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2005 OIM Problems/Problem 1

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Problem

Determine all triples of real numbers $(x, y, z)$ that satisfy the following system of equations:

\[xyz = 8\] \[x^2y+y^2z+z^2x = 73\] \[x(y-z)^2+y(z-x)^2+z(x-y)^2 = 98\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

One can observe: $x(y-z)^2+y(z-x)^2+z(x-y)^2 = x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)-6xyz$

A little more manipulation and we get:

$xy^2+yz^2+zx^2 = 73$

Subtract this from $x^2y+y^2z+z^2x = 73$

To get: $-(x-y)(y-z)(z-x) = 0$

Implying that $x = y, y = z$ or $z = x$

WLOG, assume $x = z$,

So, $y = \frac{8}{x^2}$

$x^2y+y^2x+x^3 = 73$

Implying, $\frac{64}{x^3}+x^3 = 65$

Therefore, $x = 1$ or $4$

So, $y = 8$ or $1/2$

$(x,y,z) =$ permutations of $(1,8,1)$ and $(4,4,\frac{1}{2})$

~creativeRaven

See also

OIM Problems and Solutions

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