The Art and Craft of Problem Solving

154 CHAPTER 5 ALGEBRA

Example 5.2.21 (AIME 198 6) The polynomial 1 -x+x^2 -..J + ... +xI^6 _xI^7 may

be written in the form ao +aly+a 2 y^2 +a 3 y^3 + ... +aI 6 yI^6 +a 17 yI^7 , where y =x+ 1

and the aiS are constants. Find the value of a 2.

Solution: Using our active knowledge of the factorization formulas, we recognize

immediately that

(^18 1)
1 _x+�_x^3 + ... +xI^6 xI^7 =
x -
.
x-(-l)
(Alternatively, we could have used the formula for the sum of a geometric series.)
Substituting y = x + 1, we see that the polynomial becomes^6
(Y
(^1) �I (^8) - 1
= � (y^18 (\
(^8) )
yI^7 + C;)yI^6 _ ... + 1- 1 )

= _y^17 + (\

(^8) )

yI^6 - ... + C�)i - C!)y

+ C�) ·

Thus

The following problem appeared in the 1972 IMO. Its solution depends on sym­

metry and the careful extraction of squares, but more than anything else, on confidence

that a reasonably elegant solution exists. It is rather contrived, but quite instructive.

Example 5.2.22 Find all solutions (XI,X2,X 3 ,X 4 ,X 5 ) of the system of inequalities

(xi -X 3 X 5 )(X� -X 3 X 5 ) � 0

(�-X 4 XI)(.xi -X 4 XI) � 0

(x� - X 5 X 2 )(� - X 5 X 2 ) � 0

(x� -XIX 3 )(X� -XIX 3 ) � 0

(x� -X 2 X 4 )(xi -X 2 X 4 ) � 0

where Xl ,X 2 ,X 3 ,X 4 ,X 5 are real numbers.

Solution: This problem is pretty intimidating, but notice that it is cyclically sym­

metric: each inequality has the form

(xT -Xi+ 2 Xi+ 4 ) (xT+I -Xi+ 2 Xi+ 4 ),

where the indices are read modulo 5. For example, if i = 3, then the inequality becomes

(x� -X 5 X 2 )(X� -X 5 X 2 ) � O.

When the left-hand sides are multiplied out, we will get a total of 20 terms: all

m = 10 "perfect square" terms of the form qX] (i -I-j) as well as 10 "cross-terms,"

five of the form -qxi+IXi+ 3 and five of the form -qxi+ 2 Xi+ 4.

(^6) If you are not familiar with the binomial theorem, read Section 6.1.