2019 AIME II Problems/Problem 15
Problem
In acute triangle points
and
are the feet of the perpendiculars from
to
and from
to
, respectively. Line
intersects the circumcircle of
in two distinct points,
and
. Suppose
,
, and
. The value of
can be written in the form
where
and
are positive relatively prime integers. Find
.
Solution
Let
Therefore $AB= \frac{b}{k} , AC= \frac{a}{k}
By power of point, we have$ (Error compiling LaTeX. Unknown error_msg)AP*BP=XP*YP , AQ*CQ=YQ*XQ400= \frac{ab}{k} - a^2$$ (Error compiling LaTeX. Unknown error_msg)525= \frac{ab}{k} - b^2
a^2= \frac{ab}{k} - 400$$ (Error compiling LaTeX. Unknown error_msg)b^2= \frac{ab}{k} - 525$(1)
Or$ (Error compiling LaTeX. Unknown error_msg)k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525}
Let
Then, $a=\sqrt{u-400},b=\sqrt{u-525},k=\frac{\sqrt{(u-400)(u-525)}}{u}
In triangle$ (Error compiling LaTeX. Unknown error_msg)APQ25^2= a^2 + b^2 - 2abk
625= \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk
\frac{ab}{k} - abk =775
u$$ (Error compiling LaTeX. Unknown error_msg)u- \frac{(u-400)(u-525)}{u} =775$The quadratic term is cancelled out after simplified
Which gives$ (Error compiling LaTeX. Unknown error_msg)u=1400a= \sqrt{1000} , b=\sqrt{775}
AB*AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} *\frac{ab}{u} } = \frac{u^2}{ab}
= \frac{1400 * 1400}{ \sqrt{ 1000* 875 }} = 560 \sqrt{14}
So the final answer is 560 + 14 = 574
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
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