and r is the remainder with 0 r b .
There exists a unique pair (q, r) such that a qb r or
a
q , which can be written as a qb . This is equivalent to saying that a is
b
divisible by b, ba, and b divides a.
If r = 0,
Example 4. When 1270 is divided by an integer n, the quotient is 74. Find the divisor and
the remainder.
Solution:
Let the divisor be x and the remainder be r.
1270 = 74x + r (0 r < x) r = 1270 74x .
1270
1270
x
Therefore 0 1270 74x < x
75
74
Since x is integer, x = 17.
r = 1270 74 17 = 12.
The divisor is 17 and the remainder is 12.
.
16
14
6
x 17
15
37
Example 5. If 112 is the remainder when a 4-digit positive integer is divided by 131, and
98 is the remainder when the 4-digit positive integer is divided by 132, find the 4-digit
positive integer.
Solution:
Let the 4-digit positive integer be n.
n 132a 98 131a a 98 .
(Note that when n 132a 98 131a a 98 is divided by 131, the remainder will be a +
98).
So a + 98 = 131b +112
a = 131b +14.
Therefore n = 132(131b + 14) + 98 = 132 131b + 1946.
Since n is a 4-digit number, b must be zero and n = 1946.
Example 6. Find the greatest integer that will divide 2613, 2243, 1503, and 985 and
leave the same remainder.
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50 AMC Lectures
Chapter 8 Divisibility
Solution:
Let the divisor be b and the remainder be r.
2613 = bq1 + r,
2243 = bq2 + r,
1503 = bq3 + r,
985 = bq4 + r,
(1)
(2)
(3)
(4)
(1) – (2): 370 = b(q1 – q2).
(3) – (2): 518 = b(q3 – q4).
Since both 370 and 518 are divisible by b, b is the common factor of 370 and 518. The
common factors of 370 and 518 are 2, 37, and 74, so the greatest divisor is 74 (and the
remainder is 23).
Example 7. Both x and y are integers. x + 9y is divisible by 5. What is the remainder
when 8x + 7y is divided by 5?
Solution:
Method 1:
8x 7 y 5(2 x 5 y) 2( x 9 y) .
Since x + 9y is divisible by 5, the remainder when 8x + 7y is divided by 5 is 0.
Method 2: (Sam Chen’s way)
Let x = 1 and y = 1. Therefore x + 9y = 10 which is divisible by 5.
8x + 7y = 8 1 + 7 1 = 15 which is also divisible by 5.
The remainder when 8x + 7y is divided by 5 is 0.
Example 8. (2001 AMC 12) A charity sells 140 benefit tickets for a total of $2001. Some
tickets sell for full price (a whole dollar amount), and the rest sell for half price. How
much money is raised by the full-price tickets?
(A) $782
(B) $986
(C) $1158
(D) $1219
(E) $1449
Solution:
Let n be the number of full-price tickets and p be the price of each in dollars. Since n
tickets will be sold for full price and the charity sells a total of 140 tickets, the number of
tickets sold for half price is equal to 140 – n.
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50 AMC Lectures
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p
2001
2
so p(n + 140) = 4002.
Thus n + 140 must be a factor of 4002 = 2 ∙ 3 ∙ 23 ∙ 29.
Since 0 n 140, we have 140 n + 140 280.
The only factor of 4002 that is in the required range for n + 140 is 174 = 2 ∙ 3 ∙ 29.
Therefore, n + 140 = 174, so n = 34 and p = 23. The money raised by the full-price tickets
is 34 ∙ 23 = 782 dollars.
np (140 n)
Example 9. (1985 AMC 12) Find the least positive integer n for which
zero reducible fraction.
(A) 45
(B) 68
(C) 155
(D) 226
n 13
is a non5n 6
(E) none of these
Solution:
Method 1:
n 13
If
is reducible, then there is some prime number p that divides both n – 13 and 5n
5n 6
+ 6. That same p will also divide any linear combination of these terms; in particular, p
divides (5n + 6) – 5(n – 13) = 71. Thus p = 71 and n – 13 = 71k for some integer k. Since
n > 0 and n – 13 0, n must be at least 84.
Method 2:
n 13
n 13
is reducible and nonzero if its reciprocal
exists and is reducible. By long
5n 6
5n 6
n 13
71
71
. Thus it is necessary and sufficient that
division,
=5+
be
5n 6
n 13
n 13
reducible. Since 71 is a prime, n – 13 = 71, or n = 84, is the smallest solution.
Example 10. (2001 AMC 12) How many positive integers not exceeding 2001 are
multiples of 3 or 4 but not 5?
(A) 768
(B) 801
(C) 934
Solution: (B).
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(D) 1067
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50 AMC Lectures
Chapter 8 Divisibility
For integers not exceeding 2001, there are 2001/3 = 667 multiples of 3 and 2001/4 =
500 multiples of 4. These 667 + 500 = 1167 numbers includes the 2001/12 = 166
multiples of 12 twice, so there are in total 1167 166 = 1001 multiples of 3 or 4. We also
have to subtract the number of multiples of 5, so from this group of numbers, we subtract
the 2001/15 = 133 multiples of 15 and the 2001/20 =100 multiples of 20, since these
are all multiples of 5 and 3, as well as multiples of 5 and 4. However, in this subtraction
segment, we have subtracted 2001/60 = 33 multiples of 60 two times, so we must reinclude these. The number of integers satisfying the conditions is 1001 133 100 + 33
= 801.
Theorem 4: (Residue Classes) For any given positive integer m, when it is divided by n,
the remainder must be one the following: 0, 1, 2, …, n – 1. All integers can be classified
into a unique class according to their remainder when divided by n. A complete set of
residue classes has n classes.
For example, when a positive integer is divided by 2, the remainder can only be 0 or 1.
Therefore all positive integers can be classified as even (with a remainder of 0 when
divided by 2) or odd numbers (with a remainder of 1 when divided by 2).
Any integer can be classified into three residue classes when divided by 3:
3k
3k + 1
3k + 2
k is an integer.
(remainder is 0)
(remainder is 1)
(remainder is 2)
Any integer can be classified into 6 classes when divided by 6:
6k
6k + 1
6k + 2
6k + 3
6k + 4
6k + 5
k is an integer.
(remainder is 0)
(remainder is 1)
(remainder is 2)
(remainder is 3)
(remainder is 4)
(remainder is 5)
Example 11. (1968 AMC 12) Let p be the product of any three consecutive positive odd
integers. The largest integer dividing all such P is:
(A) 15
(B) 6
(C) 5
(D)3
(E) 1
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50 AMC Lectures
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Solution: (D).
Let the three consecutive odd integers be 2k – 1, 2k + 1, and 2k + 3.
The required product P may be written as P = (2k – 1)(2k + 1)(2k + 3), where k is any
positive integer.
The integer k has exactly one of the following three properties:
(i) it is divisible by 3 (i. e. k = 3m, m an integer)
(ii) it leaves 1 as remainder when divided by 3 (i.e. k = 3m + 1)
(iii) it leaves 2 as remainder when divided by 3 (i.e. k = 3m + 2).
In case (i), when k is divisible by 3, 2k + 3 is also divisible by 3. In other words, the last
factor of P is divisible by 3.
In case (ii), the second factor, 2k + 1 = 2 (3m + 2) – 1 = 6m + 3, is divisible by 3.
In case (iii), the first factor, 2k – 1 = 2(3m + 2) – 1 = 6m + 3, is divisible by 3.
Thus, for any integer k, P is divisible by 3. To check to make sure that 3 is the largest
integer dividing P and there exists no larger number, take P1 = 1∙ 3 ∙ 5, P2 = 7∙ 9 ∙ 11, and
observe that 3 is the greatest common divisor of P1 and P2.
Theorem 5. (The Fundamental Theorem of Arithmetic)
Each composite natural number can be expressed in one and only one way as a product of
primes.
For an integer n greater than 1, the unique prime factorization of n is n p1a p2b p3c pkm
, where a, b, c,…, and m are nonnegative integers, p1, p2, …, pk are prime numbers.
Theorem 6. The number of divisors of n is:
d (n) (a 1)(b 1)(c 1)(m 1)
Theorem 7. The sum of divisors of n is:
( n) (
p m1 1
p1a 1 1 p 2b1 1
)(
) ( k
)
p1 1
p2 1
pk 1
Example 12. Let N = 2 4 36 510 7 9 .
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50 AMC Lectures
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(a) How many factors does N have?
(b) How many odd factors does N have?
(c) How many even factors does N have?
(d) How many perfect square factors does N have?
(e) How many odd perfect square factors does N have?
(f) How many even perfect square factors does N have?
(g) How many factors of N are multiples of 7?
(h) How many factors of N are not multiples of 7?
Solution:
(a) 2 4 36 510 7 9
(b) 2 4 36 510 7 9
(4 + 1)(6 + 1)(10 + 1)(9 + 1) = 3850.
(6 + 1)(10 + 1)(9 + 1) = 770.
36 510 7 9
(c) 2 4 36 510 7 9 = 2(23 × 36 510 7 9 ) .
The number of even factors of N is the same as the number of factors of
23 × 36 510 7 9 , which is (3 + 1)(6 + 1)(10 + 1)(9 + 1) = 3080.
(d) 2 4 36 510 7 9 = (24 36 510 78 )7 (22 33 55 74 )2 7
The number of perfect square factors of N is the same as the number of factors of
22 33 55 74 , which is (2 + 1)(3 + 1)(5 + 1)(4 + 1) = 360.
(e) 2 4 36 510 7 9
36 510 7 9
(33 55 7 4 ) 2 7
The number of odd perfect square divisors of N is the same as the number of divisors for
33 55 7 4 which can be calculated as (3 + 1)(5 + 1)(4 + 1) = 120.
(f) 2 4 36 510 7 9 = 2(2 33 55 74 )2 2 7
The number of even perfect square divisors of N is the same as the number of divisors for
2 33 55 74 which can be calculated as (1 + 1)(3 + 1)(5 + 1)(4 + 1) = 240.
(g) 2 4 36 510 7 9 7(24 × 36 510 78 ) (4 + 1)(6 + 1)(10 + 1)(8 + 1) = 3465.
(h) 2 4 36 510 7 9 (24 × 36 510 ) (4 + 1)(6 + 1)(10 + 1) = 385.
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50 AMC Lectures
Chapter 8 Divisibility
Example 13. How many perfect squares are divisors of the product 1! 2! 3! 4! 5!
6! 7! 8! 9!? (AMC 12 A 2003).
Solution: 672.
1! 2! 3! 4! 5! 6! 7! 8! 9!
=1 2 (3 2)(4 3 2)(5 4 3 2) (6 5 4 3 2)(7 6 5 4 3
2)(8 7 6 5 4 3 2) (9 8 7 6 5 4 3 2)
= 230 313 55 73
= (230 312 54 72 ) 31 51 71
= (215 36 52 71 )2 31 51 71 .
The number of perfect square divisors = (15 + 1)(6 + 1)(2 + 1)(1 + 1) = 672.
Example 14. What is the smallest positive integer with six positive odd integer and
twelve even integer divisors? (2000 AIME II).
Solution:
6 can be written as 3 × 2 = (2 + 1)(1 + 1) and 12 can be written as 3 × 2 × 2 = (2 + 1)(1 +
1)(1 + 1).
Since we are looking for the smallest positive integer with six positive odd integer
factors, two of the factors of this number are the two smallest odd prime numbers, or 3
and 5. The smallest such number is 32 51. 32 51 will be a factor of this integer.
Now that we have covered all the odd factors in this number, we need to find out how
many factors of the even prime factor 2 it has so that the total number of even divisors
will be 12. Since 3 × 2 × 2 = (2 + 1)(1 + 1)(1 + 1), the answer is 2(21 32 51) = 180.
Theorem 8: For any positive integer n and integers a and b, a b a n b n .
(a n bn ) (a b)(a n 1 a n 2b a n 3b2 bn 1 ) .
Theorem 9: For any even positive integer n and integers a and b, a b a n b n .
an bn = (a + b)(an 1 an2b + an3b2 …+ abn2 bn1)，n is even.
Example 15. (1971 AMC 12) The number (248 – 1) is exactly divisible by two numbers
between 60 and 70. These numbers are
(A) 61,63
(B) 61,65
(C)63,65
(D)63,67
(E)67,69
Solution:
248 1 (224 1)(224 1) (212 1)(212 1)(224 1)
= (26 1)(26 1)(212 1)(224 1) =63 ∙ 65 (212 1)(224 1).
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50 AMC Lectures
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From factoring, we have obtained that 63 and 65 are both divisors.
Example 16: Find the remainder when 255 + 1 is divided by 11.
Solution:
We know that
(an bn) = (a b)(an 1 + an2b + an3b2 +…+ bn1).
Let a = 25, b = 1, and n = 11:
255 + 1 = (25 + 1)(250 245 + 240 25 +1) = 33 ∙ N
where N is a natural number.
Therefore 255 + 1 is divisible by 11 and the remainder when it is divided by 11 is zero.
Theorem 10: For any odd positive integer n and integers a and b, a b a n b n .
((an + bn) = (a + b)(an 1 an2b + + ( 1 ) k ak bnk1+…+ bn1)).
Example 17: Find the remainder when 255 + 1 is divided by 11.
Solution:
We know that (an + bn) = (a + b)(an 1 an2b + + ( 1 ) k ak bnk1+…+ bn1)).
Let a = 25, b = 1, n = 11:
255 + 1 = (25 + 1)(250 245 + 240 25 +1) = 33 ∙ N.
where N is a natural number.
Therefore 255 + 1 is divisible by 11 and the remainder when it is divided by 11 is zero.
Theorem 11: For any positive integers m and n and integers a and b, a m 1 a mn 1 .
Theorem 12: 1k 2k nk is always divisible by 1 2 n for odd positive integer
k.
Proof:
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50 AMC Lectures
Chapter 8 Divisibility
n(n 1)
.
2
When n is even,
1 2 n
n
n
(1)
1k 2k nk = [(1k n k ) [2k (n 1) k ] [( ) k ( 1) k ] .
2
2
Since k is an odd positive integer, a b a k b k . Every term in (1) is divisible by n + 1.
Therefore 1k 2k nk is divisible by n + 1.
Similarly,
n
n
n
1k 2k nk = [1k (n 1) k ] [2k (n 2) k ] [( 1) k ( 1) k ] ( ) k n k
2
2
2
n
So 1k 2k nk is also divisible by .
2
n
Since (n, n + 1) = 1, and ( , n + 1) = 1, therefore
2
n
n(n 1)
.
1k 2k nk is divisible by their product (n 1)
2
2
When n is odd, we can similarly show that 1k 2k nk is divisible by both n and
n 1
n 1 n(n 1)
, and thus also divisible by their product n
.
2
2
2
Example 18. What is the remainder when 15 + 25 + + 95 is divided by (1 + 2 + + 9)?
Solution:
Method 1:
Let S = 15 + 25 + + 95.
2S = (05 95 ) (15 85 ) (95 05 )
= (0 9)m1 (1 8)m2 (9 0)m10
= 9(m1 m2 m10 ).
Therefore 9 2S .
2S = (15 95 ) (25 85 ) (95 15 )
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50 AMC Lectures
Chapter 8 Divisibility
= (1 9)n1 (2 8)n2 (9 1)n9 ,
= 10(n1 n2 n10 ).
Therefore 10 2S .
Since 9 and 10 are relatively prime, so 90/2S or 45/S.
The remainder is zero.
Method 2:
Setting k equal to 5 in Theorem 12, which states that 1k 2k nk is always divisible
by 1 2 n for odd positive integer k, we see that the remainder is 0.
Theorem 13: n! is divisible by 1 + 2 + + n if n + 1 is not an odd prime.
Example 19: How many positive integers are there such that n! is divisible by 1 + 2 +
+ n if n 100?
Solution:
n! is divisible by 1 + 2 + + n if n + 1 is not an odd prime. There are 24 odd prime
numbers less than 100, so we must subtract these 24 from the initial 100 n possible
values. When n = 100, n + 1 = 101, which is an odd prime, so we must also subtract this
n value. Therefore, there are 100 – 24 – 1 = 75 positive integers.
Divisibility rules
Characteristic of number
Number
divisible by:
Last digit is even
2
The sum of the digits is divisible by 3
3
The last two digits form a number divisible by 4
4
The last digit is 0 or 5
5
The number is divisible by both 2 and 3.
6
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50 AMC Lectures
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To find out if a number is divisible by seven, take the last digit, double it,
and subtract it from the rest of the number.
If you get an answer divisible by 7 (including zero), then the original number
is divisible by seven. If you don't know the new number's divisibility, you
can apply the rule again.
7
The last three digits form a number divisible by 8
8
The sum of the digits is divisible by 9
9
The numeral ends in 0
10
To find out if a number is divisible by eleven, add every other digit, and call
that sum "x." Add together the remaining digits, and call that sum "y." Take
the difference, x – y. If the difference is zero or a multiple of eleven, then the
original number is a multiple of eleven.
11
The number is divisible by both 3 and 4.
12
Delete the last digit from the number, and then subtract 9 times the deleted
digit from the remaining number. If what is left is divisible by 13, then so is
the original number. Repeating the rule if necessary.
13
The last four digits form a number divisible by 16
16
A number is evenly divisible by 35 (relative prime)
A number is evenly divisible by 29 (not 36) (relative prime)
A number is evenly divisible by 38 (neither 46 nor 212) (relative prime)
A number is evenly divisible by 49 (neither 66 nor 312)(relative prime)
15
18
24
36
TIP: If a number is divisible by two different prime numbers, then it is divisible by the
products of those two numbers. Since 36, is divisible by both 2 and 3, it is also divisible by
6.
Example 20. (1967 AMC 12) The three-digit number 2a3 is added to the number 326 to
give the three-digit number 5b9. If 5b9 is divisible by 9, then a + b equals:
(A) 2
(B) 4
(C) 6
(D) 8
(E) 9
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50 AMC Lectures
Chapter 8 Divisibility
Solution:
5 + b + 9 = 14 + b.
From our divisibility rules, in order for 5b9 to be divisible by 9, the sum of its digits must
also be divisible by 9. Since b is a digit, b must be equal to 4. 2a3 + 326 = 549, so 2a3
equals 243, and a equals 4.Therefore a + b = 4 + 2 = 6.
Example 21. Six-digit number 81ab93 is a multiple of 99. Find the digits a and b.
Solution:
Since 99 = 9 11, 81ab93 is divisible by 9 and 11.
From our divisibility rules, in order for 81ab93 to be divisible by 9, the sum of its digits
must be a multiple of 9, so
8 + 1 + 9 + 3 + a + b = 21 + a + b is a multiple of 9.
From our divisibility rules,
(8 + a + 9) – (1 + b + 3) = 13 + a – b must also be a multiple of 11.
We have the following system of equations
a b 9k 21,
a b 11h 13
where h and k are integers, 0 a, b 9, and a and b are integers.
Since 0 a + b 18, 0 9k – 21 18 and 3 k 4.
Since –9 a – b 9, –9 11h – 13 9 and 1 h 2.
1
a 2 (9k 11h 34)
b 1 (9k 11h 8)
2
k and h have the same parity. The only possibility is:
k 3
k 4
or
.
h 1
h 2
a 2
a 12
or
b 4
b 3
Since 0 a, b 9, a = 2, b = 4.
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Example 22. If N = 2,x78 is a four-digit number divisible by 17, what is the value of x?
Solution: 2.
Method 1:
N = 2078 + 100x = (122 17) + 17 6x + 4 – 2x = 17(122 + 6x) + (4 – 2x)
Since N is divisible by 17, 17(122 + 6x) is already divisible by 17, so 4 – 2x must be
divisible by 17. In order for 4 – 2x to be divisible by 17, x must be 2.
