# Difference between revisions of "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 $m+n$.

## Solution

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

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, $\frac{r^2}{4.5} = 2 \implies r = 3$. We can use the Pythagorean theorem to find $ON$, so we have $$NB^2 + ON^2 = OB^2 \implies \left(\frac{5}{2}\right)^2 + ON^2 = 3^2 \implies ON = \sqrt{\frac{11}{4}}.$$ Likewise, $\triangle{PBO} \sim \triangle{PNO}$ because both are right triangles, and $\angle{BPO} \cong \angle{NPO}$. Hence, since $\triangle{BNO} \sim \triangle{BPO}$ as well, we have that $\triangle{BNO} \sim \triangle{PNO}$. It follows that $NP = \sqrt{\frac{11}{4}}\left(\frac{\sqrt{{\frac{11}{4}}}}{\frac{5}{2}}\right) = \frac{11}{10}$. We add this to $BN$ to get $BP$, so $BP = \frac{5}{2} + \frac{11}{10} = \frac{36}{10} = \frac{18}{5}$. Our answer is $18 + 5 = \boxed{023}$.

 2015 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions