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 19:30, 16 February 2023
Contents
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:
\begin{align*}
x^2+4y-xy-4x&=0 \\
x(x-4)-y(x-4)&=0 \\
(x-4)(x-y)&=0
\end{align*}$$ (Error compiling LaTeX. Unknown error_msg)
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.