Difference between revisions of "2023 AIME II Problems/Problem 4"

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== Problem ==
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Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system of equations
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<cmath>\begin{align*}
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xy + 4z &= 60 \
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yz + 4x &= 60 \
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zx + 4y &= 60.
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\end{align*}</cmath>
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Let <math>S</math> be the set of possible values of <math>x.</math> Find the sum of the squares of the elements of <math>S.</math>
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==Solution 1==
 
==Solution 1==
  

Revision as of 15:36, 16 February 2023

Problem

Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

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