Let $ABC$ be an acute triangle with incenter $I$. 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 $M_A$ over the line $IM_B$ lies on the circumcircle of the triangle $IM_BM_C$.
Problem
Source: Swiss MO 2023/1
Tags: geometry, geometric transformation, reflection, circumcircle
bora_olmez
12.03.2023 15:07
Outline from Contest: $M_AM_C$ is perpendicular bisector of $BI$ and angle chase - nothing substantive. Remark: Problems are ordered by difficulty (mod 4), only $5$ of the problems were released due to confidentiality.
gghx
12.03.2023 15:26
Rephrase with $ABC$ as $M_AM_BM_C$. Rephrased qn wrote: Given triangle $ABC$ with orthocenter $H$ and orthic triangle $DEF$, show that $A'$, the reflection of $A$ about $F$, lies on $(BHC)$. Proof: Obvious since $\angle BA'C=\angle BAC=\angle FHB$.
Taha-R
04.10.2023 02:17
Lemma 1 : In triangle $ABC$ , if T is midpoint of minor arc BC of the circumcircle of the triangle $ABC$ and I is incenter of $ ABC , TI = TB = TC$. Proof : Angle bashing . Lemma 2 : $ M_AH $ always passes through point $M_C$. Proof : It's enough to prove $ M_AB = M_AI $ and by lemma1 it's obvious. Solution : It's enough to prove $ IM_BEM_C $ is cyclic. $ \angle AM_AE = \angle IEM_A = \angle M_CM_BB $ . And the proof is complete !
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