Maths Olympiad Prep

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, 2020

Algebra Difficulty 5.5 AIME, harder Prove it United States

Problem:
Determine all triplets of real numbers (x,y,z)(x, y, z) satisfying the system of equations
x2y+y2z=1040x2z+z2y=260(xy)(yz)(zx)=540 \begin{aligned} x^{2} y + y^{2} z & = 1040 \\ x^{2} z + z^{2} y & = 260 \\ (x - y)(y - z)(z - x) & = -540 \end{aligned}

Solution

Solution:
Call the three equations (1), (2), (3).

(1)/(2) gives y=4zy = 4z.

(3) ++ (1) - (2) gives
(y2z2)x=15z2x=240 \left(y^{2} - z^{2}\right)x = 15z^{2}x = 240
so z2x=16z^{2} x = 16.

Therefore
z(x+2z)2=x2z+z2y+4z2x=815z(x2z)2=x2z+z2y4z2x=495 \begin{aligned} & z(x + 2z)^{2} = x^{2} z + z^{2} y + 4z^{2} x = \frac{81}{5} \\ & z(x - 2z)^{2} = x^{2} z + z^{2} y - 4z^{2} x = \frac{49}{5} \end{aligned}
so x+2zx2z=97\left|\frac{x + 2z}{x - 2z}\right| = \frac{9}{7}.

Thus either x=16zx = 16z or x=z4x = \frac{z}{4}.

If x=16zx = 16z, then (1) becomes 1024z3+16z3=10401024z^{3} + 16z^{3} = 1040, so (x,y,z)=(16,4,1)(x, y, z) = (16, 4, 1).

If x=z4x = \frac{z}{4}, then (1) becomes 14z3+16z3=1040\frac{1}{4}z^{3} + 16z^{3} = 1040, so (x,y,z)=(1,16,4)(x, y, z) = (1, 16, 4).

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