Track / Stage 6 / 13 of 400 #1013 of 2000

Problem 1013

National olympiad, first round

Geometry Difficulty 6.0 Prove it

Let $\mathrm{Zij} ABC$ be an acute triangle. Let $\mathrm{Zij} H$ be the foot of the altitude from $C$ to $AB$ . Suppose that $|AH|=3|BH|$ . Let $M$ and $N$ be the midpoints of $AB$ and $AC$ , respectively. Let $P$ be a point such that $|NP|=|NC|$ and $|CP|=|CB|$ , and such that $B$ and $P$ lie on opposite sides of the line $AC$ .
Prove that $\angle APM = \angle PBA$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

There is only one configuration. Since $N$ is the midpoint of $A C$ , $|N C|=|N A|$ . It was also given that $|N P|=|N C|$ , so $N$ is the center of a circle through $A, C$ , and $P$ . From Thales' theorem, it follows that $\angle A P C=90^{\circ}$ . Since $|A H|=3|B H|$ and $M$ is the midpoint of $A B$ , it follows that $|M H|=|B H|$ . Since $\angle C H B=90^{\circ}=\angle C H M$ , triangles $C H B$ and $C H M$ are congruent, from which it follows that $|C M|=|C B|$ . It was further given that $|C P|=|C B|$ , so $C$ is the center of a circle through $P, M$ , and $B$ . Since $\angle A P C=90^{\circ}$ , $A P$ is a tangent to this circle. Using the tangent-secant angle theorem, we now get $\angle A P M=\angle P B M=\angle P B A$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.