2021 JMC 10 Problems/Problem 18
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Problem 18
If 
and 
are positive real numbers that satisfy the equation
![\[xy+yz+xz=96(y+z)-x^2 = 24(x+z) -y^2 = 54(x+y) -z^2,\]](http://latex.artofproblemsolving.com/8/6/e/86e8bf90f3dcd7b5cae5bf0aaf684d9a89898df2.png)
what is the value of 
?
Solution
Putting terms of different degrees on different sides, we can factor to obtain; ![\[\begin{cases} 96(y+z)= (x+y)(x+z) \\ 24(x+z)=(y+z)(x+y) \\ 54(x+y)=(x+z)(y+z) \end{cases}\]](http://latex.artofproblemsolving.com/d/8/e/d8e97bb77f920a0be0102645d020651172d3a136.png)
Multiplying these and simplifying yields 
and substituting gives ![\[96(y+z)=\frac{96 \cdot 24 \cdot 54}{(y+z)} \implies (y+z)^2=24 \cdot 54=1296 \implies y+z = 36.\]](http://latex.artofproblemsolving.com/3/5/f/35f8e34c391b73954d2da35981d13aa3596b3b8f.png)
Similarly, 
and 
. So 
and 
as desired.
