Olympiad Maths Prep

Track / Stage 5 / 85 of 400 #685 of 2000

Problem 685

AIME late
Algebra Difficulty 5.2 Find the answer

The numbers x,yx, y and zz are such that xy+z+yz+x+zx+y=1\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1. What values can the expression x2y+z+y2z+x+z2x+y\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} take?

Official solution

x2y+z+y2z+x+z2x+y+(x+y+z)=x2y+z+x+y2z+x+y+z2x+y+z=\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}+(x+y+z)=\frac{x^{2}}{y+z}+x+\frac{y^{2}}{z+x}+y+\frac{z^{2}}{x+y}+z=

=x2+xy+xzy+z+y2+yz+yxz+x+z2+zx+zyx+y==\frac{x^{2}+xy+xz}{y+z}+\frac{y^{2}+yz+yx}{z+x}+\frac{z^{2}+zx+zy}{x+y}=

(xy+z+yz+x+zx+y)(x+y+z)=x+y+z\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)(x+y+z)=x+y+z.

From this, it is clear that the desired sum is zero.

## Answer

0.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.