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

(→‎Solution 1)
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~SAHANWIJETUNGA
 
~SAHANWIJETUNGA
 +
 +
==Solution 2==
 +
 +
We index these equations as (1), (2), and (3), respectively.
 +
Taking <math>(1)-(2)</math>, we get
 +
<cmath>
 +
\[
 +
\left( x - z \right) \left( y - 4 \right) = 60 .
 +
\]
 +
</cmath>
 +
 +
Denote <math>x' = x - 4</math>, <math>y' = y - 4</math>, <math>z' = z - 4</math>.
 +
Thus, the above equation can be equivalently written as
 +
<cmath>
 +
\[
 +
\left( x' - z' \right) y' = 0 . \hspace{1cm} (1')
 +
\]
 +
</cmath>
 +
 +
Similarly, by taking <math>(2)-(3)</math>, we get
 +
<cmath>
 +
\[
 +
\left( y' - x' \right) z' = 0 . \hspace{1cm} (2')
 +
\]
 +
</cmath>
 +
 +
By taking <math>(3) - (1)</math>, we get
 +
<cmath>
 +
\[
 +
\left( z' - y' \right) x' = 0 . \hspace{1cm} (3')
 +
\]
 +
</cmath>
 +
 +
From <math>(3')</math>, we have the following two cases.
 +
 +
Case 1: <math>x' = 0</math>.
 +
 +
Plugging this into <math>(1')</math> and <math>(2')</math>, we get <math>y'z' = 0</math>.
 +
Thus, <math>y' = 0</math> or <math>z' = 0</math>.
 +
 +
Because we only need to compute all possible values of <math>x</math>, without loss of generality, we only need to analyze one case that <math>y' = 0</math>.
 +
 +
Plugging <math>x' = 0</math> and <math>y' = 0</math> into (1), we get a feasible solution <math>x = 4</math>, <math>y = 4</math>, <math>z = 11</math>.
 +
 +
Case 2: <math>x' \neq 0</math> and <math>z' - y' = 0</math>.
 +
 +
Plugging this into <math>(1')</math> and <math>(2')</math>, we get <math>\left( x' - y' \right) y' = 0</math>.
 +
 +
Case 2.1: <math>y' = 0</math>.
 +
 +
Thus, <math>z' = 0</math>. Plugging <math>y' = 0</math> and <math>z' = 0</math> into (1), we get a feasible solution <math>x = 11</math>, <math>y = 4</math>, <math>z = 4</math>.
 +
 +
Case 2.2: <math>y' \neq 0</math> and <math>x' = y'</math>.
 +
 +
Thus, <math>x' = y' = z'</math>. Plugging these into (1), we get <math>\left( x, y, z \right) = \left( -10, -10, -10 \right)</math> or <math>\left( 6, 6, 6 \right)</math>.
 +
 +
Putting all cases together, <math>S = \left\{ 4, 11, -10, 6 \right\}</math>.
 +
Therefore, the sum of the squares of the elements of <math>S</math> is
 +
<cmath>
 +
\begin{align*}
 +
4^2 + 11^2 + \left( -10 \right)^2 + 6^2
 +
= \boxed{\textbf{(273) }}  .
 +
\end{align*}
 +
</cmath>
 +
 +
Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:11, 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

Solution 2

We index these equations as (1), (2), and (3), respectively. Taking $(1)-(2)$, we get \[ \left( x - z \right) \left( y - 4 \right) = 60 . \]

Denote $x' = x - 4$, $y' = y - 4$, $z' = z - 4$. Thus, the above equation can be equivalently written as \[ \left( x' - z' \right) y' = 0 . \hspace{1cm} (1') \]

Similarly, by taking $(2)-(3)$, we get \[ \left( y' - x' \right) z' = 0 . \hspace{1cm} (2') \]

By taking $(3) - (1)$, we get \[ \left( z' - y' \right) x' = 0 . \hspace{1cm} (3') \]

From $(3')$, we have the following two cases.

Case 1: $x' = 0$.

Plugging this into $(1')$ and $(2')$, we get $y'z' = 0$. Thus, $y' = 0$ or $z' = 0$.

Because we only need to compute all possible values of $x$, without loss of generality, we only need to analyze one case that $y' = 0$.

Plugging $x' = 0$ and $y' = 0$ into (1), we get a feasible solution $x = 4$, $y = 4$, $z = 11$.

Case 2: $x' \neq 0$ and $z' - y' = 0$.

Plugging this into $(1')$ and $(2')$, we get $\left( x' - y' \right) y' = 0$.

Case 2.1: $y' = 0$.

Thus, $z' = 0$. Plugging $y' = 0$ and $z' = 0$ into (1), we get a feasible solution $x = 11$, $y = 4$, $z = 4$.

Case 2.2: $y' \neq 0$ and $x' = y'$.

Thus, $x' = y' = z'$. Plugging these into (1), we get $\left( x, y, z \right) = \left( -10, -10, -10 \right)$ or $\left( 6, 6, 6 \right)$.

Putting all cases together, $S = \left\{ 4, 11, -10, 6 \right\}$. Therefore, the sum of the squares of the elements of $S$ is \begin{align*} 4^2 + 11^2 + \left( -10 \right)^2 + 6^2 = \boxed{\textbf{(273) }} . \end{align*}

Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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