Maths Olympiad Prep

Library / /108 of 189

Algebra Difficulty 6.0 AIME, harder Prove it Ukraine

It is known that nonzero real numbers xx, yy, zz satisfy the condition xy+yz+zx=0xy + yz + zx = 0. What value can the expression
1x2+2yz+1y2+2zx+1z2+2xy \frac{1}{x^2+2yz} + \frac{1}{y^2+2zx} + \frac{1}{z^2+2xy}
be equal to?

Solution

Since xyz≠0xyz \neq 0, we can do the following transformation of the given expression:
1x2+2yz+1y2+2zx+1z2+2xy=1x2+2yz−xy−yz−zx+1y2+2zx−xy−yz−zx+1z2+2xy−xy−yz−zx=1x2+yz−xy−zx+1y2+zx−xy−yz+1z2+xy−yz−zx=1(x−y)(x−z)+1(y−x)(y−z)+1(z−x)(z−y)=−1(x−y)(z−x)−1(x−y)(y−z)−1(z−x)(y−z)=−(y−z)+(z−x)+(x−y)(x−y)(z−x)(y−z)=0. \begin{aligned} \frac{1}{x^2+2yz} + \frac{1}{y^2+2zx} + \frac{1}{z^2+2xy} &= \frac{1}{x^2+2yz-xy-yz-zx} + \frac{1}{y^2+2zx-xy-yz-zx} + \frac{1}{z^2+2xy-xy-yz-zx} \\ &= \frac{1}{x^2+yz-xy-zx} + \frac{1}{y^2+zx-xy-yz} + \frac{1}{z^2+xy-yz-zx} \\ &= \frac{1}{(x-y)(x-z)} + \frac{1}{(y-x)(y-z)} + \frac{1}{(z-x)(z-y)} \\ &= -\frac{1}{(x-y)(z-x)} - \frac{1}{(x-y)(y-z)} - \frac{1}{(z-x)(y-z)} \\ &= -\frac{(y-z)+(z-x)+(x-y)}{(x-y)(z-x)(y-z)} = 0. \end{aligned}

Thus, there are no zeros among the three terms on the left-hand side, and it turned out that the sum of two is equal to the third modulo, that is, they can not be the sides of a non-degenerate triangle. The resulting contradiction completes the proof.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.