42nd Balkan Mathematical Olympiad

Let $M_A$ and $M_B$ be the midpoints of arcs $\widehat{BC}$ and $\widehat{CA}$ (not containing points $A$ and $B$ respectively), and $S_C$ be the midpoint of arc $\widehat{ACB}$ of the circumcircle of $△ABC$. We will prove that lines $X_A{X}_B$ and $I_A{I}_B$ intersect on $C{S}_C$. By the Incenter/Excenter lemma, $M_{A}B=M_A{I}_A=M_{A}I=M_{A}C$, hence $CI{I}_{A}B$ is cyclic with circumcenter $M_A$. Similarly, $CI{I}_{B}A$ is cyclic with circumcenter $M_B$. Let us denote these circumcircles as ${\omega }_A$ and ${\omega }_B$ respectively. Then $$\begin{array}{rl}\angle I_{A}I{I}_B& =\angle AIB-\angle AI{I}_B-\angle BI{I}_A=\angle AIB-\angle AC{I}_B-\angle BC{I}_A\\ & =\left(90^{\circ }+\frac{1}{2}\angle ACB\right)-\frac{1}{2}\angle ACD-\frac{1}{2}\angle BCD=90^{\circ }.\end{array}$$ Therefore $X_A{I}_A$ is a diameter of ${\omega }_A$ as $M_A$ lies on line $D{I}_A$ and $\angle I_{A}I{X}_A=90^{\circ }$. Similarly, $X_B{I}_B$ is a diameter of ${\omega }_B$. Let $C{S}_C$ intersect ${\omega }_A$ and ${\omega }_B$ for a second time at $L_A\ne C$ and $L_B\ne C$ respectively. We introduce $L_A{I}_A\cap L_B{I}_B=Y$ and $L_A{X}_A\cap L_B{X}_B=Z$. By Desargues's theorem for $△X_A{I}_A{L}_A$ and $△X_B{I}_B{L}_B$ it follows that the lines $I_A{I}_B$, $X_A{X}_B$ and $L_A{L}_B$ are concurrent if and only if the points $D=X_A{I}_A\cap X_B{I}_B$, $Y=L_A{I}_A\cap L_B{I}_B$ and $Z=X_A{L}_A\cap X_B{L}_B$ are collinear. We will show this collinearity, which will finish the proof. Note that $\angle L_{A}CI=90^{\circ }=\angle L_{B}CI$, hence $M_A,M_B$ are the midpoints of $I{L}_A$ and $I{L}_B$ in ${\omega }_A$ and ${\omega }_B$ respectively. Therefore, $I{X}_A{L}_A{I}_A$ and $I{X}_B{L}_B{I}_B$ are rectangles, and so $Y{L}_{A}Z{L}_B$ is a rectangle as well. Furthermore, it is well-known that $S_C$ is the midpoint of $L_A{L}_B$. This implies that $S_C$ is the midpoint of the diagonal $YZ$ of the rectangle $Y{L}_{A}Z{L}_B$ and the condition $D\in YZ$ is therefore equivalent to $\angle D{S}_{C}C=\angle Y{S}_C{L}_A$. We have $$\angle Y{S}_C{L}_A=180^{\circ }-2\angle C{L}_A{I}_A=180^{\circ }-2\angle CB{I}_A=180^{\circ }-\angle CBD=\angle D{S}_{C}C$$ as desired, which concludes the solution.

---

Alternative solution.

In the same way as in Solution 1, we define points $M_A$ and $M_B$, and show that they are the midpoints of $I_A{X}_A$, and $I_B{X}_B$, respectively. Now, if we denote $T=I_A{I}_B\cap X_A{X}_B$, we get that line $M_A{M}_B$ is the Newton-Gauss line of the complete quadrilateral $X_A{X}_{B}T{I}_A{I}_{B}I$. Therefore, $M_A{M}_B$ bisects the segment $TI$. Let $N$ be the midpoint of $TI$, and $S_C$ be defined as in Solution 1. Then, under the homothety centered at $I$ with coefficient 2, $N$ is mapped to $T$. Additionally, this homothety maps $M_A{M}_B$ to $C{S}_C$ because point $C$ and $I$ are symmetric with respect to $M_A{M}_B$, and $M_A{M}_B\parallel C{S}_C$. As $N\in M_A{M}_B$, we get that $T\in C{S}_C$. This is a fixed line, which concludes the proof.