2023 AIME II Problems/Problem 4

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: $\begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}$

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)

Solution 3

Quadratic Formula and Vieta's Formulas

We index these equations as (1), (2), and (3), respectively. Using equation (1), we get $z = \frac{60 - xy}{4) = 15 - \frac{xy}{4}$ (Error compiling LaTeX. Unknown error_msg) We need to solve for x, so we plug this value of z into equation (3) to get: $<cmath>15x - \frac{x^2y}{4} - 4y = 60$ $\frac{y}{4} * x^2 - 15x + (60 + 4y) = 0$ </cmath> We use the quadratic formula to get possible values of x: $$ (Error compiling LaTeX. Unknown error_msg) $x = \frac{15 \pm \sqrt{15^2 - 4(\frac{y}{4})(60 + 4y)}}{\frac{y}{2}}$

2023 AIME II (Problems • Answer Key • Resources)
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2023 AIME II Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

Solution 3

See also