Method 2 (my solution):
We can write this question as
2x78 0 (mod 17) 2078 + 100x 0 (mod 17)
15x 13 (mod 17)
15x 13+ 17 = 30 (mod 17)
x = 2.
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50 AMC Lectures
Chapter 8 Divisibility
PROBLEMS
Problem 1. Show that for any even positive integer n, n(n 1)(n 2) is divisible by 24.
Problem 2. (1967 AMC 12) For natural numbers, when P is divided by D, the quotient is
Q and the remainder is R. When Q is divided by D, the quotient is Q and the remainder
is R. Then , when P is divided by DD, the remainder is:
(A) R + RD (B) R + RD (C) RR
(D) R
(E) R
Problem 3. Both m and n are integers. 5m + 3n is divisible by 11. What is the remainder
when 9m + n is divided by 11?
Problem 4. Mr. Mathis’ class had m boys and 11 girls. Mr. English’s class had nine boys
and n girls. Two classes raised the same amount of money during a charity sell, with each
class $(mn + 9m + 11n + 145). It was known that every student participated in the activity
and each student donated the same amount of money (a whole dollar amount) less than
$30. How much money was donated by each student?
Problem 5. (2002 AMC 12 B) For how many integers n is n/(20 − n) the square of an
integer?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 10
Problem 6. Prove that (n – 2)(n – 1)(2n – 3) is divisible by 6 if n is any positive integer
greater than 2.
Problem 7. When 200 is divided by a natural number x, the remainder is 8. How many
values of x are there?
Problem 8. How many odd perfect square factors does 2 4 36 510 7 9 have?
Problem 9. Find the remainder when n5 – n is divided by 30.
Problem 10. What is the sum of all positive two-digit integers with exactly 12 divisors?
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Problem 11. (2003 AMC 10A #25) Let n be a 5-digit number, and let q and r be the
quotient and remainder, respectively, when n is divided by 100. For how many values of
n is q + r divisible by 11?
Problem 12. (1960 AMC 12) Let m and n be any two odd numbers, with n less than m.
The largest integer which divides all possible numbers of the form m2 n2 is:
(A) 2 (B)4 (C)6 (D)8 (E)16
Problem 13. (1969 AMC 12) When the natural numbers P and P, with P > P, are
divided by the natural number D, the remainders are R and R, respectively. When PP
and RR are divided by D, the remainders are r and r, respectively. Then:
(A) r > r always
(B) r < r always
(C) r > r sometimes, and r < r sometimes
(D) r > r sometimes, and r = r sometimes (E) r = r always
Problem 14. (1970 AMC 12) The greatest integer that will divide 13,511, 13,903 and
14,589 and leave the same remainder is
(A) 28
(B) 49
(C) 98
(D) an odd multiple of 7 greater than 49
(E) an even multiple of 7 greater than 98
Problem 15. (1973 AMC 12) If p 5 is a prime number, then 24 divides p2 – 1 without
remainder
(A) never (B) sometimes only
(C) always (D) only if p = 5 (E) none of these
Problem 16. (1977 AMC 12) Determine the largest positive integer n such that 1005! is
divisible by 10n.
(A) 102
(B)112
(C) 249
(D)502 (E) none of these
Problem 17. (1985 AMC 12) Let p, q and r be distinct prime numbers, where 1 is not
considered a prime. Which of the following is the smallest positive perfect cube having n
= pq2r4 as a divisor?
(A) p8 q8 r8 (B) ( pq2 r2)3 (C) ( p2 q2 r2 )3
(D) (pqr2)3
(E) 4p3 q3 r3
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Chapter 8 Divisibility
Problem 18. (2003 AMC 12 B) 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?
(A) 3
(B) 5
(C) 11
(D) 15
(E) 165
Problem 19. (2006 AMC 12 B) Mr. Jones has eight children of different ages. On a
family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which
each of two digits appears two times. “Look, daddy!" she exclaims. “That number is
evenly divisible by the age of each of us kids!" “That's right," replies Mr. Jones, “and the
last two digits just happen to be my age." Which of the following is not the age of one of
Mr. Jones's children?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
Problem 20. Prove that any six-digit number with the form of abcabc is divisible by 13.
Problem 21. When 2011 is divided by a 2-digit positive integer, what is the greatest
possible remainder?
Problem 22. Prove that (n – 1)3 + n3 + (n + 1)3 is divisible by 9 for any integer n.
Problem 23. How many positive integers less than 1000 are divisible by 5 or 7 but not
35.
A. 285
B. 313
C. 341
D. 369
Problem 24. When 732 is divided by a natural number x, the remainder is 12. How many
values of x are there?
Problem 25. Show that there are no integers a, b, and c such that a2 + b2 – 8c = 6.
Problem 26. (AIME 1988) Compute the probability that a randomly chosen positive
divisor of 1099 is an integer multiple of 1088.
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Chapter 8 Divisibility
SOLUTIONS:
Problem 1. Solution:
Since n is even, let n = 2m.
n(n 1)(n 2) = 4m(m 1)(2m 1)
= 4m(m 1) [ (m 2) (m 1)]
= 4m(m 1) (m 2) 4m(m 1)(m 1) .
Since the product of three consecutive integers is a multiple of 6, both
4m(m 1) (m 2) and 4(m 1)m(m 1) are divisible by 24.
Therefore 4m(m 1)(m 2) 4(m 1)m(m 1) is divisible by 24.
Problem 2. Solution: (A).
We are given that P = QD + R and that Q = QD + R. Therefore, by substitution,
P = (QD + R)D + R = Q( DD) + (R + RD).
To see that R + RD is the remainder in this division by DD, we must show that R + RD
is less than the divisor DD; we know that R D 1, and R D 1 so that
R + RD (D 1) + (D 1)D = DD 1< DD.
Problem 3. Solution: 0.
3(9m + n) = 27m + 3n = (5m + 3n) + 22m
Since both 5m + 3n and 22 are divisible by 11, 3(9m + n) is also divisible by 11. The
remainder when 9m + n is divided by 11 is 0.
Method 2: (Sam Chen’s way)
Let m = 1 and n = 2. Therefore 5m + 3n = 11 which is divisible by 11.
Substituting in these values into 9m + n, we get
9m + n = 9 1 + 2 = 11 which is also divisible by 11.
The remainder when 9m + n is divided by 11 is 0.
Problem 4. Solution: $25.
We know that m + 11 = n + 9.
mn + 9m + 11n + 145 is divisible by both m + 11 and n + 9.
We have mn + 9m + 11n + 145 = (m + 11)(n + 9) + 46.
So 46 is divisible by both m + 11 and n + 9.
We know that 46 1 46 2 23
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50 AMC Lectures
Therefore we have
n 9 46
Chapter 8 Divisibility
n = 37 and m = 35
Each student donated
(m 11)(n 9) 46 (35 11)(37 9) 46
47
m 11
35 11
OR
n = 14 and m = 12
n 9 23
Each student donated
(m 11)(n 9) 46 (11 11)(14 9) 46
25 .
m 11
12 11
Since each student donated a whole dollar amount less than $30, the answer is $25.
Problem 5. Solution: (D).
n
20k 2
k 2 , for some k ≥ 0, then n 2
If
. Since k2 and k2 +1 have no common
20 n
k 1
2
factors and n is an integer, k + 1 must be a factor of 20. This occurs only when k = 0, 1,
2, or 3. The corresponding values of n are 0, 10, 16, and 18.
Problem 6. Solution:
Since (n – 2)(n – 1) is the product of two consecutive integers, it must be divisible by 2.
Therefore, we only need to show that (n – 2)(n – 1)(2n – 3) is divisible by 3.
We classify n based on the remainders 0, 1, or 2 as follows:
n = 3k, n = 3k + 1, n = 3k +2, where k is natural number.
Case I. When n = 3k,
(n – 2)(n – 1)(2n – 3) = (3k – 2)( 3k – 1)( 6k – 3) =3(3k – 2)( 3k – 1)( 2k – 1).
Case II.When n = 3k + 1,
(n – 2)(n – 1)(2n – 3) = (3k – 1) ∙ 3k ∙ ( 6k – 1) = 3 ∙ k ∙ ( 3k – 1)( 6k – 1)
Case III. When n = 3k + 2,
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Chapter 8 Divisibility
(n – 2)(n – 1)(2n – 3) =3k ∙ (3k + 1)( 6k + 1) .
Therefore (n – 2)(n – 1)(2n – 3) is divisible by 3.
Since 2 and 3 are relatively prime to each other, (n – 2)(n – 1)(2n – 3) is divisible by 6.
Problem 7. Solution: 8.
We have 200 = x ∙ q + 8 (x > 8).
Or 200 – 8 = x ∙ q
192 = x ∙ q.
6
Since 192 = 2 3, the number of factors of 196 is (6 + 1) (1 + 1) = 14.
Since x > 8, we need to exclude 1, 2, 3, 4, 6, 8.
There are 14 – 6 = 8 such numbers.
Problem 8. Solution: 120.
36 510 7 9
2 4 36 510 7 9
(33 55 7 4 ) 2 7
The number of odd perfect square divisors is the same as the number of divisors for
33 55 7 4 which can be calculated as (3 + 1)(5 + 1)(4 + 1) = 120.
Problem 9. Solution: 0.
The product of three consecutive integers, (n 1)n(n 1) , must be divisible by 6. Since n5
– n = n(n4 1) = n(n2 1)(n 1)(n 1) , therefore n5 – n is divisible by 6.
n5 – n = (n 1)n(n 1)(n2 1) = (n 1)n(n 1)[(n 2)(n 2) 5]
= (n 1)n(n 1)(n 2)(n 2) 5(n 1)n(n 1)
We know that (n 2)(n 1)n(n 1)(n 2) is the product of 5 consecutive integers.
Therefore n5 – n is divisible by 5.
Since 5 and 6 are relatively prime, n5 – n is divisible by their product 5 × 6 = 30, and so
the remainder is zero when n5 – n is divided by 30.
Problem 10. Solution: 402.
Since 12 = 4 × 3 = (3 + 1)(2 + 1), we have: 23 × 32 = 72.
12 = 6 × 2 = (5 + 1)(1 + 1), from which we get 96.
12 = 2 × 2 × 3 = (1 + 1)(1 + 1)(2 + 1) yields 22 × 3 × 5 = 60, 22 × 3 × 7 = 84, and 32 × 2 ×
5 = 90.
There are five such integers: 60, 72, 84, 90, and 96, and their sum is 402.
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50 AMC Lectures
Chapter 8 Divisibility
Problem 11. Solution: 8181.
n abcde 10000a 1000b 100c 10d e
q 100a 10b c and r 10d e
q r 100a 10b c 10d e 0
a bc d e 0
(mod 11)
ace bd
(mod 11)
(mod 11)
(1)
Note that equation (1) is the rule of divisibility for 11, so we can solve this problem by
finding N, the number of all five-digit numbers that are divisible by 11.
99999 10000
N
9090 909 8181 .
11 11
Method 2:
Note that n = 100q + r = q + r + 99q. Hence q + r is divisible by 11 if and only if n is
divisible by 11. Since 10,000 n 99, 999, there are 99999/11 9999/11 = 9090
909 = 8181 such numbers.
Problem 12. Solution: (D).
m = 2r + 1 where r = 0, ± 1, ±2, . . . ;
n = 2s + 1 where s = 0, ±1, ±2, . . .;
m2 n2 4r 2 4r 1 4s 2 4s 1
= 4(r – s)(r + s + 1), a number certainly divisible by 4.
If r and s are both even or both odd, r – s is divisible by 2, and r + s + 1 is not.
If r and s are one even and one odd, then r + s + 1 is divisible by 2, and r – s is not.
Thus m2 – n2 is divisible by 4 · 2 = 8.
Problem 13. Solution: (E).
Let Q, Q and Q be the quotients in the respective divisions of P, P and RR by D, so
that P = QD + R, P QD R , RR QD r .
Multiplication of the first two equations followed by replacement of RR from the third
gives
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50 AMC Lectures
Chapter 8 Divisibility
PP = (QD R)(QD R)
= (QQD QR QR) D RR)
= (QQD QR QR Q) D r .
Since r < D and division is unique, the remainder r in the division of PP by D is equal
to r as stated in choice (E).
Problem 14. Solution: (C).
(Official Solution):
If three integers a, b, and c have the same remainder r upon division by an integer d, then
a = ad + r, b = βd + r, and c = γd + r,
where α, β, γ are the quotients. The differences a – b = (α – β)d, a – c = (α – γ)d, and b
– c = (β – γ)d are exactly divisible by d.
Moreover, since (a – b) – (a – c) + (b – c) = 0, any common divisor d of two of the
differences is a divisor of the third. Hence the GCD (greatest common divisor) of any
pair of the differences is the greatest integer leaving the same remainder when divided
into all three of the original numbers a, b, and c.
In the present problem, we seek GCD of the two differences
13,903 – 13,511 = 392 = 72 · 23
and
14,589 – 13,903 = 72 · 2 which, by inspection, is 72 · 2 = 98.
Problem 15. Solution: (C).
(Official Solution):
Since the factors p – 1 and p + 1 of p2 – 1 are consecutive even integers, both are
divisible by 2 and one of them by 4, so that their product is divisible by 8. Again (p – 1),
p and (p + 1) are three consecutive integers, so that one of them (but not the prime p) is
divisible by 3. Therefore the product (p – 1) (p + 1) = p2 – 1 is divisible by both 3 and 8,
and hence by 24.
Problem 16. Solution: (E).
(Official Solution):
Let 2k3l5m ∙ ∙ ∙ be the factorization of 1005! into powers of distinct primes; then n is the
minimum of k and m. Now 201 of the integers between 5 and 1005 are divisible by 5;
forty of these 201 integers are divisible by 52; eight of these forty integers are divisible by
53; and one of these eight integers is divisible by 54. Since 502 of the numbers between 2
and 1005 are even, k > 502; so
n = m = 201 + 40 + 8 + 1 = 250.
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Problem 17. Solution: (D).
(Official Solution):
If n = pq2r4 is a divisor of cube c, then c must have p, q and r as primes in its
factorization. Moreover, the exponents of p, q and r in the factorization must be multiples
of 3 and they must be at least as great as 1, 2 and 4, respectively (1, 2, 4 being the
exponents of p, q, r in n). Thus p3q3r6 = (pqr2)3 is the smallest such cube.
Problem 18. Solution: (D).
(Official Solution):
Among five consecutive odd numbers, at least one is divisible by 3 and exactly one is
divisible by 5, so the product is always divisible by 15. The cases n = 2, n = 10, and n =
12 demonstrate that no larger common divisor is possible, since 15 is the greatest
common divisor of 3∙5∙7∙9∙11, 11∙13∙15∙17∙19, and 13∙15∙17∙19∙21.
Problem 19. Solution: (B).
(Official Solution):
The 4-digit number on the license plate has the form aabb or abab or baab, where a and
b are distinct integers from 0 to 9. Because Mr. Jones has a child of age 9, the number on
the license plate is divisible by 9. Hence the sum of the digits, 2(a + b), is also divisible
by 9. Because of the restriction on the digits a and b, this implies that a + b = 9.
Moreover, since Mr. Jones must have either a 4-year-old or an 8-year-old, the license
plate number is divisible by 4. These conditions narrow the possibilities for the number to
1188, 2772, 3636, 5544, 6336, 7272, and 9900. The last two digits of 9900 could not
yield Mr. Jones's age, and none of the others is divisible by 5, so he does not have a 5year-old. Note that 5544 is divisible by each of the other eight non-zero digits.
Problem 20. Solution:
abcabc a 105 b 104 c 103 a 102 b 10 c
= a (105 102 ) b (104 10) c (103 1)
= a (100000 + 100) + b (10000 + 10) + c (1000 + 1)
=1001 (100a + 10b + c) = 13 11 7 abc .
13 abcabc .
Problem 21. Solution: 91.
Since 2011 ÷ 99 = 20 R 31, we are looking for R such that:
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50 AMC Lectures
Chapter 8 Divisibility
2011 9? 20 R
99 20 31
98 20 51
97 20 71
96 20 91
95 20 111
Note that 111 is a 3-digit divisor and is not counted.
The greatest possible remainder when 2011 is divided by a 2-digit positive integer is 91.
Problem 22. Solution:
(n – 1)3 + n3 + (n + 1)3 = 3n3 + 6n = 3n(n2 + 2)
= 3n[(n – 1)(n + 1) + 3] = 3(n – 1) ∙ n ∙ (n + 1) + 9n
Since 3(n – 1) ∙ n ∙ (n + 1) is a multiple of 3 ∙ 3! = 18 and 9n is a multiple by 9,
( n – 1)3 + n3 + (n + 1)3 is divisible by 9.
Problem 23. Solution: 285.
999
There are
numbers divisible by 5.
5
999
There are
numbers divisible by 7.
7
999
There are
numbers divisible by 5 or 7.
5 7
The number of positive integers less than 1000 that are divisible by 5 or 7 but not 35 is
999 999 999 999
5 + 7 5 7 5 7 = 285.
Problem 24. Solution: 20.
Because the remainder when 732 is divided by x is 12, we can subtract 12 from 732 to
find a number that leaves a remainder of 0 when divided by 12. 732 – 12 = 720 and 720 =
24 32 5.
There are (4 + 1) (2 + 1) (1 + 1) = 30 factors of 720. Among them, we must subtract
the factors that are less than 12, because they will leave a remainder less than 12. There
are 10 factors: 1, 2, 3, 4, 5, 6, 8, 9, 10, and 12. The answer is 30 – 10 = 20.
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50 AMC Lectures
Chapter 8 Divisibility
Problem 25. Solution:
In order to prove that there are no integers a, b, and c such that a2 + b2 – 8c = 6, we can
prove that a2 + b2 = 8c + 6 is not possible, or in other words, when a2 + b2 is divided by
8, the remainder could not be 6.
We classify all the integers into four kinds: 4k, 4k + 1, 4k + 2, 4k + 3.
Thus, all squares can be written in one of the following forms: 16k 2 , 8k(2k + 1) + 1, 16k
(k + 1) + 4, and 8k(2k + 3) + 9, respectively.
The remainders of these squares are 0, 1, 4, 1, respectively when divided by 8.
The sum of any combination of two remainders cannot be 6, so when a2 + b2 is divided
by 8, the remainder can not be 6. Therefore, there are no integers a, b, and c such that a2
+ b2 – 8c = 6.
Problem 26. Solution:
9
.
65
1099 = 299 · 599. The number of divisors of 1099 is (99 + 1)(99 + 1) = 100 × 100. This will
be on the denominator of our fraction.
1099 = 1088 + 11= 1088(1011). The number of divisors that are multiples of 1088 is equal to
the number of factors of (1011). Since 1011 = 211 ×511, the number of factors is (12 × 12).
The probability that a randomly chosen positive divisor of 1099 is an integer multiple of
12 12
9
.
1088 is
100 100 65
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Chapter 9 Geometry Congruent Triangles
BASIC KNOWLEDGE
(1). If two sides and the enclosed angle of one triangle are equal to two sides and the
enclosed angle of another triangle, the triangles are congruent by S(ide).A(ngle).S(ide).
SAS SAS
a = c; b = d, =
(2). If two angles and an enclosed side of one triangle are equal to two angles and the
enclosed side of another triangle, the triangles are congruent by A(ngle).S(ide).A(ngle).
ASA ASA
a = c; 1 = 4; 2 = 3.
We are able to determine that two triangles are
congruent (AAS AAS) if 1 = 4, 2 = 3, e
= f or b = d or a = c.
