2022 CGMO

Proof. We need the following lemma. Lemma: In triangle ABC, let I be the incenter. The incircle $\odot I$ is tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. Let $A_{1}I$ intersect $B_1{C}_1$ at $L$. Then, $AL$ passes through the midpoint $M$ of $BC$. Proof of lemma: Draw a line through $L$ parallel to $BC$, intersecting $AC$ at $X$ and $AB$ at $Y$. Connect $I{B}_1$, $I{C}_1$, $IX$, and $IY$. Since $A_{1}L\perp BC$ and $XY\parallel BC$, we know that $A_{1}L\perp XY$. Thus, $\angle ILX=\angle I{B}_{1}X=90^{\circ }$, and $I,L,X,B_1$ are concyclic. Similarly, $I,L,C_1,Y$ are concyclic. Therefore, $\angle IX{B}_1=\angle IL{B}_1=\angle IY{C}_1$. Combining this with $\angle I{B}_{1}X=\angle I{C}_{1}Y=90^{\circ }$ and $I{B}_1=I{C}_1$, we have $△I{B}_{1}X\cong △I{C}_{1}Y$. Thus, $IX=IY$. Also, note that $IL\perp XY$, so $L$ is the midpoint of $XY$. Combined with $XY\parallel BC$, we conclude that $AL$ passes through the midpoint $M$ of $BC$. The lemma is proven.

Returning to the original problem, let the incircle $\odot I$ of triangle $ABC$ be tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. By the lemma, $L$ is on segment $B_1{C}_1$. Let $S$ be the midpoint of $BI$, and let $T$ be a point on $AS$ such that $\angle TBS=\angle BAS$. We will now prove that points $B$, $T$, $L$ are collinear.

Since $\angle TBS=\angle BAS$, we know $△BST\sim △ASB$. Thus, $B{S}^2=SA\cdot ST$. Also, $BS=SI$, so $I{S}^2=SA\cdot ST$. This implies $△IST\sim △ASI$, and therefore, $$\angle BTI=\angle BTS+\angle ITS=\angle ABI+\angle AIB=180^{\circ }-\angle BAI.$$ Let $H$ be the orthocenter of triangle $AIB$. By the properties of the orthocenter, $\angle BHI=\angle BAI$. Thus, points $B$, $T$, $I$, $H$ are concyclic. Therefore, $\angle STH=\angle BTH-\angle BTS=\angle BIH-\angle ABI=90^{\circ }$, and combined with $\angle A{C}_{1}H=90^{\circ }$, we have points $A$, $C_1$, $T$, $H$ concyclic, with $AH$ being the diameter of the circle. Thus, $$\begin{array}{rl}\angle IT{C}_1& =\angle HT{C}_1-\angle HTI=180^{\circ }-\angle HA{C}_1-\angle HBI\\ & =180^{\circ }-(90^{\circ }-\angle ABI)-(90^{\circ }-\angle ABI-\angle BAI)\\ & =2\angle ABI+\angle BAI.\end{array}$$ Moreover, note that $$\angle IL{B}_1=\angle I{C}_{1}L+\angle LI{C}_1=\frac{1}{2}\angle BAC+\angle AB{A}_1=\angle BAI+2\angle ABI.$$ So $\angle IT{C}_1=\angle IL{B}_1$, so $I$, $T$, $C_1$, $L$ are concyclic. From this, we deduce $\angle ITL=\angle I{C}_{1}L=\angle BAI$. Therefore, points $B$, $T$, $L$ are collinear, and $\angle LBI=\angle TBI=\angle BAS$. Since $AS$ is the median of triangle $BIJ$, we have $AS\parallel BJ$, and $\angle BAS=\angle ABJ$. Therefore, $\angle ABJ=\angle LBI$. The conclusion is proven. $\square$