Competition problem — Geometry — Maths Olympiad Prep

Library / /24 of 52

Geometry Difficulty 8.1 Shortlist Prove it Romania

Let $ABC$ be an isosceles triangle, $AB = AC$ , and let $M$ and $N$ be points on the sides $BC$ and $CA$ , respectively, such that the angles $BAM$ and $CNM$ are equal. The lines $AB$ and $MN$ meet at $P$ . Show that the internal angle bisectors of the angles $BAM$ and $BPM$ meet at a point on the line $BC$ .

Bogdan Enescu

Solution

Denote $I$ the intersection of the bisector of $\angle BAM$ with $BC$ and denote $D$ the reflection of $A$ about $BC$ . Then $\angle BMD = \angle BMA = \angle CMN$ , so $P, M, D$ are collinear. On the other hand, $DI$ is the bisector of $\angle BDM$ – the reflection of $\angle BAM$ – and $BI$ is the bisector of $\angle ABD$ , therefore $I$ is the incenter of triangle $PBD$ , whence the conclusion.

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 reproduced verbatim; metadata (topic, difficulty) added by this project.