JBMO Shortlist

Geometry Difficulty 5.3 AIME, harder Prove it JBMO

Problem:

Let $ABC$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $ABC$ and distinct from $A$ , $B$ and $C$ . If the lines through $P$ and parallel to $BC$ , $CA$ , $AB$ intersect the lines $CA$ , $AB$ , $BC$ at $M$ , $N$ and $Q$ respectively, prove that $M$ , $N$ and $Q$ are collinear.

Solution

Solution:

Without any loss of generality, let $P$ be in the minor arc of the chord $AC$ as in Figure 1. Since $\angle PNA = \angle NPM = 60^{\circ}$ and $\angle NAM = \angle PMA = 120^{\circ}$ , it follows that the points $A$ , $M$ , $P$ and $N$ are concyclic. This yields
$\angle NMP = \angle NAP$

Figure 1: Exercise G1.
Similarly, since $\angle PMC = \angle MCQ = 60^{\circ}$ and $\angle CQP = 60^{\circ}$ , it follows that the points $P$ , $M$ , $Q$ and $C$ are concyclic. Thus
$\angle PMQ = 180^{\circ} - \angle PCQ = 180^{\circ} - \angle NAP \stackrel{(2)}{=} 180^{\circ} - \angle NMP .$
This implies $\angle PMQ + \angle NMP = 180^{\circ}$ , which shows that $M$ , $N$ and $Q$ belong to the same line.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.