We are not able to determine that AAS AAS
based on the following criteria: 1 = 4; 2 = 3, a = d, or e = c.
We are able to determine that two triangles are congruent (SAA SAA) if e = f or b = d
or a = c; 1 = 4; 2 = 3.
We are not able to determine that two triangles are congruent (SAA SAA) based on the
following criteria: e = f, 1 = 3; 2 = 4; or b = d 1 = 3; 2 = 4.
(3). If three sides of one triangle are equal to three sides of another triangle, the triangles
are congruent by S.S.S.
SSS SSS
(4). If the hypotenuse and leg of one right triangle is equal to the hypotenuse and leg
of another triangle, respectively, the two right triangles are congruent.
Right triangles HL HL
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Chapter 9 Geometry Congruent Triangles
(5). If two sides and the angle that faces the longer side of the triangle are equal to two
sides and the angle that faces the longer side of another triangle, the two triangles are
congruent by SSA. The small “s” denotes the side length of the shorter side.
SSA SSA (a = c, b = d, b > a, b > e, d > f, 1 = 4).
(Note the positioning of S, s, and A. For example, we are not able to determine whether or
not two triangles are congruent or not based on the following information: a = c, b = d, b
> a, 1 = 4).
(6). If two angles and the angle bisector of the third angle of one triangle are equal to two
angles and the angle bisector of the third angle of another triangle, then the two triangles
are congruent to each other.
AD = A1D1; ABC = A1B1C1; ACB = A1C1B1.
ABC A1B1C1.
Proof:
Since B = B1 and C = C1, then BAC = B1A1C1.
Hence BAD = B1 A1 D1.
In ∆ ABD and ∆ A1 B1 D1, AD = A1 D1, B = B1 , BAD = B1A1D1
Therefore∆ ABD ∆ A1 B1 D1 (A, A, S)
AB = A1 B1.
In ABC and A1B1C1, AB = A1 B1, B = B1, C = C1.
Therefore ABC A1B1C1 (A, A, S).
(7). If two sides and the median on the third side of one triangle are equal to two sides
and the median of the third side of another triangle, then the triangles are congruent.
AB = A1B1, AC = A1C1, BD = DC, B1D1 = D1C1, AD = A1 D1, ABC A1B1C1.
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Chapter 9 Geometry Congruent Triangles
Proof:
Extend AD to E such that DE = AD,
and then connect CD.
Since AD = DE, BD = DC, ADB =
CDE, so ABD ECD (S, A, S)
Hence CE =AB, BAD = CED.
Similarly, extend A1D1 to E1 such that
D1 E1 = A1 D1 and connect C1E1, We can show that A1B1D1 ∆ E1C1D1 (S, A, S) .
C1E1 = A1B1, B1A1D1 C1E1D1.
In AEC and ∆ A1E1C1, we see that AEC ∆ A1E1C1 (S, S, S).
(AE = A1E1, AC = A1C1, CE = AB = A1B1 =C1 E1).
So AEC = A1E1C1, EAC = E1A1C1 BAD = AEC = A1E1C1 = B1A1D1.
BAC = BAD +DAC = B1A1D1 + D1A1C1 = B1A1C1 .
Therefore ABC A1B1C1 (S, A, S).
Example 1: As shown in the figure to the right, OA = OB. Point C is on OA. Point D is
on OB. OC = OD. AD and BC meet at E. How many pairs of triangles are
congruent?
(A) 2 pairs (B) 3 pairs
(C) 4 pairs (D) 5 pairs.
Solution: (C).
△AOD △BOC (OA = OB, OD = OC, AOD =BOC).
So we have A =B.
Since AEC =BED and AC = OA OC = OB OD = BD, △AEC △BED.
Therefore we have AE = BE, CE = DE.
Taking OE = OE under consideration, we have △OEA △OEB and △OEC △OED.
There are 3 pairs of congruent triangles.
Example 2: D and E are points on AB and AC of equilateral triangle ABC,
respectively. AD = CE. BE and CD meet at F. Find BFC.
Solution:
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Chapter 9 Geometry Congruent Triangles
Examining the figure to the right, F = 1 + A + 2. We can easily see that
△ADC △CEB (AD = CE, AC = BC, and CAD = BCE = 60°).
Therefore 1 = 3. BFC = 3 + A + 2 = B + A = 120.
Example 3: In △ABC, BAC = 120, ADBC at D.
AB + BD = CD. Find C.
(A) 20
(B) 25
Solution: (A).
Find point E on DC such that DE = DB. Connect AE.
In Rt△ABD and Rt△AED, BD = ED, AD = AD.
Therefore Rt△ABD Rt△AED, AB = AE, B = AED.
Since AB + BD = CD, EC = CD – DE = CD – BD = (AB + BD) – BD = AB = AE.
Hence C = CAE, and B = AED = 2C .
We also have B +C = 180– BAC = 60
So C = 20.
Example 4: (1981 AMC) In ABC, M is the midpoint of side BC, AN bisects BAC, BN
AN and is the measure of BAC. If sides AB and AC have
lengths 14 and 19, respectively, then length MN equals
5
(A) 2
(B)
2
5
1
sin ( )
(E)
2
2
2
Solution: (B).
In the adjoining figure, BN is extended past N and meets AC at E. Triangle BNA is
congruent toENA, since BAN =ENA, AN = AN and ANB =ANE. There N is the
midpoint of BE, and AB = AE = 14. Thus EC = 5. Since MN is
the line joining the midpoints of sides BC and BE of CBE,
1
5
its length is (EC) = .
2
2
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Example 5: In △ABC, C = 90. The angle bisector of BAC meets BC at D. If
CD = 15, AC = 30, find AB.
Solution:
Draw DE so that DEAB.
△ADC △ADE.
Let BE = x, BD = y.
In Rt△ABC, AB2 = AC2 + BC2.
In Rt△BDE , BD2 = BE2 + ED2.
(x 30) 2 30 2 (y 15) 2
2
2
2
x 15 y
(1) – (2): y 2 x 15
x 2 152 (2 x 15)2 .
Simplifying: x( x 20) 0
AB = AE + BE = 30 + 20 = 50.
(1)
(2)
x = 20.
Example 6: Triangle ABC is an equilateral triangle. D is the on
AB. Extend AC to E and connect DE so that BD = CE. Prove: GD
= GE.
Proof:
Draw DH//CE.
Since DH = DB, E =HDG, DGH =CGE. Therefore DHG
GCE. DG = GE.
Example 7: In ABC, AB > AC, Prove C >B.
Proof:
Method 1:
Draw AF, the angle bisector of BAC to meet BC at F.
We are given that AB > AC. Find E, a point on AB so that AE =
AC, and connect EF.
∆AEF ∆ACF, so AEF = C.
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We also know that AEF = B + EFB, so AEF > B and therefore C > B.
Method 2:
Draw AF, the angle bisector of BAC to meet BC at F.
We are given that AB > AC. We find E, a point on the extension
of AC such that AE = AB. Connect EF.
∆ABF ∆AEF. By this congruity, E = B.
We also know that ACF = E + EFC, so ACF > E =B,
and therefore C > B.
Note: See “Chapter 11 Eight Methods To Draw Auxiliary Lines”
for two more proofs.
Example 8: In an equilateral triangle ABC, we extend BA to D, and BC to E. If DC =
DE, Prove AD = AC + CE.
Proof:
Method 1:
Extend CE to F so that EF = BC.
Since DC = DE, DCE =DEC, DCB =DEF.
Examining DCB and DEF, we see that DC = DE, DCB
=DEF, BC = EF.
Therefore DCB DEF, so DB = DF.
Since ABC is equilateral, B = 60 and AC = BC = AB. Since DB
= DF, B = 60, DBF is an equilateral triangle.
Therefore DB = FB, AD = CF or AD = CE + EF = CE + AC.
Method 2.
Draw EM//AC. Then CE = AM.
In ACD and DME, DAC = 120, DME = 120, DC = DE,
MDE = 180 – 60 −DEC, ACD =180 − 60 −DEC.
Hence MDE = ACD and ACD DME. (ASA).
Therefore DM = AC and AD = AM + DM = CE + AC.
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Method 3:
Draw DFBE. Then BDF = 30, BC + CF = 1/2 (BA + AD) = 1/2 (BC + AD).
Therefore AD + BC = 2BC + 2CF
AD = BC + 2CF
We also know that 2CF = CE (DCE is an isosceles triangle. DF is the
perpendicular bisector of CE).
Therefore AD = BC + CE = AC + CE.
Method 4:
Extend AC to P such that CP = CE and connect EP.
We also know that ECP = 60. Therefore ∆CEP is an equilateral
triangle.
Since ∆CDE is an isosceles triangle, if we connect DP, DP will be the
perpendicular bisector of CE and DP will also bisect the angle EPC.
Since EPC = 60, FPC = 30.
Since B = 60, BDF = 30. Therefore ∆DAP is an isosceles triangle
with AD = AP.
We know that AP = AC + CP = AC + CE. Therefore AD = AC + CE.
Example 9: As shown in the figure to the right, ABCD is a square. M is
any point on BC. AN is the angle bisector of DAM and AN meets DC at
N. Prove: DN + BM = AM.
Solution:
Extend CB to E such that BE = DN.
AB = AD, BE = DN, ABE = 90 = ADN
∆ABE ∆AND.
EAB = NAD, AEB = AND.
EAM = EAB + BAM = NAD + BAM
=MAN + BAM = BAN
= AND = AEB.
AM = EM.
EM = EB + BM = DN + BM.
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AM = DN + BM.
Example 10: As shown in the figure below, ABC, AC = BC. ACB = 90. D is a point
on AC. AEBD at E. BE meets AC at D. AE = ½ BD. Prove: BD is
the angle bisector of ABC.
Solution:
Extend AE to F such that EF = AE. Connect CF.
FAC = 90 – ADE = 90 – BDC = DBC
AF = 2AE = BD. AC = BC. ∆ACF ∆BCD.
ACF = BCD = 90.
That is, B, C, and F are collinear.
Therefore BE is both the median and the height of triangle ABF on
AF.
Hence AB = AF.
We know that BE is also the angle bisector of ABF, so BD is the
angle bisector of ABC.
Example 11: ABC is an equilateral triangle. O is a point inside ABC. AOB = 115.
BOC = 125. Find all the angles of a triangle formed by the three sides OA, OB, and
OC.
Solution:
Rotating ∆AOB 60 along B, we get ∆CDB.
OB = DB, and OBD = 60, so
∆OBD is an equilateral triangle with OD = OB.
Since OA = CD, ∆OCD is a triangle with three sides OA, OB, OC.
ODC = BDC 60= AOB 60 =55,
COD = COB 60= 125 60 = 65.
OCD = 180 ODC COD = 60.
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Example 12: (2011 China Middle School Math Contest) As shown in the figure, in
△ABC, ∠BAC＝60°, AB＝2AC. Point P is inside △ABC. PA＝
3 ， PB＝5，and PC＝2. Find the area of △ABC.
Solution:
We know that ∠BAC＝60°, and AB＝2AC.
Denote the midpoint of AB as D. We have AC＝AD＝BD, so
△ACD is an isosceles triangle with one angle of 60° and therefore
△ACD is equilateral.
So △BCD is an isosceles triangle with one angle of 120°, and
△ABC is a right triangle with one angle of 30°.
Let P1, P2, and P3 be the reflections of P about BC, AC, and AB,
respectively.
△APB △AP3B. △APC △AP2C, and △BPC △BP1C.
Therefore AP＝ AP3＝ AP2＝ 3 , BP＝ BP3＝ BP1＝ 5, CP＝
CP1＝ CP2＝ 2.
We know that ∠ABC＝30°, ∠BAC＝ 60°, ∠ACB ＝ 90°.
Therefore ∠P3BP1 ＝ 60°, ∠P3AP2 ＝ 120°, ∠P1CP2 ＝
180°.
Therefore the area of pentagon
AP2P1BP3 ＝2 × the area of △ABC ＝ SAP2 P3 (isosceles
triangle with the base length 3 and one angle of 120°) ＋
SBP1 P3 (equilateral triangle with the side length of 5) ＋
SP1 P2 P3 (right triangle of sides 3, 4, and 5)
＝3 3 /4＋25 3 /4＋6＝7 3 ＋6.
The area of triangle ABC ＝ 3＋7 3 /2.
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PROBLEMS
Problem 1. As shown in the figure, AB//CD, AC//DB. Two diagonals AD and BC meet at
O. AEBC at E. DFBC at F. How many pairs of congruent
triangles are there?
(A) 4
(B) 6
Problem 2. As shown in the figure, in ABC, AB = BC = CA. AD = BE = CF. None of
D, E, F is the midpoint of any side. AE, BF, and CD intersect at M,
N, and P, respectively. If we call three congruent triangles a group,
how many groups are there?
(A) 2
(B) 3
Problem 3. As shown in the figure, AG is the centroid of isosceles triangle ABC (AB =
AC). How many pairs of congruent triangles are there?
(A) 3
(B) 6
Problem 4. In △ABC, ACB = 60, BAC = 75, ADBC at D. BEAC at E. AD and
BE meet at H. Find CHD.
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Problem 5. In △ABC, AC = BC = 5, ACB = 80. O is a point inside △ABC. If OAB =
10, OBA = 30, find the length AO.
Problem 6. In △ABC , ADBC at D. BEAC at E. AD and BE meet at F. If BF = AC,
find ABC.
(A) 40
(B) 45
Problem 7. As shown in the figure, DOAB, OA = OD, OB = OC. Find ECO + B.
Problem 8. In a triangle ABC, BD = CE, DM = ME. Prove triangle ABC is an isosceles
triangle.
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Problem 9. In the triangle ABC, AB = AC, A = 100. BD is the angle bisector of
ABC. Prove AD + DB = BC.
Problem 10. AC = BD, AD AC, BD BC. Prove AD = BC.
Problem 11. As shown in the figure, B, C, and D are collinear. AB = BC = CA. CD = DE
= EC. If CM = r, find CN.
Problem 12. As shown in the figure, ABC, C = 90, D is a point on AB. Draw
DEBC at E. BE = AC. BD = ½. DE + BC = 1. Prove: ABC = 30.
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Problem 13. ABC is a right isosceles triangle with A = 90. Taking two points E and
F on the side BC that are not the same points as B and C such that EAF = 45. Prove:
the triangle formed by three sides EF, BE, and CF is a right triangle.
Problem 14. As shown in the figure, ABC. A = 20. AB = AC. D is a point of AB. AD
= BC. Find BDC.
Problem 15. As shown in the figure, ∆ABC has AC = BC, C = 20. M is on AC and N is
on BC. BAN= 50,ABM = 60. Find NMB.
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Problem 16. As shown in the figure, convex pentagon ABCDE with B =E, C =D,
and BC = DE. M is the middle point of CD. Prove: AM CD.
Problem 17. (2006 AMC 12 B) Isosceles 4ABC has a right angle at C. Point P is inside
4ABC, such that PA = 11, PB = 7, and PC = 6. Legs AC and BC have length
s a b 2 , where a and b are positive integers. What is a + b?
(A) 85
(B) 91
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SOLUTIONS
Problem 1. Solution: (C).
△ABC △DCB, △ACD △DBA, △AOB △DOC,
△AOC △DOB, △AOE △DOF, △AEC △DFB. △AEB
△DFC.
Problem 2. Solution: (D)
(ABE, BCF, CAD); (ABM, BCN, CAP); (APD, BME,
CNF); (ABF, BCD, CAE); (AMF, BND, CPE).
Problem 3. Solution: (C).
△AGD △AGE, △DGB △EGC,
△BGF △CGF, △AGB △AGC,
△AFB △AFC, △AEB △ADC, △DBC △ECB.
Problem 4. Solution: 45.
Since ADBC at D, CAD = 60.DAC = 30.
Since BEAC, BAC = 75. , BAD = 45, EBC = 30=DAC.
Therefore ABD = 45=BAD .AD = BD.
Rt△ADC Rt△BDH
DC = DH CHD = 45.
Problem 5. Solution: 5.
Connect CO. Draw AD, the angle bisector of CAO to meet the
extension of BO at D.
We know that ACB = 80, AC = BC so CAB = CBA = 50.
Since OAB = 10, CAD = DAO = 20, DAB = 30.
In △DAB, DBA = 30. So AD = BD, ADB = 120.
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Therefore △ACD △BCD (S S S).
Hence CDA = CDB = ADB = 120.
△ADC △ADO (ASA). AO = AC = 5.
Problem 6. Solution: (B).
Since ADBC, CAD = 90– C.
Since BEAC, CBE = 90– C .
Therefore CBE = CAD.
We also know that AC = BF. So Rt△CAD Rt△FBD and AD = BD.
ABC = 45.
Problem 7. Solution: 180.
Method 1: Rt△AOC Rt△DOB (OA = OD, OC = OB, AOC =
DOB = 90°).
A = D.
Therefore ECO + B = 90 + A + 90– D = 180.
Method 2: Rt△AOC Rt△DOB.
B = ACO
So, B + ECO = ACO + ECO = 180.
Method 3: Rt△AOC Rt△DOB. So A = D. In △AOC and △DEC,
AOC= DEC = 90. Therefore four points C, O, B, and E cyclic. So
B + ECO = 180.
Problem 8. Proof:
Draw DGBC, extending BC, and then draw EI to meet the extension of BC at I such that
EIBC.
Since DGM MEI (DM = ME, GDM =MEI, alternate
interior angles of parallel lines of DG and EI, DMG =EMI) , then
DG = EI.
Now we can prove DBG CEI, and B = ECI =BCA, i.e.
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triangle ABC is an isosceles triangle.
Problem 9. Proof:
Draw DE so that DE//CB. Draw DF so that BF = BD.
Let angle EDB be angle 3. Since DE//CB, alternate
interior angles 2 and 3 are equivalent to each other.
We are given that 1 = 2, 1 = 3. Thus, BE = ED.
Because AB = AC, so ABC = ACB, and ED//BC.
6 = ABC, 7 = C.
AED is an isosceles triangle, so AE = AD. BE = CD. ED = CD.
We know A = 100, so 2 = 20, 4 = 80, 5 = 80.
AED FDC (AAS), so FC = AD, AD + DB = BC.
Problem 10. Proof:
Method 1:
Extend DA and BC to meet at point F. Focusing on triangles AED
and BEC, since AED = BEC due to the congruity of vertical
angles and DAE = EBC = 90, ADE = BCE. We can rewrite
the names of these congruent angles as FDB = FCA. We also
know that FBD = FAC = 90, and we are given that BD is
congruent to AC. Therefore, by ASA, triangle FBD is congruent to
triangle FAC. Thus, FB = FA and FD = FC. Since FD = FA + AD
= FC = FB + BC, and we have already established that FB = FA,
thus AD = DC.
Method 2:
Extend DA and BC to meet at point F. Focusing on triangles AED and
BEC, since AED = BEC due to the congruity of vertical angles
and DAE = EBC = 90, ADE = BCE. We can rewrite the
names of these congruent angles as FDB = FCA. We also know
that FBD = FAC = 90, and we are given that BD is congruent to
AC. Therefore, by ASA, triangle FBD is congruent to triangle FAC.
Thus, FB = FA and FD = FC. Since FB = FA, triangle FAB is an
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isosceles triangle and so FAB = FBA. Right angles FAC and FBD are equivalent
to one another, so
FAC FAB = FBD FBA, or
BAC = ABD. Thus, triangle BAE is isosceles, and AE = EB. Therefore, triangle AED
and BEC are congruent due to ASA.
Thus AD = DC.
Method 3:
Connect DC. Applying the Pythagorean Theorem to triangles
ADC and BDC, we get:
AD 2 AC 2 DC 2
BC 2 BD 2 DC 2
Since we are given that AC = BD, and we know that DC = DC, AD = BC.
