Difference between revisions of "2023 AIME I Problems/Problem 2"
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+ | ==Solution 3 (quick)== | ||
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+ | We can let <math>n=b^{4x^2}</math>. Then, in the first equation, the LHS becomes 2x and the RHS becomes 2x^2. Therefore, x must be 1 (x can't be 0). So now we know <math>n=b^4</math>. So we can plug this into the second equation to get. This gives <math>b\cdot4=5</math>, so <math>b= \frac{5}{4}</math> and <math>n= b^4=\frac{625}{256}</math>. Adding the numerator and denominator gives <math>\boxed{881}</math>. | ||
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+ | Honestly this problem is kinda misplaced. | ||
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+ | ~yrock | ||
Revision as of 18:33, 23 May 2024
Contents
[hide]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 2x and the RHS becomes 2x^2. Therefore, x must be 1 (x can't be 0). 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.