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 relatively prime integers. 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 $AP*BP=XP*YP , AQ*CQ=YQ*XQ$ 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

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

So the final answer is 560 + 14 = 574

By SpecialBeing2017

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2019 AIME II Problems/Problem 15

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