Competition problem — Geometry — Maths Olympiad Prep

Library / /12 of 69

Geometry Difficulty 5.4 AIME, harder Prove it Mongolia

Let $M$ be the midpoint of side $AC$ of triangle $ABC$ . Let $P$ be a point inside triangle $ABC$ such that $\angle BAP = \angle BCP$ . The line $CP$ intersects side $AB$ at point $R$ . Let $Q$ be the foot of the perpendicular drawn from $B$ to line $CP$ . If $Q$ lies inside triangle $ABC$ and $QM = \frac{PC}{2}$ then prove that $RQ = QP$ .
(Khulan Tumenbayar)

Solution

Let $K$ , $N$ , and $L$ be the midpoints of segments $BP$ , $BC$ , and $PC$ , respectively. Since $AB$ is parallel to $MN$ and $AP$ is parallel to $ML$ , we have $\angle BAP = \angle NML$ .

Considering the parallelogram $KNCL$ , $\angle BCP = \angle BCL = \angle NKL$ . Moreover, since $\angle NML = \angle NKL$ , we have $K$ , $N$ , $L$ , and $M$ lie on the same circle.

Since $\angle BQP = 90^\circ$ , we can conclude that $QK = \frac{BP}{2}$ . $Q$ , $K$ , $N$ , $L$ , and $M$ all lie on the same circle, because $QK = NL$ .

Since $QM = \frac{CP}{2} = KN$ , $QK$ is parallel to $AB$ . Consequently, we can deduce that $RQ = QP$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.