Competition problem — Geometry — Maths Olympiad Prep

Library / /11 of 34

Geometry Difficulty 5.4 AIME, harder Prove it Mongolia

Given a triangle $ABC$ , let the bisectors of the angles $\angle BAC$ , $\angle CBA$ , $\angle ACB$ intersect the circumcircle of the triangle $ABC$ at the points $M$ , $N$ , $K$ respectively. Let $P$ be the intersection of the segments $AB$ and $MK$ and $Q$ be the intersection of the segments $AC$ and $MN$ . Prove that the lines $PQ$ and $BC$ are parallel.

(Battsengel B.)

Solution

From $\vec{AN} = \vec{NC}$ , we see that $MQ$ is a bisector of $\triangle AMC$ .

By the angle bisector theorem, we have
$\frac{AQ}{QC} = \frac{AM}{MC}. \qquad (1)$
Similarly, we have
$\frac{AP}{PB} = \frac{AM}{MB}. \qquad (2)$
Since $\vec{BM} = \vec{MC}$ , we have $MB = MC$ . Thus from (1) and (2), we get
$\frac{AQ}{QC} = \frac{AP}{PB}.$
By Thales' theorem, we have $PQ \parallel BC$ .

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.