Difference between revisions of "2016 AIME II Problems/Problem 15"
Line 8: | Line 8: | ||
<cmath>\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{215}.</cmath> Conveniently, <math>\sum_{i=1}^{216} 1-a_i=215</math> so we find | <cmath>\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{215}.</cmath> Conveniently, <math>\sum_{i=1}^{216} 1-a_i=215</math> so we find | ||
<cmath>\left(\sum_{i=1}^{216} 1-a_i\right)\left(\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}\right)=1=\left(\sum_{i=1}^{216} x_i\right)^2.</cmath> This is the equality case of the Cauchy-Schwarz Inequality, so <math>x_i=c(1-a_i)</math> for some constant <math>c</math>. Summing these equations and using the facts that <math>\sum_{i=1}^{216} x_i=1</math> and <math>\sum_{i=1}^{216} 1-a_i=215</math>, we find <math>c=\frac{1}{215}</math> and thus <math>x_2=c(1-a_2)=\frac{1}{215}\cdot \left(1-\frac{1}{4}\right)=\frac{3}{860}</math>. Hence the desired answer is <math>860+3=\boxed{863}</math>. | <cmath>\left(\sum_{i=1}^{216} 1-a_i\right)\left(\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}\right)=1=\left(\sum_{i=1}^{216} x_i\right)^2.</cmath> This is the equality case of the Cauchy-Schwarz Inequality, so <math>x_i=c(1-a_i)</math> for some constant <math>c</math>. Summing these equations and using the facts that <math>\sum_{i=1}^{216} x_i=1</math> and <math>\sum_{i=1}^{216} 1-a_i=215</math>, we find <math>c=\frac{1}{215}</math> and thus <math>x_2=c(1-a_2)=\frac{1}{215}\cdot \left(1-\frac{1}{4}\right)=\frac{3}{860}</math>. Hence the desired answer is <math>860+3=\boxed{863}</math>. | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/mjtM-Coav4k | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2016|n=II|num-b=14|after=Last Question}} | {{AIME box|year=2016|n=II|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:02, 19 April 2023
Contents
[hide]Problem
For let and . Let be positive real numbers such that and . The maximum possible value of , where and are relatively prime positive integers. Find .
Solution
Note that Substituting this into the second equation and collecting terms, we find Conveniently, so we find This is the equality case of the Cauchy-Schwarz Inequality, so for some constant . Summing these equations and using the facts that and , we find and thus . Hence the desired answer is .
Video Solution
~MathProblemSolvingSkills.com
See Also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.