Difference between revisions of "2023 AIME II Problems/Problem 4"
(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> <...") |
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+ | == Problem == | ||
+ | Let <math>x,y,</math> and <math>z</math> be real numbers satisfying the system of equations | ||
+ | <cmath>\begin{align*} | ||
+ | xy + 4z &= 60 \\ | ||
+ | yz + 4x &= 60 \\ | ||
+ | zx + 4y &= 60. | ||
+ | \end{align*}</cmath> | ||
+ | Let <math>S</math> be the set of possible values of <math>x.</math> Find the sum of the squares of the elements of <math>S.</math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
Revision as of 16:36, 16 February 2023
Problem
Let and
be real numbers satisfying the system of equations
Let
be the set of possible values of
Find the sum of the squares of the elements of
Solution 1
We first subtract the 2nd equation from the first, noting that they both equal .
Case 1: Let
The first and third equations simplify to:
From which it is apparent that and
are solutions.
Case 2: Let
The first and third equations simplify to:
We subtract the following equations, yielding:
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