Method 4:
Connect DC. We have AC=BD (given), DC=DC. We know
that triangle ACD and Triangle BDC are right triangles. So by
(4) Right triangles are congruent if HL HL. Therefore,
AD=BC.
Problem 11. Solution: r.
Since AC = BC, CD = CE.
ACD = 120 = BCE
∆ACD ∆BCE.
DAC = EBC. Or NAC = MBC.
ACN = 60 = BCM, and AC = BC, ∆NAC ∆MBC.
Since ACN = 60 = BCM, and AC = BC, ∆NAC ∆MBC.
Therefore CN = CM = r.
Problem 12. Proof:
Extend BC to F such that DE = CF. Connect FA and extend FA to G
such that
GA = FA. Connect BG.
We know that BE = AC, DE = FC, BED = 90 = ACF, so ∆ BED
∆ACF. So BD = AF, DBE = FAC.
Since DBE + BAC = 90, so BAF = BAC + FAC = 90.
Since GA = FA, GAB = FAB = 90, and BA is a common side,
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1
, so FG = 2AF = 1.
2
We know that BF = BC + CF = BC + DE = 1, so FG = BF.
Then BF = FG = GB, and ∆BFG is an equilateral triangle.
F = 60, so ABC = 90 – F = 30.
then ∆GAB ∆FAB. Thus GB = FB. AF = BD =
Problem 13. Solution: 90.
Draw FAD = FAC. Denote point D such that AD = AC. Connect DF
and DE.
FAD = FAC, AD = AC, AF = AF.
∆FAD ∆FAC.
FAD = FAC, FD = FC.
We know that DAE = EAF FAD = 45 FAD = 45 FAC =
BAE, and AD = AC = AB, AE = AE, so ∆DAE ∆BAE.
EDA = EBA, DE = BE.
Therefore, ∆DEF is the triangle with sides EF, BE, CF.
EDF = EDA +FDA = EBA + FCA = 90.
Problem 14. Solution: 30.
Construct equilateral triangle ∆ACE using AC as a side.
Connect DE. Then DAE = 80 = B.
Since AE = AC = AB, AD = BC, so ∆EAD ∆ABC.
AED =BAC = 20. DEC = 60 AED = 40.
However, DE = AE = EC.
1
So EDC (180 DEC ) 70.
2
We know that ADE = 80.
Therefore BDC 180 EDC ADE 30.
Problem 15. Solution: 30.
Draw MP//AB to meet BC at P. Connect PA. PA meets MB at Q. Connect
QN.
Since CAB = CBA = 80 and BAN = 50, so ANB = 50. AB = BN,
CMP = CAB = 80 = CBA = CPM. CM = CP. Since CB = CA and
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C = C, ∆CBM ∆CAP. CAP = CBM = 20, QAB = QBA = 60. Therefore
∆QAB is an equilateral triangle and PMQ = MPQ = 60.
Thus, ∆MQP is an equilateral triangle .
Since AB = BN and QB = AB, BN = BQ.
1
Since QBN = 20, BNQ (180 QBN ) 80 .
2
However, QPN = C + PAC = 40, so PQN = BNQ QPN = 40 = QPN.
Therefore QN = PN. We also know that MQ = MP, and MN = MN. Hence ∆MPN
1
∆MQN. NMQ = NMP = PMQ = 30.
2
NMB = 30.
Problem 16. Solution:
Extend BC and ED so that they meet at F. Connect AC and AD.
Since BCD = EDC, FCD = FDC.
FC = FD.
Since BC = DE, FB = FE. FBE = FEB.
Since ABC = AED, ABE = AEB.
AB = AE.
Since BC = ED, and ABC = AED, ∆ABC ∆AED.
AC = AD.
M is the midpoint of CD.
AMCD.
Problem 17. Solution: (E).
Let D, E, and F be the reflections of P about AB, BC, and CA,
respectively. FAD = DBE = 90°, and ECF = 180°. Thus the area of
pentagon ADBEF is twice that of ABC, so it is s2.
Observe that DE 7 2 , EF = 12, and FD 11 2. Furthermore,
(7 2 )2 122 98 144 = 242 (11 2 )2 , so DEF is a right triangle.
Thus the pentagon can be tiled with three right triangles, two of which are
isosceles, as shown.
1
1
It follows that s 2 (7 2 112 ) 12 7 2 85 42 2 ,
2
2
so a + b = 127.
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BASIC KNOWLEDGE
1. FORMULAS
1. S
1
1
1
aha = bhb = chc
2
2
2
(1)
2. Let ha b sin C, hb c sin A, and hc a sin B. Equation (1) becomes:
1
1
1
S bc sin A = ac sinB = ab sinC
(2)
2
2
2
3. S s(s a)(s b)(s c)
1
s = (a + b + c)
2
(3)
4. S = s ∙ r
(4)
1
(a + b + c). The figure to the right shows triangle ABC and its inscribed circle
2
O of radius r.
where s =
abc
(5)
4R
Figure to the right shows the triangle ABC and its circumcircle O of radius R.
5. S
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2. THEOREMS
Theorem 1: The ratio of the areas of any two triangles is:
1
AB H
S ABC
AB H
2
S A1 B1C1 1 A B H
A1B1 H1
1 1
1
2
Theorem 2: If two triangles have the same base, the ratio of the areas is the ratio of the
heights.
S ABD H
S ABC
h
Theorem 3: If two triangles have the height, the ratio of the areas is the ratio of the
bases.
SABC
AB
S
AB S ADC AD
; ABC
;
.
S ADC AD S DBC DB S BDC DB
Theorem 4(a): If AB//CD, then SABC = SABD and SAED = SBEC
(Same base and same height).
4(b): If SAED = SBEC, then AB//CD.
Theorem 5:
S AEC AD
S BEC DB
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Theorem 6:
Chapter 10 Area And Area Method
S AED AD
,
S BED DB
Theorem 7: If AE = n and EC = m, then
S AED n
S
n
and AEB .
S EDC m
S EBC m
Theorem 8: If AE = n and EC = m, then
S ABD n
S BDC m
Theorem 9: If DE = n and DC = m, then
S AEB n
S ABC m
Theorem 10: The ratio of the areas of two similar triangles is the same as the ratio of the
squares of their corresponding sides.
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EXAMPLES
Example 1: The area of triangle ABC is 16 cm2. D and E are
midpoints of AB and AC, respectively. F is a point on BC such that
BF = 3 cm. Find the area of triangle DEF.
Solution:
Let G be the midpoint of BC. Connect GD and GE.
Then DEG ABC.
1
Therefore SDEG SABC .
4
Since DE//BC, SDEF SDEG .
1
Therefore the shaded area = S ABC = 4 cm2.
4
Note that F can be any point on BC.
Example 2: As shown in the figure, ABCD is a rectangle. AD = 6, CD = EF = 3, and CG
= 2. Find the shaded area.
Solution:
Let the shaded area be S.
Since CD = EF =AB, ABEF is a parallelogram.
We see that rectangle ABCD has the same base and the same
height as the parallelogram ABEF, so we know that their areas
are the same.
S ABCD S AGCD SABG
(1)
S ABEF S SABG
(2)
(1) – (2): S S AGCD .
We also know that
S AGCD = S ABCD – SABG = 3 ∙ 6 –
1
∙ 3 ∙ (6 – 2) = 12, so the shaded area is also 12.
2
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Example 3: In square ABCD, AB = 1. BD and AC are arcs of radius 1. Two shaded areas
are the same. Find the difference of the unshaded areas.
(A)
– 1.
2
(B) 1 –
.
4
(C)
– 1.
3
(D) 1 –
.
6
Solution:
The areas of S1 and S form a quarter circle:
S1 = – S
4
4
The areas of S2 and S is obtained by subtracting the quarter circle from the square:
S2 + S = 1 –
S2 = 1 – S – .
4
4
Therefore, the difference of the unshaded areas is:
S1 – S2 = ( – S) – (1 – S – ) =
– 1.
4
4
2
S1 + S =
Example 4: AM is the median of ABC. D is any point on MC such that ME//AD. Show
that BDE and quadrilateral AEDC have the same areas.
Solution:
Since BM = MC, triangles ABM and AMC have the same area.
Since ME//AD, triangles OAE and OMD have the same area.
Therefore triangles ABM and BDE have the same area,
and triangles AMC and quadrilateral AEDC have the same area.
Thus, BDE and quadrilateral AEDC have the same area.
Example 5: As shown in the figure, BF and AC cut the rectangle ABCD into 4 regions.
The area of ECF is 4 cm2 and the area of CEB is 6 cm2. Find the
area of quadrilateral AEFD.
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Solution: 11.
Connect AF. We know that the area of AEF is 6 cm2 (Theorem 4(a)) and the ratio of the
area of AEF to the area of AEB is 4/6 = 2/3, so the area of AEB is
3
6 9 and the area of ABC =ADC is 9 + 6 = 15.
2
The area of quadrilateral AEFD = 15 – 4 = 11.
Example 6: (1984 AMC) In the obtuse triangle ABC, AM = MB, MD BC, EC BC. If
the area of ABC is 24, find the area of BED.
Solution: 12.
Connect MC (Figure 1).
Since MD and EC are parallel, the colored areas in Figure 2 are the same. The area of
BED is the same as the area of BMC (Figure 3), which is half of the area of ABC
(Figure 4). The answer is 24/2 = 12.
Figure 1
Figure 2
Figure 3
Figure 4
Example 7: As shown in the figure to the right, trapezoid ABCD has the
area of S. Two diagonals meet at O. The areas of AOD = S1 and
BOC = S2. Show that S ( S1 S2 )2 .
Solution:
Since BC//AD, triangle AOB and DOC have the same area S3.
Applying Theorem 7 to triangles AOB and DOC, we have
S3 BO
S1 OD
Applying Theorem 7, to triangles BOC and COD, we have
S 2 BO
S3 OD
207
(1)
(2)
50 AMC Lectures
Therefore, we have
Chapter 10 Area And Area Method
S3 S 2
S1S2 S32
S1 S 3
S1S2 S3
(3)
So ( S1 S2 )2 S1 S2 2 S1S2 S1 S2 2S3 S .
Example 8: (1980 AMC) In triangle ABC, CBA = 72, E is the midpoint of side AC,
and D is a point on side BC such that 2BD = DC; AD and BE
intersect at F. The ratio of the area of ∆BDF to the area of
quadrilateral FDCE is
1
1
1
2
(A)
(B)
(C)
(D)
(E) none of these
5
4
3
5
Solution: (A).
The official solution of this problem is given in “Chapter 18 Eight Methods To Draw
Auxiliary Lines”. Our first solution to this problem is given in “Chapter 9 Similar
Triangles”. Here we are providing a second solution of our own.
Connect CF.
We know that E is the midpoint of AC, so AE = EC.
SBAE SBEC and SFAE SFEC .
We also know that 2BD = DC, so 2SBFD SDFC .
Now, we are ready to label each area as shown in the figure
to the right with a, b, and c.
We have: 2(b a) 2a 2c
c = 3a
b c a 2a c
b=c
Therefore the ratio of the area of ∆BDF to the area of quadrilateral FDCE is
SBDF
a
a
a 1
.
S FDCE 2a c 2a 3a 5a 5
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50 AMC Lectures
Chapter 10 Area And Area Method
Example 9: (1991 AMC) If ABCD is a 2 × 2 square, E is the midpoint
of AB , F is the midpoint of BC , AF and DE intersect at I, and BD
and AF intersect at H, then the area of quadrilateral BEIH is
1
2
7
8
3
(A)
(B)
(C)
(D)
(E)
3
5
15
15
5
Solution: (C).
We know that ∆ADH ~ ∆FBH, so since AD = 2BF, DH = 2HB. Connect BI. Let the area
of ∆BEI be x and the area of ∆BHI be y. We label each region as follows:
Therefore
x + 3y = 1
2x + 2.25y = 1
Solving for x and y: x = 1/5 and y = 4/15.
The area of quadrilateral BEIH is x + y = 7/15.
Note: The AMC gives three solutions for this problem. The solution presented above is
the fourth solution given by us.
Example 10: (2012 ARML Individual) In triangle ABC, C is a right angle and M is on
AC. A circle with radius r is centered at M, is tangent to AB, and is tangent to BC at C. If
AC = 5 and BC = 12, compute r.
Solution:
Method 1 (Official Solution):
Let N be the point of tangency of the circle with AB
and draw MB, as shown to the right. Because BMC
and BMN are right triangles sharing a hypotenuse,
and MN and MC are radii, BMC BMN. Thus BN
= 12 and AN = 1. Also ANM ACB because the
right triangles share A, so NM/AN = CB/AC.
Therefore r /1 = 12/5, so r = 12/5.
Method 2 (our solution):
ABC is a 5 – 12 – 13 right triangle with AB = 13. Drop a perpendicular to AB from M to
meet AB at N. Connect BM.
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50 AMC Lectures
SBMC SBAM SABC
Chapter 10 Area And Area Method
12 r 13 r 12 5
2
2
2
BC MC AB MN BC AC
2
2
2
12
r .
25r 12 5
5
Example 11. (1985 AIME) As shown in the figure on the right, ∆ 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 ∆ ABC.
Solution: 315.
Method 1 (Official Solution):
Let the areas of two unknown triangles be x and y.
By Theorem 9:
40 40 y 84
=
(1)
30 30 35 x
35 35 30 40
=
(2)
x
x 84 y
84 x 35
84
=
(3)
y 40 30
y
Solving we get: x = 70, y = 56.
The area is then 30 + 35 + 70 + 84 + 56 + 40 = 315.
Method 2 (Our solution):
We label the points D and E as shown in the figure.
Looking at triangles ADC and BDC, by the Theorem 5, we have
84 m 40
(1)
n 35 30
Looking at triangles ADB and DEB, by the Theorem 3, we have
AD 40 30
2.
DE
35
Looking at triangles ADC and DEC, by the Theorem 3, we have
m 84 AD
2
84 + m = 2n
(2)
n
DE
210
50 AMC Lectures
Chapter 10 Area And Area Method
2n
4
6n = 4n +140 n =70.
n 35 3
Substituting n = 70 into (2), we get m = 56.
Therefore the area of ABC is 84 + 40 + 30 + 35 + 126 = 315.
(Our solution may be a little neater and quicker than the first solution)
Substituting (2) into (1), we get:
Method 3 (Our Solution):
For any triangle, we know that following relationship is true:
pqr = mnk
1 1 1 1 1 1
p q r m n k
(1)
(2)
(see Example 18 for more details about these relationships)
For our case, let the areas of two unknown triangles be m and n.
We have
84 40 35 = 30mn
(3)
1
1
1
1 1 1
(4)
84 40 35 m n 30
or mn = 3920
(5)
1 1 1
1
1
1
9
(6)
m n 84 40 35 30 280
1 1 mn
9
From (6), we have
(7)
m n
mn
280
Substituting (5) into (7), we get
9
9
mn
mn
3920 126 .
280
280
Therefore the area of ABC is 84 + 40 + 30 + 35 + 126 = 315.
Example 12: Let P be an interior point of 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 = 29 and d = 6.
Solution:
Let SABC = S, SAPF S1 , SBPF S2 .
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50 AMC Lectures
Chapter 10 Area And Area Method
SBPD S3 , SCPD S4 , SCPE S5 , SAPE S6 .
Therefore
S1 S2
d
6
S
cd c6
S3 S 4
6
S
a6
S5 S 6
6
S
b6
6
6
6
+
+
= 1.
a6 b6
c6
Solving we get abc = 36 (a + b + c) + 432.
(1)
(2)
(3)
Example 13: A wire of length of a is used to form an equilateral triangle with the area of
b. Let P be any point inside the equilateral triangle. Find the sum of the distances from P
to each side.
(A)
2b
.
a
(B)
4b
.
a
(C)
6b
.
a
(D)
8
a
Solution:
Method 1:
a
.
3
a
As shown in the figure, AB = BC = CA = .
3
Let x, y, z be the distances from P to AB, BC, and AC, respectively.
The length of each side of the triangle is
We know that SABC = b.
SPBC + SPCA + SPAB = SABC
1
a
1
a
1
a
× x + × y+ × z=b
2
3
2
3
2
3
a
( x + y + z) = b
6
6b
x+y+z=
.
a
Method 2 (our solution):
212
50 AMC Lectures
Chapter 10 Area And Area Method
3 a
3
( )
a
2 3
6
where h is the height of the equilateral triangle ABC.
We know that x + y + z = h
We also know that the area of the equilateral triangle ABC is
3 a 2
3 2
3 6b
b
( )
a
2
4 3
36
6
a
3
6b
6b
Substituting (2) into (1), we have x + y + z =
.
a 2 a
6
a
a
(1)
(2)
Example 14: O is the circumcenter (the center of circumcircle O of radius R about
triangle ABC, where the perpendicular bisectors of the sides intersect) of triangle ABC.
1
1
1
2
Show that
+
+
= .
AD
CF
BE
R
Proof:
AO
S
S
S
S ACO
= ABO = ACO = ABO
(1)
AD
S ABC
S ABD
S ACD
BO
S
S BAO
= BCO
(2)
BE
S ABC
CO
S
SCBO
= CAO
(3)
CD
SABC
(1) + (2) + (3):
AO
BO
CO
2( SABO SBCO SCAO )
2S ABC
+
+
=
=
= 2.
SABC
AD
BE
CD
S ABC
OA = OB = OC = R,
R
R
R
R
R
R
2
+
+
= 2,
+
+
= .
AD BE
CF
AD BE
CF
R
Example 15: As shown in the figure, P is any point inside the triangle
ABC. Line segment AP, BP, CP meet BC at D AC at E, and AB at F,
AF BD CE
respectively. Prove that
∙
∙
= 1.
BF CD AE
Solution:
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50 AMC Lectures
Chapter 10 Area And Area Method
Taking a closer look at CAF and CFB, since they share the same vertex C, the ratio of
their areas is the same as the ratio of their bases (AF and FB).
AF
APC
Therefore we have
=
(1)
BF
BPC
BD
APB
Similarly we have
=
(2)
CD
APC
CE
BPC
and
=
(3)
AE
APB
Multiplying (1), (2), and (3) together we get:
AF BD CE
APC APB BPC
∙
∙
=
∙
∙
=1
BF CD AE
BPC APC APB
Note that the conclusion is still valid if P is outside the triangle ABC. This is called
“Ceva's Theorem”.
The converse is also true: If points D, E and F are chosen on BC, AC and AB respectively
AF BD CE
so that
∙
∙
= 1, then AD, BE and CF are concurrent. See Lecture
BF CD AE
“Geometry Menelaus and Ceva” for details.
Example 16: a 2 b 2 , 4a 2 b2 , a 2 4b2 are three sides of a triangle, where a, b,
and c are positive numbers. Find the area of the triangle.
Solution:
We observe that the lengths of the three sides are
(2a)2 b2 , and
a 2 b2 ,
a 2 (2b)2 .
We can construct a rectangle ABCD such that
EF = a 2 b 2 , EC = (2a)2 b2 , and CF = a 2 (2b)2 .
Therefore, the area of the triangle will be:
SCEF = S ABCD – SAEF – SBCE – SCDF = 4ab –
214
1
3
ab – ab – ab = ab.
2
2
50 AMC Lectures
Chapter 10 Area And Area Method
Example 17: Find the area of triangle ABC if three medians are known.
Solution:
2
We know that AP = AD.
3
1
1
So APF = ABP = ABC.
