Maths Olympiad Prep

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Problem 1157

AIME late
Geometry Difficulty 5.1 Prove it Ukrainian National Mathematical Olympiad - Third Round · Ukraine

Is it possible to construct a triangle with sides xx, yy, zz satisfying the condition:
3x2y2+3y2z2+3z2x2=x4+y4+z4? 3x^2y^2 + 3y^2z^2 + 3z^2x^2 = x^4 + y^4 + z^4?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Rewrite the equation as:
2x2y2+2y2z2+2z2x2−x4−y4−z4=−(x2y2+y2z2+z2x2). 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 = -(x^2y^2 + y^2z^2 + z^2x^2).
The left-hand side can be decomposed as:
(x+y+z)(x+y−z)(y+z−x)(z+x−y)=−(x2y2+y2z2+z2x2).(x+y+z)(x+y-z)(y+z-x)(z+x-y) = -(x^2y^2 + y^2z^2 + z^2x^2).
Hence, the left-hand side is negative, therefore, at least one of the multipliers is negative too. The first one is always positive, then, without loss of generality, we can assume the second one is negative, i.e. x+y−z<0x+y-z<0, which contradicts the triangle inequality.

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