Let $ABC$ be a non-isosceles triangle with circumcircle $k$, incenter $I$ and $C$-excenter $I_C$. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of arc $\widehat{ACB}$ on $k$. Prove that $\angle IMI_C + \angle INI_C = 180^{\circ}$.
Problem
Source: Bulgaria JBMO TST 2023, Day 3, Problem 3
Tags: geometry, incircle, excircle, similarity, circumcircle, incenter
VicKmath7
17.06.2023 21:32
Here is a solution with mixtilinear circles, probably not the intended one.
Let the incircle touch $AB$ at $D$ and the excircle touch $AB$ at $E$; let $T_1=NI \cap (ABC)$ be the $C$-mixtilinear incircle touchpoint and let $T_2=NI_c \cap (ABC)$ be the $C$-mixtilinear excircle touchpoint. Then $CT_1, CE$ are isogonal and $CT_2, CD$ are isogonal with respect to $\angle ACB$. Moreover, $IM \parallel CE$ and $I_cM \parallel CD$. Thus $\angle INI_c=\angle T_1NT_2=\angle T_1CT_2=\angle DCE=180^{o}-\angle IMI_c$.
Helixglich
17.06.2023 21:52
I hate C-centered problems . I'm so sad I read the problem in the last 30 min and didn't have time to think. We do complex numbers Let $\triangle ABC$ and $\overline{ND}$ be the unit circle and the Real-axis. $$|A|=|B|=|C| = 1$$$A = a^2$, $B = \bar{A} = \frac{1}{a^2}$, $C = c^2$ From this we get the following: $$N = -1\text{ , } I_c = -1 + ac +\frac{c}{a}\text{ , }I=-1-ac-\frac{c}{a}\text{ , }M = \frac{a^4+1}{2a^2}$$We want to show the following: $$\frac{I_c-M}{I-M}\times\frac{I_c-N}{I-N}\in \mathbb{R}$$$\Longrightarrow$ $\frac{-1+ac+\frac{c}{a}-\frac{a^4+1}{2a^2}}{+1+ac+\frac{c}{a}+\frac{a^4+1}{2a^2}}\times \frac{-2+ac+\frac{c}{a}}{+2+ac+\frac{c}{a}} = \frac{-2a^2+2a^3c+2ac-a^4-1}{+2a^2+2a^3c+2ac+a^4+1}\times \frac{-2a+a^2c+c}{2a+a^2c+c}\overset{Algebruh}{=}\frac{(a^2c-2a+c)(-a^2+2ac-1)}{(a^2c+2a+c)(a^2+2ac+1)}\in \mathbb{R}$ $\overline{\frac{(a^2c-2a+c)(-a^2+2ac-1)}{(a^2c+2a+c)(a^2+2ac+1)}}=\frac{(\frac{1}{a^2c}-\frac{2}{a}+\frac{1}{c})(-\frac{1}{a^2}+\frac{2}{ac}-1)}{(\frac{1}{a^2c}+\frac{2}{a}+\frac{1}{c})(+\frac{1}{a^2}+\frac{2}{ac}-1)}\overset{\times a^6c^4}{=}\frac{(+1-2ac+a^2)(-c+2a-a^2c))}{(1+2ac+a^2)(+c+2a+a^2c)}$ Which concludes the proof In the Algebruh step I just noticed that $a=\textit{i}$ is a zero.
Orestis_Lignos
17.06.2023 22:29
Here's a more suitable solution for a junior contest. Let $T$ be the midpoint of the minor arc $AB$, $M'$ be the reflection of $M$ across $CI$ and $T'$ be the reflection of $M$ across $T$. We have the following two Claims. Claim 1: Points $I,M',T',I_C$ are concyclic. Proof: We may note that $IMI_CT'$ is a parallelogram, and so $IM'=IM=T'I_C$. Moreover, $M'T' \parallel II_C$, as both lines are perpendicular to $MM'$. Hence, $IM'T'I_C$ is an isosceles trapezoid, which implies the desired result $\blacksquare$ Claim 2: Points $I,N,I_C,T'$ are concyclic. Proof: Note that $TT' \cdot TN=TM \cdot TN=TB^2=TI \cdot TI_C,$ as desired $\blacksquare$ To the problem, by Claims 1 and 2 points $I,N,I_C,M'$ are concyclic, and so $\angle IMI_C+\angle INI_C=\angle IM'I_C+\angle INI_C=180^\circ,$ as desired.
Jalil_Huseynov
17.06.2023 23:48
Lol. $T=CI \cap k$. As $TI^2=TM\cdot TN=TI_C^2$, we are done.
wizixez
01.01.2025 13:28
Easy problem for p3 Just note that $2m=a+b,I_c=ca+cb-ab,-I=xy+yz+zx,n=xy$