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

m (→‎Solution 2)
m (→‎Solution 1)
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==Solution 1==
 
==Solution 1==
  
−
We first subtract the 2nd equation from the first, noting that they both equal <math>60</math>.  
+
We first subtract the second equation from the first, noting that they both equal <math>60</math>.  
 +
<cmath>\begin{align*}
 +
xy+4z-yz-4x&=0 \\
 +
4(z-x)-y(z-x)&=0 \\
 +
(z-x)(4-y)&=0
 +
\end{align*}</cmath>
  
−
<cmath>xy+4z-yz-4x=0</cmath>
+
Case 1: Let <math>y=4</math>.
−
<cmath>4(z-x)-y(z-x)=0</cmath>
 
−
<cmath>(z-x)(4-y)=0</cmath>
 
−
 
 
−
Case 1: Let <math>y=4</math>
 
  
 
The first and third equations simplify to:
 
The first and third equations simplify to:
−
<cmath>x+z=15</cmath>
+
<cmath>\begin{align*}
−
<cmath>xz=44</cmath>
+
x+z&=15 \\
 +
xz&=44
 +
\end{align*}</cmath>
 +
from which it is apparent that <math>x=4</math> and <math>x=11</math> are solutions.
  
−
From which it is apparent that <math>x=4</math> and <math>x=11</math> are solutions.
+
Case 2: Let <math>x=z</math>.
−
 
 
−
Case 2: Let <math>x=z</math>
 
  
 
The first and third equations simplify to:
 
The first and third equations simplify to:
−
<cmath>xy+4x=60</cmath>
+
<cmath>\begin{align*}
−
<cmath>x^2+4y=60</cmath>
+
xy+4x&=60 \\
 +
x^2+4y&=60
 +
\end{align*}
  
 
We subtract the following equations, yielding:
 
We subtract the following equations, yielding:
 +
</cmath>\begin{align*}
 +
x^2+4y-xy-4x&=0 \\
 +
x(x-4)-y(x-4)&=0 \\
 +
(x-4)(x-y)&=0
 +
\end{align*}<math></math>
  
−
<cmath>x^2+4y-xy-4x=0</cmath>
+
We thus have <math>x=4</math> and <math>x=y</math>, substituting in <math>x=y=z</math> and solving yields <math>x=-6</math> and <math>x=10</math>.
−
<cmath>x(x-4)-y(x-4)=0</cmath>
 
−
<cmath>(x-4)(x-y)=0</cmath>
 
−
 
 
−
We thus have <math>x=4</math> and <math>x=y</math>, substituting in <math>x=y=z</math> and solving yields <math>x=-6</math> and <math>x=10</math>
 
  
−
Then, we just add the squares of the solutions (make sure not to double count the 4), and get: <math>4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}</math>
+
Then, we just add the squares of the solutions (make sure not to double count the 4), and get: <math>4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}</math>.
  
 
~SAHANWIJETUNGA
 
~SAHANWIJETUNGA

Revision as of 18:30, 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 second equation from the first, noting that they both equal $60$. \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}

Case 1: Let $y=4$.

The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} 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: \begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*} We subtract the following equations, yielding:\begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}$$ (Error compiling LaTeX. Unknown error_msg)

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)

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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