2
6
Take M, the midpoint of AP and connect FM:
1
1
1
MP ma , FP mc , FM mb
3
3
3
1
1
SABC
SMPF = SAPF =
2
12
By Heron’s Formula, we have
1
1
m(m ma )(m mb )(m mc )
SABC = SMPF =
12
9
1
where m = (ma mb mc ) .
2
4
m(m ma )(m mb )(m mc ) .
Therefore, SABC =
3
Example 18: In triangle ABC, AD, BE, and CF meet at O. Let the areas of six small
triangles be p, q, r, m, n, and k, respectively, as shown in the figure. Show that
pqr = mnk
1 1 1 1 1 1
p q r m n k
(1)
(2)
Solution:
We prove (1) first:
From Example 15, we proved Ceva’s Theorem that stated that:
AF BD CE
1
(3)
FB DC EA
By Theorem 6, we have
AF p
FB k
BD q
DC m
(4)
(5)
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50 AMC Lectures
Chapter 10 Area And Area Method
CE r
EA n
(6)
Substituting (4), (5), and (6) into (3):
p q r
1 pqr = mnk
k m n
Now we will prove (2):
By Theorem (5), we have:
q BD p k
m CD n r
Similarly we get
qn + qr = mp + mk
(7)
rp + rk = nq + nm
pq + pm = kn + kr
(8)
(9)
(7) + (8) + (9):
qr + rp + pq = kn + mk + nm
(10)
Dividing left hand side by pqr and right hand side by mnk:
1 1 1 1 1 1
.
p q r m n k
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50 AMC Lectures
Chapter 10 Area And Area Method
PROBLEMS
Problem 1: D and C trisect the arc of the half circle as shown in the figure. Find the
shaded area if the area of the half circle is 9.
Problem 2: As shown in the figure, right triangle ABC with BC = 20 cm. BDC is a half
circle with the diameter BC. The difference between two shaded areas I and II is 23. Find
AC in terms of .
Problem 3: The larger one of the two squares in the figure below has the side length of
10 cm. Find the shaded area.
Problem 4: As shown in the figure, in ABC, D is any point on AB. DE//BC and meets
BC at F. The area of ADE is S1. The area of DBF is S2. Find the area of
parallelogram DFCE.
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50 AMC Lectures
Chapter 10 Area And Area Method
Problem 5: In trapezoid ABCD, The area of ABC is 253 cm2. The area of AOB is 42
cm2 larger than the area of COD. Find the area of trapezoid ABCD.
Problem 6: The area of ABCis 1. As shown in the figure, D and E trisect BC. F and G
trisect CA. Find the area of quadrilateral PECF.
Problem 7: As shown in the figure, P is a point inside the equilateral triangle ABC. PQ
AB, PR BC, PS AC. PQ = 6, PR = 8, PS = 10. Find the area of ABC.
(A) 190 3
(B) 192 3
(C)194 3
(D)196 3
Problem 8: In triangle ABC, AD and BE are medians and they meet at M. Prove that MD
1
= AD.
3
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50 AMC Lectures
Chapter 10 Area And Area Method
Problem 9: As shown in the figure, square ABCD is inscribed in circle O. Find the side
of the square and the area of the circle if the shaded area is (2 + 4) cm2.
Problem 10: As shown in the figure, BC = CE，AD = CD. Find the ratio of the area
triangle ABC to the area of triangle CDE.
Problem 11: In, RtABC C = 90. D is a point on the hypotenuse
AB. DF AC at F. DE BC at E. SADF = 9, SDBE = 16. Find the area
of rectangle DECF.
Problem 12: As shown in the figure, SABC 4, AE ED, AF
1
AC .
4
Find the shaded area.
Problem 13: Triangle ABC is divided into four parts with the areas of three parts shown
in the figure. Find the area of the quadrilateral AEFD.
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50 AMC Lectures
Chapter 10 Area And Area Method
Problem 14: (2002 AMC12 A) Triangle ABC is a right triangle with ACB as its right
angle, m ABC = 60◦, and AB = 10. Let P be randomly chosen inside triangle ABC, and
extend BP to meet AC at D. What is the probability that BD > 5 2 ?
(A)
2 2
2
(B)
1
3
(C)
3 3
3
(D)
1
2
(E)
5 5
5
Problem 15: (1986 AIME) In ABC shown below, AB = 425, BC = 450 and CA = 510.
Moreover, P is an interior point chosen so that the segments DE, FG and HI are each of
length d, contain P, and are parallel to the sides AB, BC and CA, respectively. Find d.
Problem 16: As shown in the figure, D is any point on BC of triangle ABC. DE//AB and
meets AC at E. DF//AC and meets AB at F. Show that
S2AEF = SBDF SDCE
Problem 17: As shown in the figure, ABCD is a convex quadrilateral. AD = CD, DAB
S
= 90, BCD = CDA = 120. Find ABD .
S BDC
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50 AMC Lectures
Chapter 10 Area And Area Method
Problem 18: As shown in the figure, ABC is a right isosceles triangle with the right angle
1
at A. Find AE if EC = 2013 cm, and DC = BC.
3
Problem 19: Prove the Heron formula for the area of any triangles with three sides
known:
1
S s(s a)(s b)(s c) , where s = (a + b + c).
2
Problem 20: (2012 ARML Team #7) Given noncollinear points A, B, C, segment AB is
trisected by points D and E, and F is the midpoint of segment AC. DF and BF intersect
CE at G and H, respectively. If [DEG] = 18, compute [FGH].
221
50 AMC Lectures
Chapter 10 Area And Area Method
SOLUTIONS TO PROBLEMS
Problem 1: Solution: 3.
Connect OD and OC.
Since AD = DC = CB, CD//AB.
SOCD SACD .
Therefore the shaded area is 1/3 of the area of the half circle:
9/3 = 3.
50 23
.
10
Let AC = x, SABC be the area of triangle ABC, and S BDC be the area of half circle BDC.
From the diagram to the right, we see that
Problem 2: Solution:
SABC S III S II
(1)
S BDC S III S I
(2)
(1) – (2): SABC SBDC SII SI 23
1
= ∙ 20x –
∙ 102 = 10x – 50π.
2
2
50 23
Or 10x – 50π = 23
x=
.
10
Problem 3: Solution: 50.
We connect BD and we know that DB//GE, so the area of triangle GHD is the same as the
area of triangle BEH (Theorem 4(a)).
The area of the shaded region, triangle EDG, is the same as the area of
triangle BEG, which is 10 10 2 = 50.
Problem 4: Solution: 2 S1S2 .
Let SABC = S.
We have
S1
AD 2
S
BD 2
(
) and 2 (
) .
S
AB
S
AB
S2
AD BD
S1
+
=
= 1.
S
S
AB
222
50 AMC Lectures
S = ( S1 +
Chapter 10 Area And Area Method
S 2 )2
SDFEC = ( S1 +
S 2 )2 – S1 – S2 = 2 S1S2 .
Problem 5: Solution: 464.
Since CD//AB, ABC and ABD have equivalent areas.
We label each area as shown in the figure to the right with a, b, x,
and y.
We can write the following equations:
x + a = 253
(1)
x y = 42
(2)
(1) – (2): a + y = 211
The area of trapezoid ABCD = a + a + x + y = (x + a) + (a + y) = 253 + 211 = 464 cm2.
Problem 6: Solution:
Since EC =
1
.
6
1
1
1
BC, SAEC = SABC = .
3
3
3
Because
1
1
1
CF = CA, SBCF = SABC = .
3
3
3
Connect PC and let SPEC = x, SPFC = y.
We have SPBC = 3 x, SPCA = 3y.
1
1
Therefore x + 3y = and 3x + y = .
3
3
Adding these two equations together we get 4x + 4y =
Hence SPECF =
2
3
1
.
6
Problem 7: Solution: 192 3 .
Let the length of the sides of equilateral triangle ABC be a.
Connect PA, PB, PC.
The area of triangle ABC can be expressed as:
223
x+y=
1
.
6
50 AMC Lectures
Chapter 10 Area And Area Method
1
1
a(PQ + PR + PS) = a(6 + 8 + 10) = 12a.
2
2
The area of the triangle can also be expressed as:
3 2
SABC =
a
4
3 2
We have: 12a =
a
4
Solving for a: a = 16 3 .
Therefore SABC = 12a = 12 × 16 3 = 192 3 .
SABC = SPAB + SPBC + SPCA =
Note that h, the height of an equilateral triangle ABC equals the sum of the distances from
P, any point inside the triangle ABC, to each side.
Problem 8: Solution:
1
.
3
Connect CM.
Applying Theorem 5 onto AMB and AMC, we have
AMB BD
= 1 and
AMC CD
AMB AE
= 1.
CMB CE
Therefore we get:
1
ABC
3
MD BMC
1
By Theorem 9, we have:
= .
AD ABC
3
AMB = CMB = AMC =
Problem 9: Solution: 8π.
Connect BD. Let the side of the square ABCD be x. Then the radius of the circle is
Let the area of the circle be SC, and the shaded area be SS.
1
1
2 2 1 2 2
S s Sc + S ABCD 2π + 4 = (
x) + (
x)
4
4
4 2
4 2
x2 = 16
x = 4 cm.
224
2
x.
2
50 AMC Lectures
Chapter 10 Area And Area Method
The area of the circle is
2
4) 2 = 8π (cm2)
Sc = (
2
Problem 10: Solution: 2:1.
Connect BD. ABD and BDC have the same area, because AD = CD.
BDC and CDE have the same area because BC = CE. The ratio of the
area triangle ABC to the area of triangle CDE is then 2:1.
Problem 11: Solution: 24.
Method 1:
Let SABC S , SADF S1 , and SDBE S2 .
We know that ADF ABC and DBE ABC.
S1 DF
S ADF DF 2
2
S ABC
BC
BC
S
S2
Similarly, we have
(1) + (2):
S1
S
+
S
S2
S
=
BE
BC
(1)
(2)
DF BE BC
1
BC
BC
Therefore S ( S1 S 2 )2 ( 9 16 )2 49 .
The area of rectangle DECF is S S1 S2 49 9 16 24
Method 2:
By the formula derived from problem 4, SDECF = 2 S1S2 = 2 9 16 2 3 4 24 .
Problem 12: Solution: 1.
Method 1: Since E is the midpoint of AD, AEB and EBD have the same area.
The shaded area given is the same as the area of triangle AFB.
S AFB
AF 1
SAFB
1
S AFB
1
SABC SAFB 3
4 SAFB 3
S FCB FC 3
S AFB 1
The area of the shaded region is 1.
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50 AMC Lectures
Chapter 10 Area And Area Method
Method 2:
Connect CE. Let x denote the area of AEF and y denote the area of AEB.
Then the area of EBD is y and the area of CEF is 3x.
S
AF 1
Since AFB
, 3(x + y) = 3x + y + SCDE
S FCB FC 3
SCDE = 2y.
Since SABC 4 , (x + y) + 3x + y + 2y = 4
y=1
The area of the shaded region is 1.
4x +4y = 4
x+
Problem 13: Solution: 22.
Connect AF.
S ABF
S
5 x y
AFD
10
8
S BFC S FDC
S
S
8 y x
Similarly, we have ACF AEF
S BCF S EFB
10
5
Solving the equations for x and y, we get x =10, and y =12. The area of AEFD is 22.
Problem 14: Solution: (C).
Since AB is 10, we have BC = 5 and AC = 5 3 .
Choose E on AC so that CE = 5. Then BE = 5 2 .
For BD to be greater than 5 2 , P has to be inside triangle ABE. The probability that P is
inside triangle ABE is
1
EA BC
Area ofABE
EA
5 3 5
3 1 3 3
2
=
=
=
=
=
.
1
Area ofABC
AC
3
5
3
3
CA BC
2
Problem 15: Solution:
One solution is given in “Chapter 21 Similar Triangles”.
Here is our solution to the problem using the area method:
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50 AMC Lectures
Chapter 10 Area And Area Method
S CDE
d2
=
S ABC
AB 2
(1)
S BHI
d2
=
AC 2
S ABC
(2)
S AGF
d2
=
BC 2
S ABC
(3)
S PIF
( AB d ) 2
=
AB 2
S ABC
(4)
S PEH
( BC d ) 2
=
AB 2
S ABC
(5)
S PGD
( AC d ) 2
=
AC 2
S ABC
(6)
(1) + (2) + (3) – [(4) + (5) + (6)]:
SABC = SCDE + SBHI + SAGF – ( SPIF + SPHE + SPGD )
1
1
1
d(
+
+
)=2
d = 306.
AB
BC
AC
Problem 16: Solution:
Let BD = m, DC = n, triangle ABC be and three triangles be 1 , 2 , and 3 .
S
S 1
DF // AC
S
= 23 =
=k
1 3
2
DE // AB
n
m
( m n) 2
S 1 m 2 k
S 3 n 2 k
2
S (m n) k
We know that 2 S 2 = S AFDE S S1 S 3 2mnk .
So S 2 = mnk and S2 2 m2n2 k 2 = m2k n2k S 1 S 3 .
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50 AMC Lectures
Chapter 10 Area And Area Method
4
3
Extend AD and CD so that they meet at E.
Since ADC = BCD = 120, EDC = ECD = 60.
Therefore, CDE is an equilateral triangle, so ED = CD = AD.
This also means that SABD = SBDE .
Problem 17: Solution:
Let AD = ED = a. We have AE = 2a, AB 2 3a and
3 2
a .
4
1
SBDE a 2 3a = 3a 2 .
2
S
4
1
S
Therefore SCDE SBDE . So ABD = BDE .
SBDC 3
4
S BDC
SCDE
Problem 18: Solution: 2013.
Let DAE = and DAB = .
Since BE AD,ABE = and AEB = ..
AE
AE
DC ADC AC AD sin
sin
=
=
=
AB
AC
BD ADB AB AD sin
sin
1
We know that DC = BC, so BD = 2DC and AC = 2AE.
3
Therefore AE = EC= 2013 cm.
Problem 19: Proof:
Let BD = x. Then DC = a – x.
Using the Pythagorean Theorem twice on triangles ABD and ADC
yields
c 2 x 2 ha 2 b2 (a x)2 .
a 2 c 2 b2
Solving for x in terms of a, b, and c: x
.
2a
Therefore
1
ha2 c 2 x 2 =
[ 4a 2c 2 (a 2 c 2 b2 )2 ]
2
4a
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Chapter 10 Area And Area Method
1
1
( 2ac a 2 c 2 b2 ) ( 2ac a 2 c 2 b2 ) =
[ (a c)2 b2 ][( b2 (a c)2 ]
2
2
4a
4a
1
4
=
(a + b + c)(a – b + c)(a + b – c)( – a + b + c) = 2 s(s – a)(s – b)( s – c )
2
4a
a
=
Or
ha2 a 2
= s(s – a)( s – b)( s – c ).
4
Take the square root on both sides to give:
ha a
s( s a)(s b)(s c) .
2
Problem 20: Solution:
Our solution:
Method 1:
We first connect DC and EF. Since E and F are the
midpoints of DA and CA, respectively, DC//EF. Therefore,
[DEG] = [CFG].
Next, we connect GA. Triangle CFG and triangle AGF
share the same height, and their bases are equal to each
other since F is the midpoint of CA, so they have the same
area. Similarly, [DEG] = [AGF].
Thus, the areas of triangles DEG, CFG, AGF, and EAG equal 18.
[DFA] = [DEF] + [EFA] = 18 + 18 + 18 = 54. Since E is the midpoint of DA, [DEF] =
1
[EFA] = [ DFA] = 27. Triangles DEF and BDF share the same height and have equivalent
2
bases, so [BDF] = [DEF] = 27.
As we determined already, [AGF] = 18 and [DGA] = [DEG] + [EAG] = 18 + 18 =36.
Triangles AGF and DGA share the same height, therefore the ratio of their bases:
FG 1
.
GD 2
Let [FGH] = x. By the ratio above, we see that [DGH] = 2x.
Therefore [BDF] = [BDH] + 2x + x.
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Chapter 10 Area And Area Method
We have determined that the area of triangle [BDF] is 27, so the equation above becomes
27 = [DEH] + 2x + x
(1)
We also know that
[DEH] = [DGH] + [DGE] = 18 + 2x.
Substituting this into equation (1), we get
27 = 18 + 2 x + 2x + x
The area of FGH is 9/5.
x = 9/5.
Method 2:
We draw FK so that FK//AB and FK meets CE at K.
Let [ FGH ] x and [ FKH ] y .
Since FK//AB and F is the midpoint of CA,
1
1
1
KF EA DE BE .
2
2
4
Because FK//AB, triangles FKG and DEG are similar. Since the ratio of the area of two
similar triangles is equal to the square of the ratio of their corresponding sides,
2
18 9
[ FKG ] FK
1
[ FKG ]
4 2
[ DEG] DE
4
9
(1)
x y
2
Because FK//AB, triangles FKH and BEH are similar.
Since the ratio of the area of two similar triangles is equal
to the square of the ratio of their corresponding sides,
2
[ FKH ] FK
1
[ BEH ] BE 16
(2)
From method 1, [BDF] = 27, so [ BEH ] [ BDF ] [ HGF] [ DGE] 27 x 18 .
Substituting this into equation (2), we get
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50 AMC Lectures
[ FKH ]
Chapter 10 Area And Area Method
1
(27 x 18)
16
Or
y
1
(27 x 18)
16
(3)
Solving the system of equations (1) and (3) for x: x = 9/5.
Note: ARML has two official solutions (one used Menelaus's Theorem, and one used the
method of mass points) for this problem.
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
BASIC KNOWLEDGE
Auxiliary lines: An auxiliary line is a line or line segment added to a diagram to help in
solving a geometry problem. It is often shown as a dashed line in the diagram.
In this lecture, we introduce eight commonly used methods to draw auxiliary lines.
1. Construct congruent triangles using the angle bisector.
(1). BD is the angle bisector of ABC. P is the point on BD.
Drawing PEAB, PFBC, we get PE = PF. BPE BPF.
Theorem 1: Any point on the bisector of an angle is equidistant
from the sides of the angle.
(2). AB > AC, 1 = 2. Find the point E on AB so that AE = AC, and connect DE. ADE
ADC.
(3). Extend AC to F such that AF = AB. Connecting DF, we get
ADF ADB.
(4). Extend CA to E so that CE = CB. Then CDE CDB,
E =B.
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Theorem 2: The Angle Bisector Theorem: The bisector of an angle of a triangle divides
the opposite side into segments whose measures are proportional to
the measures of the other two sides of the triangle.
(We have four methods to prove this theorem in Chapter 12 Angle
Bisector and Median).
Example 1. As shown in the figure below, in ABC, CAB =
2ABC. CD is the angle bisector of ACB. Show that BC = AC + AD.
Solution:
Method 1:
Draw DE so that CE = AC. Since 1 = 2, and CD = CD, we have
ACD ECD. Therefore CED =CAB, and AD = DE.
Since CAB = 2ABC, CED = 2ABC, and CED = DBE
+EDB.
Hence EDB = EBD DE = EB.
Therefore BC = CE + EB = AC + DE = AC + AD.
Method 2:
Extend CA to E such that AE =AD. Therefore E =ADE, and
CAB = 2E.
Since CAB = 2ABC, E = ABC.
Since 1 = 2 and CD = CD, CED CBD.
Therefore CE = CB. BC = CE = CA + AE = CA + AD.
Method 3:
(Figure 3). Extend CA to E so that CE = CB. Then CDE CDB, E =B.
Since A = 2B, and A =E +ADE, so 2B = E +ADE or
2E = E +ADE, or E =ADE, i.e, triangle ADE is an
isosceles triangle with AD = AE. Therefore BC = CA + AE = AC +
AD.
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2. Construct the median on the hypotenuse of a right triangle.
