Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 4.8 AIME Prove it Switzerland

Problem:

Let $ABC$ be an acute triangle with incentre $I$ . On its circumcircle, let $M_{A}$ , $M_{B}$ and $M_{C}$ be the midpoints of minor arcs $BC$ , $CA$ and $AB$ respectively. Prove that the reflection of $M_{A}$ over the line $IM_{B}$ lies on the circumcircle of the triangle $IM_{B}M_{C}$ .

Solution

Solution:

Let $X$ be the reflection of $M_{A}$ over the line $IM_{B}$ . We wish to prove that $X$ , $M_{C}$ , $I$ , $M_{B}$ lie on a circle. By WUM, observe that $A$ , $I$ , $M_{A}$ , $B$ , $I$ , $M_{B}$ and $C$ , $I$ , $M_{C}$ are collinear. Therefore, by symmetry, we get
$\angle IXM_{B} = \angle IM_{A}M_{B} = \angle AM_{A}M_{B} = \angle ABM_{B}$
On the other hand, note that
$\angle IM_{C}M_{B} = \angle CM_{C}M_{B} = \angle CBM_{B} = \angle ABM_{B}$
Thus, we get $\angle IXM_{B} = \angle ABM_{B} = \angle IM_{C}M_{B}$ , so by the converse of the inscribed angle theorem, $X$ , $M_{C}$ , $I$ , $M_{B}$ lie on a circle, as required.

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.