# 2015 AIME II Problems/Problem 11

## Problem

The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ and $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Diagram

$[asy] unitsize(30); draw(Circle((0,0),3)); pair A,B,C,O, Q, P, M, N; A=(2.5, -sqrt(11/4)); B=(-2.5, -sqrt(11/4)); C=(-1.96, 2.28); Q=(-1.89, 2.81); P=(1.13, -1.68); O=origin; M=foot(O,C,B); N=foot(O,A,B); draw(A--B--C--cycle); label("A",A,SE); label("B",B,SW); label("C",C,NW); label("Q",Q,NW); dot(O); label("O",O,NE); label("M",M,W); label("N",N,S); label("P",P,S); draw(B--O); draw(C--Q); draw(Q--O); draw(O--C); draw(O--A); draw(O--P); draw(O--M, dashed); draw(O--N, dashed); draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5)); draw(rightanglemark(O,N,B,5)); draw(rightanglemark(B,O,P,5)); draw(rightanglemark(O,M,C,5)); [/asy]$

### Solution 1

Call the $M$ and $N$ foot of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ and let $OQ = k$. Notice that $\triangle{OMB} \sim \triangle{OQB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. Then, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. However, since $O$ is the circumcenter of triangle $ABC$, $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$. Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle NBO$, we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$. Thus, $BP = \frac{18}{5}$. $m + n=\boxed{023}$.

### Solution 2

Notice that $\angle{CBO}=90-A$, so $\angle{BQO}=A$. From this we get that $\triangle{BPQ}\sim \triangle{BCA}$. So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$, plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$, so $BP=\dfrac{18}{5}$, and $m+n=\boxed{023}$.