Theorem 3: The measure of the median on the hypotenuse of a right triangle is one-half
the measure of the hypotenuse (AM = MB = MC).
The converse of this statement, if the median to a side of a triangle
is one-half of the measure of that side, or AM = MB = MC, then
triangle ABC is a right triangle, is also true.
Example 2. In ABC, B = 2A. AB = 2BC. Show that AB2 = AC2 + BC2.
Solution:
Since AB > BC, C A .
Draw CD to meet AB at D such that ACD A .
∆ADC is an isosceles triangle and AD = DC.
CDB ACD A 2A B
So ∆BCD is also an isosceles triangle. Therefore DC = BC.
AB AD DB 2BC
AD DB 2DC 2 AD
AD DB DC .
Therefore DC is the median of right triangle ABC with C =90.
By the Pythagorean Theorem, we have AB2 = AC2 + BC2.
3. Draw a line connecting the midpoints of triangle or trapezoid.
Theorem: If a line contains the midpoint of one side of a triangle (AB) and is parallel to
a second side (AC) of the triangle, then it will bisect the third side (BN = NC) of the
triangle.
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Theorem 4: The line segment whose endpoints are the midpoints of two sides of a
triangle is parallel to the third side of the triangle and has a measure equal to one-half of
1
the measure of the third side ( MN AC ).
2
Example 3: In ABC, point E is the midpoint of AC. D is on BC
and BD = 1/3 BC. Show that AD bisects BE.
Solution:
Let the point of intersection of BE and AD be P. Connect EF
such that F is the midpoint of DC.
Since E is the midpoint of AC and F is the midpoint of DC,
AD//EF.
1
Therefore EF AD .
2
In BEF, PD//EF, BD = DF. Therefore PD bisects BE, or in
other words, AD bisects BE.
Example 4: As shown in the figure, in triangle ABC, M is the
1
midpoint of BC. AN AC . Connect BN and BN meets AM at
3
P. Show that BP = 3PN.
Solution:
Method 1:
Draw a line through M parallel to BN to meet AC at D.
In triangle CBN, since MB = MC and MD//BN, DC = DN and
2MD = BN.
(1)
In triangle AMD, AN = ND = DC, PN//MD, so
2PN = MD
Substituting (2) into (1):
4PN = BN
4PN = BP + PN
BP = 3PN.
(2)
Method 2:
Draw a line segment through M parallel to AC to meet BN at D.
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In triangle CBN, since MB = MC and MD//CN, 2DM = CN and
DM = AN.
We also see that ∆ANP ~ ∆MDP. So DP = PN.
BN = BD + DN = 2DN = 4PN = 3PN + PN = BP + PN.
BP = 3PN.
4. Draw two heights of trapezoid from the short base to the long base.
In trapezoid ABCD, AB//DC. Draw AE and BF such that
AEDC, BFDC.
As shown in the figure to the right,
AE = BF, AB = EF.
DF + CE = DC + EF = DC + AB.
Example 5: (AMC) Figure ABCD is a trapezoid with
AB//DC; AB = 5; BC 3 2 , BCD = 45 and CDA =
60. The length of DC is
2
3
3
(E) 8 3 3
(A) 7
(B) 8
(C) 9
1
2
(D) 8 3
Solution: (D).
Drop perpendiculars from A and B to DC, intersecting DC at
F and E, respectively.
BEC is an isosceles right triangle, so BE = EC = 3. Since
ABEF is a rectangle, FE = 5 and AF = 3. AFD is a 30 60 90 triangle, so DF = AF
/ 3 = 3.
DC = DF + FE + EC = 3 + 5 + 3 = 8 + 3 .
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
5. Draw a parallel line through a point to a given line to construct proportional line
segments
If DE//BC, then ABC ADE.
AD AE DE
AB AC BC
AD AE
DB EC
AD DB
AE EC
Theorem 5: If a line is parallel to one side of a triangle it divides the other two sides of
the triangle proportionally. The converse is also true.
Theorem 6: If a line is parallel to one side of a triangle and intersects the other two
sides, it determines (with segments of these two sides) a triangle similar to the original
triangle.
Theorem 7: If a line segment is divided into congruent (or proportional) segments by
three or more parallel lines, then any other transversal will similarly contain congruent
(or proportional) segments determined by these parallel lines.
Example 6: As shown in the figure, in triangle ABC, M is the midpoint of BC.
1
AN AC . Connect BN and BN meets AM at P. Show that BP = 3PN.
3
Solution: This problem is the same as Example 7. Here, we use
two new, different ways to solve it.
Method 1:
Draw a line through N parallel to AM to meet CM at D.
AN 1 MD 1
,
.
Since
AC 3 MC 3
MD 1
.
We know that BM = MC, so
BM 3
PN 1
We also know that MP // DN, so
BP = 3PN.
BP 3
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Method 2:
Draw a line through N parallel to BC to meet AM at D.
AN 1
Since,
.
AC 3
Since MB = MC, 3DN = MC = MB.
DN PN 1
Since PDN PMA,
BP = 3PN.
MC PB 3
Example 7. In triangle ABC, if AD = AE, show that
Proof:
DB BF
.
CE CF
Method 1.
Draw CG//AB. 1 =4.
BD BF
CG CF
Since AD = AE, so 1 = 2, 3 = 4, CE = CG.
BD BF
So
CE CF
Method 2:
Draw DG//AC.
DG GF
or
EC CF
DG EC
GF CF
(1)
Since ADE = AED = GDF, so DF is the exterior angle
bisector of DBG.
Then
DB DG
BF GF
Considering (1) and (2), we get
(2)
DB BF
.
CE CF
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
6. Draw the radius passing through the tangent point
B is the tangent point and O is the center. Connect OB. We have
OB AB.
Theorem 8: The radius of a circle is only perpendicular to a
tangent line at the point of tangency.
Theorem 9: If a line is tangent to a circle, it is perpendicular to a
radius at the point of tangency.
Theorem 10: A line perpendicular to a radius at a point on the circle is tangent to the
circle at that point.
Theorem 11: A line perpendicular to a tangent line at the point of tangency with a circle
contains the center of the circle.
Example 8: In ABC, in which AB = 12, BC = 18, and AC = 25, a semicircle is drawn so
that its diameter lies on AC, and so that it is tangent to AB
and BC. If O is the center of the circle, find the measure of
AO.
Solution:
Draw radii OD and OE to the points of contact of tangents AB and BC, respectively. OD
= OE (radii), and BDO =BEO = 90°. Since DB = BE, right BDO right BEO, and
DBO = EBO.
In ABC, BO bisects B so that
Let AO = x; then
AB BC
.
AO OC
12
18
and x = 10 = AO.
x 25 x
Example 9: (AMC) Two circles are externally tangent. Lines PAB and PAB are
common tangents with A and A on the smaller circle and B and B on the
larger circle. If PA = AB = 4, then the area of the smaller circle is
(A) 1.44 (B) 2 (C) 2.56 (D) 8
(E) 4
Solution:
Connect PO, BO, AO1. Since PA = AB, OB PB, O1A PB,
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50 AMC Lectures
PB = PA + AB = 8,
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
OB = 2r.
PO = 2OO = 2(R + r) = 2(2r + r) = 6r
By the Pythagorean Theorem in right triangle POB, PO 2 OB2 PB 2 .
64 + 4r2 = 36r2
r2 = 2
The area of the smaller circle is 2.
7. When two circles are tangent, draw the common tangent line or connect the centers.
AB = O1C.
AB O1C.
AB O2B.
O1O2= r1 + r2.
O2C= r2 r1.
∆ O1CO2 is a right triangle.
O1C (r2 r1 )2 (r2 r1 )2
AB is the common tangent line. r1 and r2 are the radius of circle O1 and O2, respectively.
Example 10: (2006 12 B) Circles with centers O and P have radii 2 and 4, respectively,
and are externally tangent. Points A and B are on the circle centered at O, and points C
and D are on the circle centered at P, such that AD and BC are
common external tangents to the circles. What is the area of
hexagon AOBCPD?
(A) 18 3
(B) 24 2
(C) 36 (D) 24 3 (E) 32 2
Solution: (B).
Through O draw a line parallel to AD intersecting PD at F. Then AOFD is a rectangle and
OPF is a right triangle. Thus DF = 2, FP = 2, and OF = 4 2 .
The area of trapezoid AOPD is 12 2 , and the area of
hexagon AOBCPD is 2 × 12 2 = 24 2 .
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8. When four points are concyclic, draw the circle.
Theorem 12: A quadrilateral is cyclic (i.e. may be inscribed in a
circle) if one side subtends congruent angles at the two opposite
vertices.
If BAC = BDC, points A, B, C, and D are concyclic.
Theorem 13: The opposite angles of a cyclic (inscribed) quadrilateral
are supplementary.
If the opposite angles of a quadrilateral are
supplementary, the quadrilateral can be inscribed by a circle.
If B + D = 180, points A, B, C, and D are concyclic.
Theorem 14: If two chords of a circle intersect, the product of the
measures of the segments of one chord equals the product of the
segments of the other chord.
If AE EC = BE ED, points A, B, C, and D are concyclic.
Theorem 15: If two secants intersect outside the circle, the product of the measures of
one secant and its external segment equals the product of the measures of the other
secant and its external segment.
If PC PA = PD PB, points A, B, C, and D are concyclic.
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Example 11: (1997 AMC) Triangle ABC and point P in the same plane
are given. Point P is equidistant from A and B, angle APB is twice angle
ACB, and AC intersects BP at point D. If PB = 3 and PD = 2,
then AD · CD =
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Solution: (A).
Method 1 (Official Solution):
Construct a circle with center P and radius PA. Point C then lies on the circle, since the
angle ACB is half angle APB. Extend BP through P to get diameter
BE. Since A, B, C, and E are concyclic, using the Power of Point
formula, we have:
AD · CD = ED · BD = (PE + PD)(PB − PD) = (3 + 2)(3 − 2) = 5.
Method 2 (Our solution):
Extend BP to E such that PE = PB. Since PA = PB = PE, points A, B, and E are
concyclic. Construct this semicircle with center P as shown in the
figure to the right.
1
So we have AEB APB ACB . We also know that
2
ADE BDC (vertical angles). Therefore AED DCB.
AD ED
AD · CD = ED · BD = (PE + PD)(PB − PD)
BD DC
= (3 + 2)(3 − 2) = 5.
Example 12: In ∆ABC, the angle bisector of A meets BC at D. Show that
AD 2 AB AC BD CD.
Solution:
Construct a circle that circumscribes the triangle as shown in the
figure. Extend AD to meet the circle at E and connect BE.
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Since E = C, 1 = 2, ∆ABE ~∆ ADC
AB AC AD AE
(1)
BD DC AD DE
(2)
(1) – (2):
AB AC BD CD AD AE AD DE AD( AE DE)
AD AD AD 2
Therefore AD 2 AB AC BD CD.
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
PROBLEMS
Problem 1. In ABC, B = 2A. AB = 2BC. Show that AB2 = AC2 + BC2.
Problem 2: In ∆ABC, B 2C . AD BC at D. M is the midpoint of BC. If AB = 10
cm, find MD.
Problem 3. ABCD is a trapezoid with AD//BC. M is the middle point of BD and N is the
1
middle point AC. Show that MN ( BC AD ) .
2
Problem 4: Show that in trapezoid ABCD, if the two diagonals AC = BD, the trapezoid
is isosceles.
Problem 5: The measure of the longer base of a trapezoid is 97. The measure of the line
segment joining the midpoints of the diagonals is 3. Find the measure of the shorter base.
(Note that the figure is not drawn to scale.)
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Problem 6: The measures of the sides of a right triangle are 60, 80, and 100. Find the
measure of a line segment, drawn from the vertex of the
right angle to the hypotenuse, that divides the triangle into
two triangles of equal perimeters.
Problem 7: The perimeter of trapezoid ABCD is 60. AD // BC, AB = CD, B = 60. Find
the greatest value of the area ABCD.
Problem 8: (1959 AMC) In triangle ABC, BD is a median. CF intersects BD at E so BE
= ED. Point F is on AB. Then, if BF = 5, BA equals:
(A) 10
(B) 12
(C) 15 (D) 20
(E) none of these
Problem 9: Triangle ABC is an equilateral triangle. D is the on AB.
Extending AC to E and connect DE so that BD = CE. Prove: GD =
GE.
Problem 10: Given any ∆ABC, AE bisects BAC, BD bisects ABC, CP BD, and CQ
AE, prove that PQ is parallel to AB.
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Problem 11: Given ABC, AB = AC, E is the midpoint of AB. Extend AB to D such that
BD = BA. Prove: CD = 2CE.
Problem 12: As shown in the figure, in trapezoid ABCD, A + B = 90. Find EF if AB
= a, CD = b. E is the midpoint of AB and F is the midpoint of DC.
1
1
(A) (a b). (B) (a b). (C) a 2 b 2 . (D) a 2 b 2 .
2
2
Problem 13. In ABC, AE is the diameter of the circumcircle. AD is the altitude on BC.
Show that AB × AC = AD × AE.
Problem 14: B is the trisection point of the side AC of AFC. Draw a line through B to
ED AB 2
. Show that
meet the extension of CF at E, to meet AF at D such that
DB BC 1
AD 7
.
DF 2
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
SOLUTIONS
Problem 1. Solution:
Method 1:
Draw CD, the angle bisector of C. By the angle bisector formula,
AC BC
(1)
AD BD
Draw DE to meet AC at E such that CED = B.
We get ∆CDE ∆CDB.
Therefore CED = B = 2A or CED = A + EDA
Since EDA = A, AE = ED = BD
Let AB = 2BC = c = 2a, AC = b, BC = a.
We have BD = b – a, and AD AB BD 2a (b a) 3a b .
Substituting all the values to (1):
b
a
b 2 3a 2
b(b a) a(3a b)
3a b b a
a 2 b2 c 2
b 2 a 2 4a 2
Method 2:
Draw CD, the angle bisector of ACB. Extend CB to E
such that CE = AC, and connect DE.
We have ∆ACD ∆ECD. CAD = CED =BED.
Since CBD = 2BED, then BED = BDE.
Triangle BDE is therefore isosceles, so BE = BD = b –
a.
AD = c – (b – a).
AC BC
By the angle bisector theorem,
.
AD BD
b
a
b2 ba ac ab a 2 .
b(b a) a(c b a)
c (b a) b a
We are given that c 2a .
Therefore
b2 ac a 2 a(2a) a 2 3a 2
a 2 b2 a 2 3a 2
a 2 b2 4a 2 (2a)2 c 2 .
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Problem 2: Solution: 5.
Connect MN and DN such that N is the midpoint of AB. We see that DN is the median of
right triangle ADB on the hypotenuse AB.
Therefore DN = BN and B NDB .
We also see that
NDB NMD DNM C DNM ( NM // AC)
So B = C DNM .
Since B = 2C, DNM C NMD.
1
Therefore DM DN AB and MD = 5cm.
2
You willalso see this problem in the Problems section solved using a different way.
Problem 3. Solution:
Let L be the midpoint of AB, and connect ML and NL.
Since L is the midpoint of AB, and N is the midpoint of AC, LN//BC and LM//AD.
We also know that AD//BC, so L, M, and N are collinear.
LN
1
BC
2
MN
MN ML
1
BC
2
1
1
1
1
BC ML BC AD ( BC AD )
2
2
2
2
Problem 4: Solution:
As shown in the figure, drop the perpendiculars CE at E and DF at F.
CE DF
AB // CD
AC BD
Rt∆ACE Rt∆BDF (HL)
CAF
AB AB
AC BD
∆ABC ∆ABD
(SAS)
AD = BC
Therefore, if the two diagonals AC = BD, the trapezoid is isosceles.
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Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Problem 5: Solution: 91.
Method 1: Since E and F are the midpoints of DB and
AC, respectively, EF must be parallel to DC and AB.
Since EF is parallel to DC, ∆EGF ~ ∆DGC, and
GC DC
.
GF EF
GC 94
However, since DC = 97 and EF = 3,
.
GF 3
GC GF 97 3
FC 94
Then,
, or
.
GF
3
GF 3
FA 94
GA 91
.
Since FC = FA,
or
GF 3
GF 3
GA AB
.
Since ∆AGB~ ∆FGE,
GF EF
91 AB
Thus,
, and AB = 91.
3
3
1
Method 2: Extend FE to meet AD at H. In ∆ADC, HF ( DC) .
2
97
91
Since DC = 97, HF . Since EF = 3, HE .
2
2
1
In ∆ADB, HE ( AB ) . Hence, AB = 91.
2
Problem 6: Solution: 24 5 .
Let AB = 60, AC = 80, and BC = 100. If ABD is to have the same perimeter as ACD,
then AB + BD must equal AC + DC, since both triangles share AD; that is, 60 + BD =
80 + 100 - BD. Therefore, BD = 60 and DC = 40.
Draw DE such that DE perpendicular to AC.
ED DC
Right EDC ~ right ABC; therefore,
.
AB BC
By substituting in the appropriate values, we have
and ED = 24.
249
ED 40
,
60 100
50 AMC Lectures
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Using the Pythagorean Theorem on EDC, we see EC = 32. AE = AC – EC = 48. Now
using the Pythagorean Theorem on AED, we get that AD = 24 5 .
Problem 7:
225 3
Solution:
.
2
Draw AE such that AE BC at E.
Since B = 60, right triangle AEB is a 30 – 60 – 90 degree
3
AB .
right triangle and AE
2
We are given that AB = CD and the perimeter of the trapezoid, so
AD BC 60 2 AB .
1
3
3
225 3
(30 AB AB 2 )
( AB 15)2
.
S∆BCA ( AD BC ) AE
2
2
2
2
Trapezoid ABCD will obtain the greatest area when AB = 15, and this value is
Problem 8: Solution: 15.
Draw DG//AB to meet CF at G. Since D is the midpoint of AC,
AF = 2DG.
Since BE = ED, EBF = EDG (alternate interior angles) and
BEF = DEG (vertical angles),
∆EFB ∆EGD and DG = BF = 5.
AF = 2DG = 10. AB = 5 + 10 = 15.
Problem 9: Proof:
Method 1: Draw DH such that DHBC. Extend BC, and then draw
EF to meet extension of BC at F such that EFBC.
Since BD = CE, B =ACB =FCE, F =DHB=90, and
BDH=FEC, by ASA, DBH CEF. Therefore, DH = FE.
Since DH =FE, HGD=FGE, F=DHB=90, and
GDH=GEF, DHG GFE. Then DG = GE.
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225 3
.
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50 AMC Lectures
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Method 2: Draw DH such that DH//BC. Then HC = DB = CE. Since CG//DH and cuts
HE into two equal sections, it also cuts DE into two equal sections, which means that DG
= GE.
There is a third method for the proof that uses a third way of drawing an auxiliary line,
shown in the figure below. See if you can prove DG = GE using
the diagram below on your own.
Problem 10: Solution:
Extend CP and CQ to meet AB at S and R, respectively.
CPB SPB, and CQA RQA (A.S.A.)
It then follows that CP = SP and CQ = RQ, or P and Q are
midpoints of CS and CR, respectively.
Therefore, in ∆CSR, PQ //SR. Thus, PQ //AB.
Problem 11: Proof:
1
AB 2 AB AE AD , so AC is the tangent of the
2
circumscribed circle of ECD.
Therefore ACE = D.
Since BCE = C – ACE and BCD = B – D, BCE = BCD.
Hence CB is the angle bisector of ECD.
Since AC 2 AB 2
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50 AMC Lectures
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Problem 12: Solution: (B).
Extend AD and BC meet at K. Since A + B = 90, triangle
ABK is a right triangle. KE and KF are the medians of right
triangles ABK and DCK, respectively.
