# 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 $BC$ at $P$ 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

### Solution 1

Call $M$ and $N$ the feet of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{QOB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. By $\frac{MB}{BO}=\frac{BO}{BQ}$, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. 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 $\triangle BOP \sim \triangle BNO$, we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$. Thus, $BP = \frac{18}{5}$. $m + n=\boxed{023}$.

### Solution 2 (fastest)

Minor arc $BC = 2A$ so $\angle{BOC}=2A$. Since $\triangle{BOC}$ is isosceles ($BO$ and $OC$ are radii), $\angle{CBO}=(180-2A)/2=90-A$. $\angle{CBO}=90-A$, 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 $m+n=\boxed{023}$.

### Solution 3

Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, $$OM=\sqrt{r^2-4}$$ Thus, $$OQ=\frac{\sqrt{4r^2+9}}{2}$$ Using $\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\triangle{BON}$ ~ $\triangle{OPN}$, ${NP=11/10}$. Therefore, $$BP=\frac{5}{2}+\frac{11}{10}=18/5$$ $18+5=\boxed{023}$.

### Solution 4

Let $\angle{BQO}=\alpha$. Extend $OB$ to touch the circumcircle at a point $K$. Then, note that $\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$. But since $BK$ is a diameter, $\angle{KAB}=90^\circ$, implying $\angle{CAB}=\alpha$. It follows that $APCQ$ is a cyclic quadrilateral.

Let $BP=x$. By Power of a Point, $$5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.$$The answer is $18+5=\boxed{023}$.

### Solution 5

$\textit{Note: This is not a very good solution, but it is relatively natural and requires next to no thinking.}$

Denote the circumradius of $ABC$ to be $R$, the circumcircle of $ABC$ to be $O$, and the shortest distance from $Q$ to circle $O$ to be $x$.

Using Power of a Point on $Q$ relative to circle $O$, we get that $x(x+2r) = 0.5 \cdot 4.5 = \frac{9}{4}$. Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \frac{81}{4}$. Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$. Now, set $\cos{ABC} = y$. Using law of cosines on $ABC$ to find $AC$ in terms of $y$ and plugging that into the extended law of sines, we get $\frac{\sqrt{4^2 + 5^2 - 2 \cdot 4 \cdot 5 x}}{\sqrt{1 - x^2}} = 2R = 6$. Squaring both sides and cross multiplying, we get $36x^2 - 40x + 5 = 0$. Now, we get $x = \frac{10 \pm \sqrt{55}}{18}$ using quadratic formula. If you drew a decent diagram, $B$ is acute and therefore $x = \frac{10 + \sqrt{55}}{18}$(You can also try plugging in both in the end and seeing which gives a rational solution). Note that $BP = 3\frac{1}{\sin{OPB}} = \frac{3}{\cos{\angle ABC - \angle QBO}}.$ Using the cosine addition formula and then plugging in what we know about $QBO$, we get that $BP = \frac{162}{2\cos{B} + \sqrt{5}\sin{B}}$. Now, the hard part is to find what $\sin{B}$ is. We therefore want $\frac{\sqrt{324 - (10 + \sqrt{55})^2}}{18} = \frac{\sqrt{169 - 20\sqrt{55}}}{18}$. For the numerator, by inspection $(a + b\sqrt{55})^2$ will not work for integers $a$ and $b$. The other case is if there is $(a\sqrt{5} + b\sqrt{11})^2$. By inspection, $5\sqrt{5} - 2\sqrt{11}$ works. Therefore, plugging all this in yields the answer, $\frac{18}{5} \rightarrow \boxed{23}$. Solution by hyxue

### Solution 6

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.7673964645097335, xmax = 9.475267639476614, ymin = -1.6884766592324019, ymax = 6.385449160754665; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.7129306199257198,2.4781596958650733), 3.000319171815248), linewidth(2) + wrwrwr); draw((0.7129306199257198,2.4781596958650733)--(3.178984115621537,0.7692140299269852), linewidth(2) + wrwrwr); draw((xmin, 1.4430262733614363*xmin + 1.4493820802284032)--(xmax, 1.4430262733614363*xmax + 1.4493820802284032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -0.020161290322580634*xmin + 0.8333064516129032)--(xmax, -0.020161290322580634*xmax + 0.8333064516129032), linewidth(2) + wrwrwr); /* line */ draw((xmin, -8.047527437688247*xmin + 26.352175924366414)--(xmax, -8.047527437688247*xmax + 26.352175924366414), linewidth(2) + wrwrwr); /* line */ draw((xmin, -2.5113572383524088*xmin + 8.752778799300463)--(xmax, -2.5113572383524088*xmax + 8.752778799300463), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.12426176956126818*xmin + 2.389569675458691)--(xmax, 0.12426176956126818*xmax + 2.389569675458691), linewidth(2) + wrwrwr); /* line */ draw(circle((1.9173376033752174,4.895608471162773), 0.7842529827808445), linewidth(2) + wrwrwr); /* dots and labels */ dot((-1.82,0.87),dotstyle); label("A", (-1.7801363959463627,0.965838014692327), NE * labelscalefactor); dot((3.178984115621537,0.7692140299269852),dotstyle); label("B", (3.2140445236332655,0.8641046996638531), NE * labelscalefactor); dot((2.6857306099246263,4.738685150758791),dotstyle); label("C", (2.7238749148597092,4.831703985774336), NE * labelscalefactor); dot((0.7129306199257198,2.4781596958650733),linewidth(4pt) + dotstyle); label("O", (0.7539479965810783,2.556577122410283), NE * labelscalefactor); dot((-0.42105034508654754,0.8417953698606159),linewidth(4pt) + dotstyle); label("P", (-0.38361543510094825,0.9195955987702934), NE * labelscalefactor); dot((2.6239558409689123,5.235819298886746),linewidth(4pt) + dotstyle); label("Q", (2.6591355325688624,5.312625111363486), NE * labelscalefactor); dot((1.3292769824200672,5.414489427724579),linewidth(4pt) + dotstyle); label("A'", (1.3643478867519216,5.488346291867214), NE * labelscalefactor); dot((1.8469115849379867,4.11452402186953),linewidth(4pt) + dotstyle); label("P'", (1.8822629450786978,4.184310162865866), NE * labelscalefactor); dot((2.5624172335003985,5.731052930966743),linewidth(4pt) + dotstyle); label("D", (2.603644633462422,5.802794720137042), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Reflect $A$, $P$ across $OB$ to points $A'$ and $P'$, respectively with $A'$ on the circle and $P, O, P'$ collinear. Now, $\angle A'CQ = 180^{\circ} - \angle A'CB = \angle A'AB = \angle P'PB$ by parallel lines. From here, $\angle P'PB = \angle PP'B = \angle A'P'Q$ as $P, P', Q$ collinear. From here, $A'P'QC$ is cyclic, and by power of a point we obtain $\frac{18}{5} \implies \boxed{023}$. ~awang11's sol