Question

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 $m + n.$

答案

Solution:

The central angle over $\overline{BC}$ is $\angle BOC = 2\angle A,$ and $OB = OC$ makes triangle $OBC$ isosceles, so $\angle OBC = 90^\circ - \angle A.$ In triangle $OBQ$ the angle at $O$ is $90^\circ,$ hence $\angle BQP = 90^\circ - \angle OBQ = 90^\circ - (90^\circ - \angle A) = \angle A.$

Triangles $BQP$ and $BAC$ share the angle at $B$ and have $\angle BQP = \angle BAC,$ so they are similar, giving $\frac{BP}{BC} = \frac{BQ}{BA}.$ Therefore $BP = \frac{BQ \cdot BC}{BA} = \frac{4.5 \cdot 4}{5} = \frac{18}{5},$ and $m + n = 18 + 5 = 23.$

2015 AIME II Problem 11

Solution:

Problem 11 in Other Years