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2007 Cyprus MO/Lyceum/Problem 13

From AoPSWiki

Problem

If $x_1,\ x_2$ are the roots of the equation $\displaystyle x^2+ax+1=0$ and $x_3,\ x_4$ are the roots of the equation $\displaystyle x^2+bx+1=0$ , then the expression $\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3}$ equals to

$\mathrm{(A) \ } a^2+b^2-2\qquad \mathrm{(B) \ } a^2+b^2\qquad \mathrm{(C) \ } \frac{a^2+b^2}{2}\qquad \mathrm{(D) \ } a^2+b^2...$

Solution

$\displaystyle (x - x_1)(x - x_2) = x^2 + ax + 1 = 0$ , so $a = -(x_1 + x_2) \displaystyle$ and $x_1 \cdot x_2 = 1$ (the same goes for $b,\ x_3,\ x_4$ ).

$\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4} = \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1 \cdot x_2} = x_1^2...$

Similarly (for $b,\ x_3,\ x_4$ ) $\frac{x_3}{x_1x_2x_4} + \frac{x_4}{x_1x_2x_3} = b^2 - 2$

So

$\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3} = a^2 + b^2 - 4 \Longrightarrow \mat...$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2007_Cyprus_MO/Lyceum/Problem_13"

Category: Introductory Algebra Problems

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