2007 Cyprus MO/Lyceum/Problem 13

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

$\mathrm{(A) \ } a^2+b^2-2\qquad \mathrm{(B) \ } a^2+b^2\qquad \mathrm{(C) \ } \frac{a^2+b^2}{2}\qquad \mathrm{(D) \ } a^2+b^2+1\qquad \mathrm{(E) \ } a^2+b^2-4$

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 $b,\ x_3,\ x_4$).


$\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4} = \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1 \cdot x_2}  = x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = a^2 - 2$

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

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

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 12
Followed by
Problem 14
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