2023 AIME II Problems/Problem 4

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Solution 1

We first subtract the 2nd equation from the first, noting that they both equal $60$.

\[xy+4z-yz-4x=0\] \[4(z-x)-y(z-x)=0\] \[(z-x)(4-y)=0\]

Case 1: Let $y=4$

The first and third equations simplify to: \[x+z=15\] \[xz=44\]

From which it is apparent that $x=4$ and $x=11$ are solutions.

Case 2: Let $x=z$

The first and third equations simplify to: \[xy+4x=60\] \[x^2+4y=60\]

We subtract the following equations, yielding:

\[x^2+4y-xy-4x=0\] \[x(x-4)-y(x-4)=0\] \[(x-4)(x-y)=0\]

We thus have $x=4$ and $x=y$, substituting in $x=y=z$ and solving yields $x=-6$ and $x=10$

Then, we just add the squares of the solutions (make sure not to double count the 4), and get: $4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}$

~SAHANWIJETUNGA