Difference between revisions of "2016 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | + | Note that | |
+ | <cmath>\begin{align*}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{align*}</cmath> | ||
+ | Substituting this into the second equation and collecting <math>x_i^2</math> terms, 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2016|n=II|num-b=14|after=Last Question}} | ||
+ | {{MAA Notice}} |
Revision as of 09:27, 17 May 2016
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 .
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 |
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