1
1
KE AB a
(1)
2
2
1
1
KF CD b
(2)
2
2
1
1
1
EF KE KF a b (a b) .
2
2
2
Problem 13. Solution:
Method 1:
Connect BE. Triangle ABE is a right triangle.
Since ACB and AEB face the same arc, ACB = AEB.
We also know that ABE =ADC = 90.
Therefore ACD and ABE are similar.
AB AD
AB × AC = AD × AE.
AE AC
Method 2:
Connect EC. Triangle AEC is a right triangle.
Since ABC and AEC face the same arc, ABC = AEC.
We also know that ADB =ACE = 90. Therefore AEC and
ABD are similar.
AB AE
AB × AC = AD × AE.
AD AC
Problem 14: Solution:
AD 7
.
DF 2
Method 1:
As shown in the figure to the right, draw DG//AG through D to
meet EC at G.
FA AC 3BC
In FAC,
(1)
FD DG DG
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50 AMC Lectures
In EBC,
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
BC EB 3
DG ED 2
Substituting (2) into (1) gives us:
(2)
AF 9
FD 2
AD 7
DF 2
Method 2:
As shown in the figure to the right, draw DG//CE through D to meet AC at G.
BG BD 1
BC 3
In BEC,
GC DE 2
GC 2
1
AC 9
Since BC AC ,
3
GC 2
In ACF,
AF AC
AF 9
AD 7
,
,
.
DF GC
DF 2
DF 2
Method 3:
As shown in the figure to the right, draw BG//AF through B to meet
CE at G.
AF AC 3
,
(1)
In ∆CFA,
BG BC 1
BG EB 3
,
(2)
In ∆EBG
DF ED 2
AF 9
AD 7
,
.
(1) (2):
DF 2
DF 2
Method 4:
As shown in the figure to the right, draw BG//CE through B to
meet AF at G.
DF ED 2
.
Since ∆EFD ~∆BGD,
DG DB 1
1
Therefore DG DF
2
AG AB 2
AD DG 2
.
.
In ∆ACF,
AF AC 3
AD DF 3
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50 AMC Lectures
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
1
AD DF
2
2
.
AD DF
3
Simplifying yields:
AD 7
.
DF 2
Method 5:
As shown in the figure to the right, draw AG//CE through A to
meet the extension of CE at G.
BE BC 1
(1)
In ∆ACG,
AG AC 3
AG AF
(2)
In ∆AFG ,
DE DF
BE
AF
(1) (2):
DE 3DF
BE 3 3
AF AF 9
,
.
Since
,
DE 2 2 3DF DF 2
AD 7
.
Hence
DF 2
Method 6:
As shown in the figure to the right, draw AG//CE through A to meet the extension of EB
at G.
BG AB 2
Since ∆ ABG~∆CBE,
and.BG = 2EB
BE BC 1
Since ∆ FED~∆AGD,
DF ED
ED
ED
2 DB
2
.
AD DG DB BG DB 2 EB DB 6DB 7
AD 7
.
Therefore
DF 2
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50 AMC Lectures
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
Method 7:
As shown in the figure to the right, draw EG//FA through E to meet the extension of CA
at G.
EG EB GB 3
In BEG:
AD DB AB 1
Therefore we have:
GB = 3AB
EG = 3AD
(1)
(2)
In CEG:
EG GC GB BC 3(2BC ) BC 7
AF AC AB BC
2 BC BC
3
7
Substituting (2) into (3): 3 AD ( AD DF )
3
9 AD 7( AD DF )
AD 7
.
Therefore
DF 2
EG
7
7
AF ( AD DF ) (3)
3
3
Method 8:
As shown in the figure below, draw EG//AB through E to meet the extension of AF at G.
EG ED DG 2
.
Since DAB DGE,
AB DB AD 1
We have EG = 2AB
(1)
DG = 2AD
(2)
Since FAC FGE:
EG FG
2 AB
2 AB
4
.
AC AF AB BC AB 1 AB 3
2
We have 3FG =4AF
(3)
Or 3(DG – DF) = 4(AD + DF)
(4)
Substituting (2) into (4):
3(2AD – DF) = 4(AD + DF)
6AD – 3DF = 4AD + 4DF
2AD = 7DF
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50 AMC Lectures
Chapter 11 Geometry Eight Methods To Draw Auxiliary Lines
AD 7
.
DF 2
There are at least 4 more ways to prove this problem, through different ways of drawing
auxiliary lines. See if you can find them!
Therefore
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
BASIC KNOLWDGE
Definition of six functions in a circle
x
y
x
y
; cos ; tan ; cot ;
r
y
r
x
r
r
sec ; csc .
x
y
sin
Definition of six functions in a unit circle (r = 1)
sin y ; cos x ; tan
sec
1
;
x
csc
y
x
; cot ;
x
y
1
.
y
Example 1: (1988 AMC) If sin 3 cos then what is
sin cos ?
1
1
2
1
3
(A)
(B)
(C)
(D)
(E)
6
5
9
4
10
Solution: (E).
We know that in a unit circle sin y , and cos x .
We are given that
(1)
y 3x
We also know that x 2 y 2 1
Substituting (1) into (2): x 2 9 x 2 1
(2)
x2
1
.
10
1
3
.
10 10
Note: We are giving a different solution to this problem than the official solution.
Therefore sin cos y x 3x 2 3
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Example 2: (1983 AMC) Triangle ABC has a right angle at C. If
2
sin A , then tan B is
3
3
5
5
5
2
(A)
(B)
(C)
(D)
(E)
5
3
3
2
5
Solution: (D).
BC 2
In the figure, sin A
.
AB 3
So for some x > 0, BC = 2x, AB = 3x and AC ( AB )2 ( BC )2 5 x.
Thus tan B
AC
5
.
BC
2
Example 3: The two legs of a right triangle are a and b. sin A
Prove: log
1 a
.
2 b
ab 1
(log a log b) .
2
6
Solution:
The hypotenuse of the right triangle is
Since we are given that sin A
a 2 b 2 . In this right triangle, sin A
a
a b2
2
.
1 a
, so
2 b
1 a
a
ab
ab .
=
a2 + b2 = 4ab (a + b)2 = 6ab
2
2
2 b
6
a b
Take the log of both sides:
ab 1
log
(log a log b) .
2
6
Example 4: Find the value of sin – cos if the terminal side of passes through point P
(– 6a, – 8a), a ≠ 0.
1
1
1
7
1
1
A.
B.
C. or
D. or
5
5
5
5
5
5
Solution: (D).
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
By the definition, we have x = – 6a, y = – 8a, r x 2 y 2 10 a .
y
x
3
1
4
= , cos . Therefore sin cos .
r
r
5
5
5
4
3
1
When a < 0, sin , cos . Therefore sin cos .
5
5
5
When a > 0, sin
Signs of six functions
Example 5: If is an obtuse angle, which one of the following is positive?
(A) cos – sin ;
(B) sin cos ; (C) tan + cot ; (D) sin – cot
Solution: (D).
Since is an obtuse angle, we know that sin > 0, cos < 0, tan < 0, cot < 0.
(A), (B), (C) are all negative. So the answer is (D).
Example 6: In ∆ABC, if sinA∙cosB < 0, what shape is the ∆ABC?
(A) Acute
(B) Right
(C) Obtuse
(D) Undetermined.
Solution: (C).
Since sinA∙cosB < 0, so sinA and cosB have opposite signs. When 0 < A < 180, sinA >
0, cosB < 0, so B is obtuse. (C) is the answer.
Cofunction
Sine and Cosine, Tangent and Cotangent, Secant and Cosecant, are cofunctions of each
other.
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Values of special angles
Trigonometric hexagon
How to read the trigonometric hexagon
(1) Reciprocal relations (Diagonals):
sin csc = 1;
cos sec = 1;
tan cot = 1.
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
(2). Vertices of the shaded equilateral triangles (Pythagorean relations)
sin2 + cos2 = 1; tan2 + 1 = sec2 ; 1 + cot2 = csc2 .
Example 7: (1988 AMC) If sin x 3 cos x the what is sin x cos x ?
(A)
1
6
(B)
1
5
(C)
2
9
(D)
1
4
(E)
3
10
Solution: (E).
Multiplying the given equation first by sin x and then by cos x yields
sin2 x = 3 sin x cos x,
1
cos 2 x sin x cos x.
3
10
3
Adding these two together gives 1 sin x cos x , so sin x cos x .
3
10
Note: This is the official solution.
Example 8: (1999 AMC 12) Let x be a real number such that sec x – tan x = 2.
Then sec x + tan x =
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
(E) 0.5
Solution: (E).
Method 1:
From the identity 1 + tan2 x = sin2 x it follows that 1 = sin2 x – tan2 x = (secx – tanx)( sec x
+ tan x)=2(sec x + tan x), so sec x + tan x = 0.5.
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Method 2:
The given relation can be written as (1 – sin x) cos x = 2. Squaring both sides yields
(1 – sin x)2 (1 – sin2 x) = 4, hence (1 – sin x) (1 + sin x) = 4. It follows that sin x
3
1 ( )
1 sin x
5 4.
and that cos x
2
2
5
5 3
Thus sec x tan x 0.5.
4 4
Example 9: Find 1 sin 2 2 1 cos 2 2 .
(A) cos2 – sin2.
Solution: (B).
(B) –cos2 – sin2.
(C) – cos2 + sin2.
(D) cos2 + sin2.
We know that 2 ( , ) , so
2
1 sin 2 2 1 cos 2 2 = cos 2 sin 2 cos 2 sin 2 .
Example 10: Find sin x if tan x
(A)
a
.
b
(B)
b
.
a
2ab
with a > b > 0 and 0 x .
2
a b
2
2
(C)
a 2 b2
.
2a
(D)
2ab
.
a b2
2
Solution: (D).
sin x
1
1
csc x
1 cot 2 x
1
a 2 b2
1
2ab
2
1
a 4 2 a 2b 2 b 4
4 a 2b 2
Example 11: Show that (tan 1)2 (1 cot )2 (sec csc )2 .
Solution:
(tan 1)2 tan2 1 2 tan sec2 2 tan .
(1 cot )2 1 cot 2 2 cot csc2 2 cot
262
2ab
.
a b2
2
3
5
50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Therefore,
(tan 1)2 (1 cot )2
sec2 2(tan cot ) csc2 sec2 2 sec csc csc2 (sec csc )2 .
Example 12: Find sin if , are acute angles with 2tan + 3sin = 7 and
tan – 6sin = 1.
3 5
3 7
3 10
1
(A)
(B)
(C)
(D) .
.
.
.
5
7
10
3
Solution: (C).
Solve the following system of equations:
2 tan 3 sin 7
tan 6 sin 1
1
.
3
1
1
1
3 10
Since is an acute angle, sin
.
2
csc
10
1 2
1 cot
1 ( )
3
We get tan = 3. Therefore cot
(3) Quotient relations (Vertices of the isosceles triangles):
tan
sin
cos
; cot
.
cos
sin
263
50 AMC Lectures
Example 13: If cos
Chapter 12 Trigonometry Six Functions
8
, find all the values of other trignometric functions.
17
Solution:
Since cosine value is negative, the angle could be either in the second quadrant or the
third quadrant.
8
15
If is in the second quadrant, sin 1 cos 2 1 ( ) 2 ,
17
17
15
sin
15
1
8
1
17
tan
17 , cot
, sec
,
cos 8
8
tan
15
cos
8
17
1
17
csc
.
sin 15
15
15
8
17
If is in the third quadrant, sin , tan , cot , sec , and
17
8
15
8
17
csc .
15
1
Example 14: Find the value of tan if sin cos , 0 < <180.
5
3
4
3
4
(A)
(B)
(C)
(D)
4
3
4
3
Solution: (C).
(sin cos )2 sin 2 cos 2 2 sin cos
2(sin 2 cos 2 ) (sin cos )2
1
49
2 ( )2 =
5
25
7
Therefore sin cos .
5
1
4
sin cos 5
cos 5
sin 4
tan
7
3
cos
3
sin cos
sin
5
5
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
1
sin cos 5
From
, we have
sin cos 7
5
3
sin 5
.
cos 4
5
4
Since 0 < < 180, we can conclude: tan .
3
Terminal side of an angle
In trigonometry, we can visualize an angle as being formed by rotating one of the sides
about the vertex while keeping the other side fixed.
For example, AOB () in the figure is formed by rotating the side OB
while OA is fixed.
OA is called the initial side and OB is called the terminal side of the
generated angle.
All angles having the same terminal side as angle (including ) can be expressed as
360k + (k is any integer) or 2k + .
An angle is called a first quadrant angle if its terminal side is in the first quadrant.
An angle is called a second quadrant angle if its terminal side is in the second quadrant,
and so on for the other two quadrants. Coterminal angles have equal functional values.
Example 15: If is an angle in first quadrant, what are the quadrants of 2 and ½ in?
Solution:
Since is an angle in first quadrant, 2n 2n
Therefore 4n 2 4n ; n
n
2
. n is integer.
.
2
4
2 is in the first or second quadrant or the terminal side is on positive y – axis.
½ is in first or third quadrant.
The reference angle
The reference angle is the acute angle made by the terminal side of the given angle and
the x-axis.
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Reduction formulas
A trigonometric reduction formula simplifies an expression in terms of more easily
calculated or readily available values. Any trigonometric function whose argument is
n 90 can be written simply in terms of . For examples: sin (360° + ) = sin ;
1
sin 390° = sin (360° + 30°) = sin 30° = 2 .
If you know the following method, you do not need to remember all the reduction
formulas.
f ( n 90) g ( ) , where
n may be any integer, positive, negative, or zero.
f is any one of the six trigonometric functions.
may be any real angle measure.
If n is even, then g is the same function as f.
If n is odd, then g is the cofunction as f.
(Remember that sine and cosine, tangent and cotangent, secant and cosecant, are
cofunctions of each other).
The second ± sign is determined by the sign of the reference angle of the original
function.
Example 16: Reduce: sin (360° + ):
Solution:
(1) Determine the final function:
360 = 90 × 4.
n = 4 which is even. The final function is the same as the original function.
sin (360° + ) = ? sin .
(2) Determine the sign:
The reference angle is on the first quadrant. We know that sin is positive in first
quadrant, so the sign is “+”.
sin (360° + ) = + sin .
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Example 17: Reduce: sin (270° + ):
Solution:
(1) Determine the final function:
270 = 90 × 3. n = 3 which is odd.
The final function is the cofunction of the original function.
sin (270° + ) = ? cos .
(2) Determine the sign:
The reference angle is on the third quadrant. We know that sin is negative in third
quadrant, so the sign is “”.
sin (270° + ) = cos .
Example 18: Reduce: cos (90° + ):
Solution:
(1) Determine the final function:
90 = 90 × 1, so n = 1 which is odd. The final function is the cofunction of the original
function.
cos (90° + ) = ? sin .
(2) Determine the sign:
The reference angle is on the second quadrant. We know that cos is negative in second
quadrant, so the sign is “”.
cos (90° + ) = sin .
Example 19: Reduce: cos ():
Solution:
(1) Determine the final function:
0 = 90 × 0. n = 0 which is even. The final function is the same as the original function.
cos () = ? cos .
(2) Determine the sign:
The reference angle is on the fourth quadrant. We know that cos is positive in the
fourth quadrant, so the sign is “+”.
cos ( ) = cos .
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Example 20: Determine the signs of the following trigonometric expressions:
(1) cos250;
(2) sin( );
4
(3) tan( – 67210);
(4) cot
11
.
3
Solutions:
(1) cos 250 = cos (3 90 + 20) = sin 20 < 0.
(2) sin( ) sin(4 90 ) sin 0;
4
4
4
(3) tan( – 67210) = tan( – 2 360+ 4750) = tan4750,
tan( – 67210) > 0;
11
5
5
11
(4) cot cot(2 ) cot , cot 0.
3
3
3
3
Example 21: Find the value of cos 570 sin 150+sin(– 330) cos(– 390).
Solution:
cos 570 sin 150+sin(– 330) cos(– 390)
= cos (360 + 180 + 30) ∙ sin (180 – 30) – sin (360 – 30) ∙ cos (360 + 30)
= – cos 30sin 30 + sin 30cos 30 = 0.
Example 22: Find the value of cos 420 tan 60 sec 45 + sin 45 cot 30 csc 450.
Solution:
cos 420 tan 60 sec 45 + sin 45 cot 30 csc 450
= cos (360 + 60) tan60sec45 + sin45cot30csc (360 + 90)
= cos60tan60sec45 + sin45cot30csc90
1
1
3 2
3 1 6 .
2
2
7
sin(5 ) cos(6 ) tan( )
2
Example 23: Find f () if f ( )
and
sin( 2 ) tan( )
2
7
1
cos(
) . is in third quadrant.
2
3
268
50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Solution:
3
)
sin cos ( cot )
2
f ( )
cos
sin
(
cot
)
sin( ) tan( )
2
7
7
3
1
cos(
) cos(
) cos( ) sin
2
2
2
3
1
2 2
.
Therefore f ( ) cos 1 sin 2 1 ( )2
3
3
sin( ) cos(2 ) tan(
269
1
sin .
3
50 AMC Lectures
Chapter 12 Trigonometry Six Functions
PROBLEMS
Problem 1: (1972 AMC 20) If tan x
2ab
, where a > b > 0 and 0 < x < 90, then sin
a b2
2
x is equal to
(A)
a
b
(B)
b
a
(C)
a 2 b2
2a
(D)
Problem 2: Find sin xcos x if sin x 4 cos x .
1
2
3
1
(A) .
(B) .
(C) .
(D
6
9
10
5
Problem 3: Find sin xcos x if
3
.
2
(B)
3
2
2ab
a b2
2
4
.
17
(C)
3
.
4
1
and .
8
4
2
3
(D) .
4
Problem 5: Find the value of tan cot if sin cos
Problem 6: Find cos(
(E)
(E)
1 tan A
3 2 2 .
1 tan A
Problem 4: Find cos – sin if sin cos
(A)
a 2 b2
2ab
1
.
3
2
1
2 ) if sin( ) .
3
6
3
Problem 7: Find the quadrant where the terminal side of is located if tan > 0 and sin
+ cos > 0.
(A) I
(B) II
(C) III
(D)
IV
Problem 8: Find the value tan(
7
).
6
Problem 9: Find tan 5 tan 15 tan 25 tan 5 tan 35 tan 45 tan 55 tan 6 tan 7 tan
85.
270
50 AMC Lectures
Chapter 12 Trigonometry Six Functions
1
1
Problem 10: is an acute angle and ≠ 45, If 2 sin cos sin cos 1, what
3
3
is the quadratic eqaution with the two roots tan and ctan?
Problem 11: Show that
1 2 sin x cos x 1 tan x
.
cos 2 x sin 2 x 1 tan x
Problem 12: If tan + sin = m and tan – sin = n, show (m2 n2 )2 16mn .
cos
1 sin
1
.
if
sin 1
cos
2
1
B.
C. 2
D. –2
2
Problem 13: Find
A.
1
2
Problem 14: Show tan 2 tan 2
cos 2 cos 2
.
cos 2 cos 2
Problem 15: Find sin2 5 + sin2 15 + sin2 25 + sin2 35 + sin2 45 + sin2 55 + sin2 65 +
sin2 75 + sin2 85.
