Let $I$ be the incenter, $A_1$ and $B_1$ midpoints of sides $BC$ and $AC$ of a triangle $\Delta ABC$. Denote by $M$ and $N$ the midpoints of the arcs $AC$ and $BC$ of circumcircle of $\Delta ABC$ which do contain the other vertex of the triangle. If points $M$, $I$ and $N$ are collinear prove that: \begin{align*} \angle AIB_1=\angle BIA_1=90^{\circ} \end{align*}
Problem
Source: Serbia JBMO TST 2022 P2
Tags: geometry, incenter, circumcircle, midpoints, mixtilinear incircle
sarjinius
01.06.2022 16:35
Let $CI \cap (ABC) = P \ne C$ and $MA \cap NB = I_C$. Note that $P$ is the midpoint of arc $AB$ not containing $C$ and $I_C$ is the $C$-excenter of $\triangle ABC$, which lies on line $CI$. By the Incenter-Excenter Lemma, points $A, I, B, I_C$ lie on a circle centered at point $P$. We see that $\angle NI_CM + \angle I_CMP = \angle BI_CA + \angle AMP = (90^{\circ} - \angle ABP) + \angle ABP = 90^{\circ}$, so $MP \perp I_CN$. Similarly, $\angle MI_CN + \angle I_CNP = \angle AI_CB + \angle BNP = (90^{\circ} - \angle BAP) + \angle BAP = 90^{\circ}$, so $NP \perp I_CM$. So $P$ is the orthocenter of $\triangle MNI_C$. If $I$ lies on $MN$, then $I = MN \cap I_CP$, so $\angle MII_C = \angle NII_C = 90^{\circ}$. Since $PI_C = PB$ and $MP \perp I_CB$, $M$ lies on the perpendicular bisector of $I_CB$, so $MI_C = MB = MC$. Together with $\angle MII_C = \angle MIC = 90^{\circ}$, we get that $I$ is the midpoint of $CI_C$. So $B_1I \parallel AI_C$ and $A_1I \parallel MI_C$. Thus, $\angle AIB_1 = 180^{\circ} - \angle B_1IC - \angle AII_C = 180^{\circ} - \angle AI_CI - \angle AII_C = \angle IAI_C = 90^{\circ}$ and $\angle BIA_1 = 180^{\circ} - \angle A_1IC - \angle BII_C = 180^{\circ} - \angle BI_CI - \angle BII_C = \angle IBI_C = 90^{\circ}$.
rafigamath
06.06.2023 11:14
We have to prove $I$ is the $C$-excenter of $\triangle CB_1A_1$
Lemma: $NB$ touches $(BIC)$.The proof is easy. Since $N$ is the midpoint of arc $BC$,$NB=NC$.Let $\angle{ACI}=\alpha$ and $\angle{NCA}=\beta$. Then $\angle{NBC}=2\alpha+ \beta$.Since $\angle{NCA}=\angle{NBA}$,and$\angle{NBC}=2\alpha+\beta$ and $I$ incenter, $\angle{NBI}=\alpha$ and we're done!Also note that $NI$ is the $I$ symmedian of $\triangle BIC$.
Let $\angle{NBI}=\alpha,\angle{IAC}=\beta$. Since $MA=MC$,$\angle{MCA}=\alpha+\beta \implies \angle{BCM}=\beta-\alpha \to \angle{BNM}=\beta-\alpha.$ And $\angle{BIM}$ is the external angle so $\angle{BIM}=\beta=\angle{A_1IC}$ because $IM \to$ symmedian. So we get that $IA_1$ is the angle bisector of $B_1A_1B$ and done!
Assassino9931
21.01.2024 01:29
Official solution at https://pregatirematematicaolimpiadejuniori.files.wordpress.com/2022/06/2022-sol.pdf is nice - firstly angle chase to show $CI \perp MN$ regardless of the collinearity of $M$, $I$, $N$ and then use the right-angled cyclic quadrilaterals which arise.