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 15:36, 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