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Algebra Difficulty 5.0 AIME Prove it Croatia

If xx, yy, zz and ww are real numbers such that
xy+z+w+yz+w+x+zw+x+y+wx+y+z=1, \frac{x}{y+z+w} + \frac{y}{z+w+x} + \frac{z}{w+x+y} + \frac{w}{x+y+z} = 1,
find
x2y+z+w+y2z+w+x+z2w+x+y+w2x+y+z. \frac{x^2}{y+z+w} + \frac{y^2}{z+w+x} + \frac{z^2}{w+x+y} + \frac{w^2}{x+y+z}.

Solution

If we multiply the condition by x+y+z+wx + y + z + w, we get:
x2+x(y+z+w)y+z+w+y2+y(x+z+w)z+w+x+z2+z(x+y+w)w+x+y+w2+w(x+y+z)x+y+z=x+y+z+w, \frac{x^2 + x(y + z + w)}{y + z + w} + \frac{y^2 + y(x + z + w)}{z + w + x} + \frac{z^2 + z(x + y + w)}{w + x + y} + \frac{w^2 + w(x + y + z)}{x + y + z} = x + y + z + w,
i.e.
x2y+z+w+x+y2z+w+x+y+z2w+x+y+z+w2x+y+z+w=x+y+z+w. \frac{x^2}{y+z+w} + x + \frac{y^2}{z+w+x} + y + \frac{z^2}{w+x+y} + z + \frac{w^2}{x+y+z} + w = x+y+z+w.
It follows that
x2y+z+w+y2z+w+x+z2w+x+y+w2x+y+z=0. \frac{x^2}{y+z+w} + \frac{y^2}{z+w+x} + \frac{z^2}{w+x+y} + \frac{w^2}{x+y+z} = 0.

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