Difference between revisions of "2023 AIME II Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
− | We first subtract the | + | We first subtract the second equation from the first, noting that they both equal <math>60</math>. |
+ | <cmath>\begin{align*} | ||
+ | xy+4z-yz-4x&=0 \ | ||
+ | 4(z-x)-y(z-x)&=0 \ | ||
+ | (z-x)(4-y)&=0 | ||
+ | \end{align*}</cmath> | ||
− | + | Case 1: Let <math>y=4</math>. | |
− | |||
− | |||
− | |||
− | Case 1: Let <math>y=4</math> | ||
The first and third equations simplify to: | The first and third equations simplify to: | ||
− | <cmath>x+z=15</cmath> | + | <cmath>\begin{align*} |
− | < | + | x+z&=15 \ |
+ | xz&=44 | ||
+ | \end{align*}</cmath> | ||
+ | from which it is apparent that <math>x=4</math> and <math>x=11</math> are solutions. | ||
− | + | Case 2: Let <math>x=z</math>. | |
− | |||
− | Case 2: Let <math>x=z</math> | ||
The first and third equations simplify to: | The first and third equations simplify to: | ||
− | <cmath>xy+4x=60 | + | <cmath>\begin{align*} |
− | + | xy+4x&=60 \ | |
+ | x^2+4y&=60 | ||
+ | \end{align*} | ||
We subtract the following equations, yielding: | We subtract the following equations, yielding: | ||
+ | </cmath>\begin{align*} | ||
+ | x^2+4y-xy-4x&=0 \ | ||
+ | x(x-4)-y(x-4)&=0 \ | ||
+ | (x-4)(x-y)&=0 | ||
+ | \end{align*}<math></math> | ||
− | + | We thus have <math>x=4</math> and <math>x=y</math>, substituting in <math>x=y=z</math> and solving yields <math>x=-6</math> and <math>x=10</math>. | |
− | |||
− | |||
− | |||
− | We thus have <math>x=4</math> and <math>x=y</math>, substituting in <math>x=y=z</math> and solving yields <math>x=-6</math> and <math>x=10</math> | ||
− | Then, we just add the squares of the solutions (make sure not to double count the 4), and get: <math>4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}</math> | + | Then, we just add the squares of the solutions (make sure not to double count the 4), and get: <math>4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}</math>. |
~SAHANWIJETUNGA | ~SAHANWIJETUNGA |
Revision as of 18:30, 16 February 2023
Contents
[hide]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 second 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 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
Solution 2
We index these equations as (1), (2), and (3), respectively. Taking , we get
Denote , , . Thus, the above equation can be equivalently written as
Similarly, by taking , we get
By taking , we get
From , we have the following two cases.
Case 1: .
Plugging this into and , we get . Thus, or .
Because we only need to compute all possible values of , without loss of generality, we only need to analyze one case that .
Plugging and into (1), we get a feasible solution , , .
Case 2: and .
Plugging this into and , we get .
Case 2.1: .
Thus, . Plugging and into (1), we get a feasible solution , , .
Case 2.2: and .
Thus, . Plugging these into (1), we get or .
Putting all cases together, . Therefore, the sum of the squares of the elements of is
~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.