Difference between revisions of "2016 AIME II Problems/Problem 15"
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− | Replace <math>\sum x_ix_j</math> with <math>\frac12\left(\left(\sum x_i\right)^2-\sum x_i^2\right)</math> and the second equation becomes <math>\sum\frac{x_i^2}{1-a_i}=\frac{1}{215}</math>. Conveniently, <math>\sum 1-a_i=215</math> so we get <math>\left(\sum 1-a_i\right)\left(\sum\frac{x_i^2}{1-a_i}\right)=1=\left(\sum x_i\right)^2</math>. This is the equality case of Cauchy so <math>x_i=c(1-a_i)</math> for some constant <math>c</math>. Using <math>\sum x_i=1</math>, we find <math>c=\frac{1}{215}</math> and thus <math>x_2=\frac{3}{860}</math>. | + | Replace <math>\sum x_ix_j</math> with <math>\frac12\left(\left(\sum x_i\right)^2-\sum x_i^2\right)</math> and the second equation becomes <math>\sum\frac{x_i^2}{1-a_i}=\frac{1}{215}</math>. Conveniently, <math>\sum 1-a_i=215</math> so we get <math>\left(\sum 1-a_i\right)\left(\sum\frac{x_i^2}{1-a_i}\right)=1=\left(\sum x_i\right)^2</math>. This is the equality case of Cauchy so <math>x_i=c(1-a_i)</math> for some constant <math>c</math>. Using <math>\sum x_i=1</math>, we find <math>c=\frac{1}{215}</math> and thus <math>x_2=\frac{3}{860}</math>. Thus, the desired answer is <math>860+3=\boxed{863}</math>. |
Revision as of 22:12, 17 March 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
Replace with
and the second equation becomes
. Conveniently,
so we get
. This is the equality case of Cauchy so
for some constant
. Using
, we find
and thus
. Thus, the desired answer is
.