2023 AIME II Problems/Problem 4
Revision as of 16:35, 16 February 2023 by Sahanwijetunga (talk | contribs) (Created page with "==Solution 1== We first subtract the 2nd equation from the first, noting that they both equal <math>60</math>. <cmath>xy+4z-yz-4x=0</cmath> <cmath>4(z-x)-y(z-x)=0</cmath> <...")
Solution 1
We first subtract the 2nd equation from the first, noting that they both equal 
.
![\[xy+4z-yz-4x=0\]](http://latex.artofproblemsolving.com/4/1/d/41de42a673c31f5a1ca95cb600947d8bde42b2fc.png)
![\[4(z-x)-y(z-x)=0\]](http://latex.artofproblemsolving.com/2/b/f/2bfeba47cf80c7f2ad7fa7cfebf03a46bae04eed.png)
![\[(z-x)(4-y)=0\]](http://latex.artofproblemsolving.com/f/1/2/f12a0d84e3439024da8bdbb59fc8151543b456d2.png)
Case 1: Let 
The first and third equations simplify to:
![\[x+z=15\]](http://latex.artofproblemsolving.com/c/f/4/cf4c2df2a5fedc96c127939686fe5e7eaa7fe3bf.png)
![\[xz=44\]](http://latex.artofproblemsolving.com/e/e/7/ee7532bfa95b5e9a79b354d631e4dda2f3c56f5b.png)
From which it is apparent that 
and 
are solutions.
Case 2: Let 
The first and third equations simplify to:
![\[xy+4x=60\]](http://latex.artofproblemsolving.com/1/c/7/1c77cbdb81faf12f89911633f6e154e4fc47c466.png)
![\[x^2+4y=60\]](http://latex.artofproblemsolving.com/3/1/0/31086e95fff556528b95ed2012be9f1aaf9a4285.png)
We subtract the following equations, yielding:
![\[x^2+4y-xy-4x=0\]](http://latex.artofproblemsolving.com/7/d/a/7dac40e4608060ab7dbd2f7d0a56d93b7113ffcc.png)
![\[x(x-4)-y(x-4)=0\]](http://latex.artofproblemsolving.com/6/c/5/6c53fb219271b714b3f891be37b4b79409cb6b40.png)
![\[(x-4)(x-y)=0\]](http://latex.artofproblemsolving.com/2/9/e/29e733edf39cffea61237b64b3cd816c24402646.png)
We thus have and 
, substituting in 
and solving yields 
and 
Then, we just add the squares of the solutions (make sure not to double count the 4), and get: 
~SAHANWIJETUNGA
