Let $I$ be the incenter of triangle $ABC$, and $M, N$ be the midpoints of arcs $ABC$ and $BAC$ of its circumcircle. Prove that points $M, I, N$ are collinear if and only if$ AC + BC = 3AB$. (A. Polyansky)
Problem
Source: 2014 Sharygin Geometry Olympiad Final Round 9.6
Tags: geometry, incenter
06.07.2019 01:56
Let $P$ be midpoint of $BC$. Lemma The condition $AC+BC=3AB$ is equivalent with $CA\cdot CB = 2\cdot CI^2$
Now, observe that $M,I,N$ are collinear if and only if $\angle MIC = \angle NIC = 90^{\circ}$, because $MN\perp CI$. Therefore, it's suffices to prove $\angle NIC=90^{\circ} \leftrightarrow CA\cdot CB = 2CI^2$.
28.06.2024 05:22
Here's a solution that probably explains where this problem comes from. We'll need a quick lemma. Lemma. Let $r$ and $r_c$ be the inradius and $C$-exradius of $\triangle ABC$. Then $AC+BC=3AB$ if and only if $r_c = 2r$. Proof. The condition $a+b=3c$ is equivalent to $2(s-c) = a+b-c = 2c$, i.e. $s-c=c$. It follows that $\tfrac{r_c}{r} = \tfrac{s}{s-c} = 2$. The reverse direction is analogous. Let $I_AI_BI_C$ be the excentral triangle of $\triangle ABC$. Recall from the Big Picture Configuration that $\odot(ABC)$ is the nine-point circle of $\odot(I_AI_BI_C)$; in particular, $M$ and $N$ are midpoints of their respective sides and $I$ lies on $\overline{CI_C}$. It follows that $M$, $I$ and $N$ are collinear if and only if $CI_C = 2CI$, which in turn holds if and only if $r_c = 2r$. By our lemma, this holds if and only if $AC+BC=3AB$ as desired.