# 2019 AIME II Problems/Problem 15

## Problem

In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

## Solution

Let $AP=a, AQ=b, cos\angle A = k$

Therefore $AB= \frac{b}{k} , AC= \frac{a}{k}$

By power of point, we have $AP\cdot BP=XP\cdot YP , AQ\cdot CQ=YQ\cdot XQ$ Which are simplified to

$400= \frac{ab}{k} - a^2$

$525= \frac{ab}{k} - b^2$

Or

$a^2= \frac{ab}{k} - 400$

$b^2= \frac{ab}{k} - 525$

(1)

Or

$k= \frac{ab}{a^2+400} = \frac{ab}{b^2+525}$

Let $u=a^2+400=b^2+525$ Then, $a=\sqrt{u-400},b=\sqrt{u-525},k=\frac{\sqrt{(u-400)(u-525)}}{u}$

In triangle $APQ$, by law of cosine

$25^2= a^2 + b^2 - 2abk$

Pluging (1)

$625= \frac{ab}{k} - 400 + \frac{ab}{k} - 525 -2abk$

Or

$\frac{ab}{k} - abk =775$

Substitute everything by $u$

$u- \frac{(u-400)(u-525)}{u} =775$

The quadratic term is cancelled out after simplified

Which gives $u=1400$

Plug back in, $a= \sqrt{1000} , b=\sqrt{775}$

Then

$AB\cdot AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} \cdot\frac{ab}{u} } = \frac{u^2}{ab} = \frac{1400 \cdot 1400}{ \sqrt{ 1000\cdot 875 }} = 560 \sqrt{14}$

So the final answer is $560 + 14 = \boxed{574}$

By SpecialBeing2017

## See Also

 2019 AIME II (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byLast 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.

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