2021 JMC 10 Problems/Problem 18

Problem 18

If $x,y,$ and $z$ are positive real numbers that satisfy the equation \[xy+yz+xz=96(y+z)-x^2 = 24(x+z) -y^2 = 54(x+y) -z^2,\] what is the value of $xyz$?

$\textbf{(A) } 1720 \qquad\textbf{(B) } 2720 \qquad\textbf{(C) } 7560 \qquad\textbf{(D) } 9600 \qquad\textbf{(E) } 15120$

Solution

Putting terms of different degrees on different sides, we can factor to obtain; \[\begin{cases} 96(y+z)= (x+y)(x+z)  \\ 24(x+z)=(y+z)(x+y) \\ 54(x+y)=(x+z)(y+z) \end{cases}\] Multiplying these and simplifying yields $(x+y)(y+z)(x+z)=96 \cdot 24 \cdot 54,$ and substituting gives \[96(y+z)=\frac{96 \cdot 24 \cdot 54}{(y+z)} \implies (y+z)^2=24 \cdot 54=1296 \implies y+z = 36.\] Similarly, $x+z=72$ and $x+y=48$. So $(x,y,z) = (42,6,30)$ and $xyz=42 \cdot 6 \cdot 30 = 7560,$ as desired.