2023 AIME I Problems/Problem 2
Contents
Problem
Positive real numbers and satisfy the equations The value of is where and are relatively prime positive integers. Find
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=jxY7BBe-4gU
Solution 1
Denote . Hence, the system of equations given in the problem can be rewritten as Solving the system gives and . Therefore, Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
We can use the property that on the first equation. We get . Then, subtracting from both sides, we get , therefore . Substituting that into our first equation, we get . Squaring, reciprocating, and simplifying both sides, we get the quadratic . Solving for , we get and . Since the problem said that , . To solve for , we can use the property that . , so . Adding these together, we get
~idk12345678
Solution 3 (quick)
We can let . Then, in the first equation, the LHS becomes and the RHS becomes . Therefore, must be ( can't be ). So now we know . So we can plug this into the second equation to get . This gives , so and . Adding the numerator and denominator gives .
Honestly this problem is kinda misplaced.
~yrock
Video Solution by TheBeautyofMath
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.