Track / Stage 8 / 115 of 180 #1815 of 1964

Problem 1815

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.4 Prove it SELECTION TESTS FOR THE 2019 BMO AND IMO · Romania · 2019

Let $ABC$ be an acute triangle such that $AB < AC$ . Let $I$ be the incentre of the triangle $ABC$ , and let the incircle touch the side $BC$ at $D$ . The line $AD$ crosses the circle $ABC$ again at $E$ . Let $M$ be the midpoint of the side $BC$ , and let $N$ be the midpoint of the circular arc $BAC$ . The line $EN$ crosses the circular arc $BIC$ at $P$ . Show that the lines $AD$ and $MP$ are parallel.

Ukraine National Olympiad, 2016

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

The internal bisectrix $AI$ and the perpendicular bisectrix $MN$ of the side $BC$ cross at the midpoint $K$ of the arc $BEC$ . It is a fact that the circle $BIC$ is centred at $K$ .

Let now the line $MP$ cross the circle $BIC$ again at $Q$ , to infer that the arcs $BL$ and $CQ$ of this circle have equal angular spans, so $L$ and $Q$ are reflexions of one another in the perpendicular bisectrix $KMN$ of the chord $BC$ .
Project $Q$ orthogonally to $Q'$ on $BC$ and refer to standard notation in the triangle $ABC$ : $a$ , $b$ , $c$ denote the lengths of the sides $BC$ , $CA$ , $AB$ , respectively, $s = (a + b + c)/2$ denotes its semiperimeter, $r$ its inradius, and $S$ its area. With reference to standard formulae, write $CQ' = BD = s - b$ and
$QQ' = DL = \frac{DB \cdot DC}{DI} = \frac{(s-b)(s-c)}{r} = \frac{s}{s-a},$
to infer that $Q$ is the $A$ -excentre of the triangle $ABC$ , so it lies on the line $AIK$ .
Finally, write $\angle PQA = \angle PQI = \angle PLI = \angle ELI = \angle EAI = \angle DAQ$ , to conclude that the lines $AD$ and $MP$ are indeed parallel.

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