Problem 16: If 5x cos y and a x sin y, then cos(2 y) is equal to
(A) 10x
(B) 25x
(C) 2(5a)x (D) 25x + a2x
(E) 25x a2x
x sec y tan 2 cos
Problem 17: (2000 NC Math Contest) If
, find what y equals.
cot
x tan y sec
cos 2
(A)
(B) sin
(C) cos
(D) sin2
(E) none of these
sin
Problem 18: (1972 AMC 30) A rectangular piece of paper 6 inches wide is folded as in
the diagram so that one corner touches the opposite side. The length in
inches of the crease L in terms of angle is
(A) 3sec2 csc
(B) 6 sin sec
(C) 3sec csc
2
(D) 6 sec csc
(E) None of these
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
1
and 0 x , then x is
5
3
4
4
3
(A)
(B)
(C)
(D)
4
3
3
4
(E) not completely determined by the given information.
Problem 19: (1978 AMC) If sin x cos x
Problem 20: (1983 AMC) If tan and tan are the roots of x 2 px q 0, cot and
cot are the roots of x 2 rx s 0, then rs is necessarily
1
p
q
p
(A) pq
(B)
(C) 2
(D) 2
(E)
pq
q
p
q
Problem 21: (1991 AMC) If f (
x
1
) for all x 0, 1 and 0 , find f (sec2 ) .
x 1 x
2
Problem 22. If is an angle in fourth quadrant, what are quadrants are 2 and
Problem 23. Show (a 2 b 2 ) 2 4a 2
and tan + tan = c.
2
in?
8ab
if c ≠ 0, sin + sin = a, cos + cos = b,
c
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
SOLUTIONS
Problem 1: Solution: (E).
The acute angle x may be taken as opposite the leg of length 2ab in a right triangle with
other leg of length (a2 – b2).Then the square of the
hypotenuse h is, by the Pythagorean Theorem,
h2 (2ab)2 (a 2 b2 )2 a 4 2a 2b2 b4 (a 2 b2 )2 .
We can see from the figure to the right and the definition of
sine that
2ab
sin x 2
.
a b2
Problem 2: Solution: (E).
sin x cos x
4 cos x cos x
4
sin x cos x
.
2
2
2
2
sin x cos x (4 cos x) cos x 17
Problem 3: Solution:
2
.
3
2
1 tan A
3 2 2 gives tan A
.
1 tan A
2
2
sin A cos A
tan A
2
2
sin A cos A 2
.
2
2
2
sin A cos A tan A 1 2
3
1
2
Problem 4: Solution: (B).
1
1 3
Since sin cos , (cos sin ) 2 cos 2 sin 2 2 sin cos 1 2 .
8
8 4
Solving
Because
4
2
, cos < sin .
Therefore cos – sin =
3
.
2
Problem 5: Solution: -3.
Since (sin cos )2 1 2 sin cos ,
1
1
1 2 sin cos . Therefore, sin cos .
3
3
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
sin cos sin 2 cos 2
1
1
3 .
cos sin
sin cos
sin cos 1
3
7
Problem 6: Solution: .
9
1
Since sin( ) ,
6
3
1
7
cos( 2 ) cos[2( )] 1 2 sin 2 ( ) 1 2 ( )2 .
3
6
6
3
9
7
2
2 ) = cos[ ( 2 )] cos( 2 )] .
Hence cos(
3
9
3
3
tan cot
Problem 7: Solution: (A).
sin
0.
We know that tan
This is the same as sin ∙ cos > 0.
cos
We also know that sin + cos > 0. Therefore we have sin > 0, and cos > 0.
The terminal side of is in the first quadrant.
Problem 8: Solution:
tan(
3
.
3
7
5
5
3
) tan(2 ) tan
tan( ) tan
.
6
6
6
6
6
3
Problem 9: Solution: 1.
tan 5 tan 15 tan 25 tan 5 tan 35 tan 45 tan 55 tan 6 tan 7 tan 85
= tan 5 tan 15 tan 25 tan 5 tan 35∙ 1 tan(90 – 35) tan(90 – 25) tan(90 – 15)
tan(90 – 5) = (tan 5 cot 5)(tan 15 cot 15)(tan 25 cot 25) (tan 35 cot 35) = 1.
Note: tan 1 tan 2 tan 3∙ ∙ ∙ tan 87 tan 88 tan 89 = 1.
Problem 10: Solution: 4 x 2 9 x 4 0.
1
1
sin cos 1 2 sin cos .
3
3
1
(sin cos ) (sin cos ) 2 .
3
274
50 AMC Lectures
is an acute angle and ≠ 45
Chapter 12 Trigonometry Six Functions
1
sin ≠ cos sin cos .
3
Squaring both sides gives us
4
sin cos .
9
sin cos
1
9
tan cot
.
cos sin sin cos 4
Since tan cot 1, so
9
x 2 x 1 0 or 4 x 2 9 x 4 0.
4
Problem 11: Solution:
1 2sin x cos x cos 2 x sin 2 x 2 sin x cos x
(cos x sin x) 2
cos 2 x sin 2 x
cos 2 x sin 2 x
(cos x sin x)(cos x sin x)
cos x sin x 1 tan x
.
cos x sin x 1 tan x
Problem 12: Solution:
Method 1:
Solving the system of equations gives:
mn
mn
tan
, sin
2
2
Or
2
cot
mn
2
csc
mn
We know that csc2 cot 2 1
Substituting (1) and (2) into (3): (m2 n2 )2 16mn .
Method 2:
Solving the system of equations gives:
mn
sin
2
(1)
(2)
(3)
(1)
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
mn
2
(1) (2):
mn
cos
mn
We know that sin 2 cos 2 1
tan
(2)
(3)
(4)
Substituting (1) and (3) into (4): (m n ) 16mn .
2
2 2
Problem 13: Solution: (A).
We know that 1 – sin2 = cos2 . Therefore (1 + sin )(1 – sin )= cos ∙ cos .
cos
1 sin
1
.
Since cos ≠ 0 and 1 – sin ≠ 0,
1 sin
cos
2
cos
1
.
Therefore
sin 1 2
Problem 14: Solution:
1
1
cos 2 cos 2
tan tan (sec 1) (sec 1)
cos 2 cos 2
cos 2 cos 2
2
2
2
Problem 15: Solution:
2
9
.
2
The expression can be rewritten as:
+ sin 225 + sin 235 + (
sin25 + sin 215
(90– 15) + sin2 (90– 5).
2 2
) + sin2(90 – 35) + sin2 (90– 25) + sin2
2
=
( sin2 5 + cos2 5 ) + ( sin2 15 + cos2 15) + (sin2 25 + cos2 25) + (sin2 35 + cos2 35)
1 9
+ .
2 2
89
.
Note: sin2 1 + sin2 2 + sin2 3 + ∙ ∙ ∙ + sin2 87 + sin2 88 + sin2 89 =
2
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Problem 16: Solution: (E).
cos (2y) = cos2 (y) – sin2 (y) = (5x )2 (a x )2 25x a 2 x .
Problem 17: Solution: (A).
We are given that
x sec y tan 2 cos
x tan y sec cot
By Cramer’s Rule:
sec
tan
y
sec
tan
2 cos
1
cot sec cot 2 tan cos sin 2 sin 1 2 sin 2 cos 2
=
=
=
=
.
tan
sec2 tan 2
sin
sin
1
sec
Problem 18: Solution: (A).
In the figure to the right, let h denote the length of the sheet.
6
cos(90 2 ) sin 2 2 sin cos
h
Solving the equation for h gives us
3
h
.
sin cos
L
1
sec
sec L h sec 3
= 3sec2 csc .
h cos
sin cos
Problem 19: Solution: (A).
1
1
1
If sin x cos x , then cos x sin x and cos 2 x 1 sin 2 x ( sin x)2 ; so
5
5
5
2
2
25 sin x 5sin x 12 0. The solutions of quadratic equation 25s 5s 12 0 are
4
4
4
1
3
s and s . Since 0 x , sin x 0, so sin x and cos x sin x .
5
3
5
5
5
4
Hence tan x .
3
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
Problem 20: Solution: (C).
By the relationships between the roots and the coefficients of quadratic equations
x 2 px q 0 and x 2 rx s 0 from Vieta’s Theorem, it follows that
p tan tan ,
r cot cot ,
q tan tan ,
s cot cot .
1
1
tan tan
1
and cot cot
,
tan tan
tan tan
tan tan
substituting in the p, q, r, and s gives us
p
1
r and s .
q
q
p
Thus rs 2 .
q
Since cot cot
Problem 21: Solution: sin2 .
x
t
x
Let t
.
x 1
t 1
1
t
1
1 cos 2 sin2 .
Let f (t ) 1 . f (sec 2 ) 1
2
sec
t 1
t
Problem 22. Solution:
3
Since is an angle in first quadrant, 2k 2k 2 . k is integer.
2
3
Therefore k k .
4
2
When k is even, is in the second quadrant.
2
When k is odd, is in the fourth quadrant.
2
3
2k 2k 2 4k 3 2 4k 4 . .
2
No matter what value k is, 2 is in the third or fourth quadrant or the terminal side is on
negative y – axis.
Problem 23. Solution:
Let (x1, y1), (x2, y2) be the coordinates of points on and respectively that are 1 unit
away from the origin.
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50 AMC Lectures
Chapter 12 Trigonometry Six Functions
From the definition of trigonometric functions, the given conditions are equivalent to:
y
y
y1 + y2 = a, x1 + x2 = b, 1 2 c
x1 x2
y y
Therefore c[(a 2 b2 )2 4a 2 ] ( 1 2 ){[( y1 y2 )2 ( x1 x2 )2 ]2 4( y1 y2 )2 }
x1 x2
Noticing that x12 y12 1, x22 y22 1 , the equation above becomes
c[(a 2 b2 )2 4a 2 ] 8(1 x1x2 y1 y2 )( x1 y2 x2 y1 ) 8( x1 x2 )( y1 y2 ) 8ab
8ab
Since c ≠ 0, so (a 2 b 2 ) 2 4a 2
.
c
279
Index
digit, 139, 142, 143, 146, 147, 148, 152, 153, 157,
158, 167, 168, 169, 170, 171, 172, 173, 177, 179,
180
Divisibility, 167
divisible, 141, 147, 150, 154, 155, 156, 157, 158, 159,
162, 164, 165, 166, 167, 168, 169, 170, 171, 172,
173, 174, 175, 176, 177, 178, 179, 180
divisor, 156, 158, 159, 172, 173, 174, 178, 179, 180,
181
A
absolute value, 64, 66, 67, 91, 92, 99, 103, 104, 127,
131
acute angle, 263, 265, 271, 273, 275
alternate interior angles, 197, 198, 250
angle, 118, 182, 183, 186, 187, 188, 189, 190, 193,
196, 198, 232, 233, 238, 242, 247, 251, 264, 265,
266, 267, 271, 272, 278
arc, 217, 252
area, 70, 71, 75, 77, 79, 82, 83, 190, 201, 205, 206,
207, 208, 209, 210, 211, 212, 213, 214, 215, 217,
218, 219, 221, 222, 223, 224, 225, 226, 229, 230,
239, 240, 245, 250
arithmetic sequence, 27
average, 43, 56
E
equation, 2, 3, 5, 9, 11, 14, 15, 16, 25, 29, 39, 40, 41,
42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56,
57, 58, 59, 60, 61, 62, 63, 68, 69, 70, 71, 74, 75, 81,
83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 98, 99,
100, 101, 102, 103, 104, 105, 106, 109, 111, 112,
114, 115, 116, 117, 118, 119, 120, 121, 122, 123,
124, 125, 126, 127, 128, 129, 130, 131, 133, 134,
135, 136, 139, 148, 154, 155, 177, 230, 261, 277,
279
equidistant, 232, 242
equilateral, 184, 186, 187, 188, 189, 190, 200, 201,
212, 213, 218, 223, 224, 228, 245, 261
equilateral triangle, 184, 186, 187, 188, 189, 190, 200,
201, 212, 213, 218, 223, 224, 228, 245, 261
even number, 156, 157
exponent, 19
expression, 2, 6, 15, 18, 20, 31, 36, 37, 38, 74, 81, 82,
143, 144, 145, 266, 276
B
base, 190, 203, 205, 236, 244
bisect, 188, 234
C
center, 213, 239, 242
chord, 241
circle, 118, 202, 206, 209, 217, 219, 222, 224, 225,
239, 240, 241, 242, 251, 257
coefficient, 134
collinear, 189, 193, 248
combination, 160, 181
common divisor, 162, 178, 179
common factor, 159, 175
congruent, 83, 182, 183, 184, 185, 191, 198, 199, 232,
237, 241
constant, 134
Converse, 111, 119
convex, 195, 220
cube, 172, 179
F
face, 252
factor, 2, 6, 9, 44, 125, 156, 157, 160, 162, 164, 175
formula, 7, 9, 21, 23, 30, 86, 135, 138, 221, 225, 242,
247, 266
fraction, 4, 25, 33, 37, 160, 181
function, 82, 115, 116, 122, 138, 266, 267
G
D
graph, 65, 67, 69, 76, 77, 81, 82, 83, 84
degree, 111, 112, 250
denominator, 4, 18, 24, 30, 31, 47, 181
diagonal, 18
diameter, 217, 239, 242, 246
difference, 37, 83, 127, 145, 148, 157, 167, 206, 217
H
hexagon, 240, 260
hypotenuse, 182, 209, 219, 234, 245, 248, 258, 273
280
Index
I
P
inequality, 8, 50, 72, 73, 100, 124, 134
integer, 6, 8, 12, 16, 17, 18, 23, 32, 36, 49, 70, 85, 86,
87, 88, 97, 101, 138, 139, 140, 141, 142, 146, 147,
148, 149, 151, 152, 153, 154, 155, 156, 157, 158,
160, 161, 162, 164, 165, 166, 167, 171, 172, 173,
175, 178, 180, 181, 265, 266, 278
integers, 9, 12, 16, 19, 25, 30, 33, 49, 77, 87, 94, 138,
139, 140, 141, 143, 145, 146, 147, 148, 149, 151,
152, 154, 155, 156, 157, 158, 159, 160, 161, 162,
164, 165, 167, 169, 171, 173, 174, 175, 176, 178,
179, 180, 181, 195
intercept, 75
intersection, 69, 70, 79, 116, 235
irrational number, 25
isosceles, 188, 190, 191, 192, 194, 198, 199, 201, 221,
233, 234, 236, 244, 247, 248, 263
isosceles triangle, 188, 190, 191, 192, 194, 198, 199,
221, 233, 234, 263
parallel, 197, 207, 220, 234, 235, 237, 238, 240, 245,
249
parallelogram, 205, 217
pentagon, 190, 195, 201
perimeter, 245, 249, 250
permutation, 148
perpendicular, 188, 209, 213, 239, 249
plane, 71, 242
point, 64, 74, 83, 84, 85, 87, 185, 186, 187, 188, 189,
192, 193, 194, 195, 198, 200, 205, 206, 208, 209,
210, 211, 212, 213, 217, 218, 219, 220, 224, 232,
235, 237, 239, 242, 244, 246, 258
polynomial, 111, 112, 124, 136
positive number, 5, 114, 127, 141, 214
prime factorization, 162
prime number, 154, 160, 162, 164, 167, 172
probability, 173, 181, 220, 226
product, 32, 33, 39, 41, 45, 46, 97, 109, 112, 113, 114,
116, 118, 119, 120, 121, 122, 123, 129, 131, 134,
142, 147, 148, 150, 152, 153, 154, 156, 157, 161,
162, 164, 166, 174, 175, 176, 178, 179, 211, 241
proportion, 30, 46, 57
Pythagorean Theorem, 199, 228, 234, 240, 250, 273
L
line, 74, 76, 84, 185, 232, 234, 235, 237, 238, 239,
240, 244, 245, 246, 251
line segment, 232, 235, 237, 244, 245
Q
M
quadrant, 264, 265, 266, 267, 268, 270, 272, 274, 278
quadrilateral, 206, 207, 208, 209, 218, 219, 220, 241
quotient, 158, 171, 172
median, 183, 189, 206, 234, 245, 248
midpoint, 185, 190, 191, 201, 205, 208, 209, 215, 221,
225, 229, 230, 234, 235, 237, 244, 246, 248, 250
multiple, 138, 145, 154, 157, 167, 169, 172, 173, 174,
180, 181
R
radius, 118, 202, 206, 209, 213, 224, 239, 240, 242
range, 29, 107, 118, 122, 124, 128, 160
ratio, 23, 203, 204, 207, 208, 214, 219, 225, 229, 230
rational number, 25, 30, 31, 143
real number, 4, 7, 8, 9, 23, 30, 53, 64, 69, 70, 72, 77,
85, 96, 99, 100, 118, 119, 120, 128, 134, 261
real numbers, 4, 7, 30, 53, 64, 70, 72, 77, 85, 96, 99,
100, 119, 120, 128, 134
reciprocal, 160
rectangle, 205, 206, 214, 219, 225, 236, 240
relatively prime, 154, 167, 176
remainder, 140, 141, 155, 156, 158, 159, 161, 162,
165, 166, 167, 171, 172, 173, 174, 176, 178, 180,
181
rhombus, 70
right angle, 195, 209, 220, 221, 245, 258
N
natural number, 162, 165, 171, 172, 173, 175
natural numbers, 171, 172
number line, 64, 74, 76, 80
numerator, 24, 30
O
obtuse angle, 259
obtuse triangle, 207
odd number, 145, 154, 156, 161, 172, 179
ordered pair, 25, 148, 152
origin, 278
281
Index
right triangle, 118, 182, 190, 194, 199, 201, 209, 217,
220, 234, 236, 240, 245, 248, 250, 252, 258, 273
root, 18, 19, 41, 53, 69, 97, 111, 112, 128, 131, 133,
136
triangle, 77, 82, 83, 182, 183, 189, 190, 192, 193, 194,
198, 199, 200, 202, 205, 207, 208, 209, 211, 212,
213, 214, 215, 218, 219, 220, 222, 223, 224, 225,
226, 227, 229, 230, 233, 234, 235, 236, 237, 238,
242, 245, 250, 252, 258
trisect, 217, 218
S
semicircle, 239, 242
set, 28, 30, 31, 77, 88, 161
similar, 22, 84, 204, 230, 237, 252
simplifying, 26
slope, 69, 84
solution, 2, 14, 16, 37, 40, 43, 44, 47, 48, 51, 55, 56,
57, 59, 60, 61, 62, 63, 71, 73, 77, 84, 85, 88, 89, 90,
97, 100, 103, 104, 105, 114, 115, 124, 125, 126,
150, 151, 160, 170, 208, 209, 210, 211, 212, 226,
229, 242, 257, 261
solution set, 84, 88
square, 1, 4, 7, 18, 25, 28, 37, 39, 45, 49, 70, 71, 75,
136, 138, 139, 140, 141, 142, 143, 144, 145, 146,
147, 148, 149, 150, 151, 152, 153, 154, 163, 164,
171, 176, 188, 206, 209, 219, 224, 229, 230, 273
square root, 4, 18, 28, 39, 45, 229
sum, 1, 7, 14, 16, 53, 61, 76, 77, 81, 82, 97, 98, 99,
101, 109, 112, 113, 114, 117, 118, 119, 120, 121,
122, 123, 126, 127, 134, 136, 139, 140, 144, 146,
147, 148, 149, 153, 162, 167, 169, 171, 176, 179,
181, 212, 224
U
union, 83
V
variable, 40, 143
vertex, 69, 214, 245, 265
vertical angles, 198, 242, 250
W
whole number, 157
X
x-axis, 70, 85, 265
x-intercept, 115
Y
y-axis, 70
T
Z
tangent line, 239, 240
term, 4, 31, 134, 141, 166
transversal, 237
trapezoid, 207, 218, 223, 234, 236, 240, 244, 245,
246, 248, 250
zero, 64, 128, 148, 156, 158, 160, 165, 167, 176, 179,
266
282
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