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 $m+n$

Solution: [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

Answer:  023