T3MO 2017

Let $D,E$, and $F$ be the feet of the altitudes from $A,B$, and $C$, respectively with respect to $△ABC$. Let $B_1{B}_2$ intersect $C_1{C}_2$ at $K$, $C_1{C}_2$ intersects $A_1{A}_2$ at $L$, and $A_1{A}_2$ intersects $B_1{B}_2$ at $M$. Let $O$ be the circumcenter of $△ABC$. ---

We have $$\begin{array}{rl}\angle (KL,DE)& =\angle (KL,BC)+\angle (BC,DE)\\ & =\angle (C_1{C}_2,C_{2}B)+\angle (BD,DE)\\ & =\angle (C_{1}A,AB)+\angle (BA,AE)\\ & =\angle (CA,AB)+\angle (AB,CA)\\ & =0\end{array}$$ Thus, $KL\parallel DE$, so $LM\parallel EF$ and $MK\parallel FD$ analogously. Therefore, there is a homothety sending $△DEF$ to $△KLM$. Call its center $X$. Consider $$\begin{array}{rl}\angle (FD,DH)& =\angle (FB,BH)\\ & =\angle (FB,BE)\\ & =\angle (FC,CE)\\ & =\angle (HC,CE)\\ & =\angle (HD,DE)\end{array}$$ Thus, $HD$ is an angle bisector of $\angle FDE$. Similarly, we can show that $HE$ and $HF$ are angle bisectors of $\angle DEF$ and $\angle EFD$, respectively. Hence, $H$ is either the incenter or an excenter of $△DEF$. Since $△ABC$ is acute, we observe that $H$ is

inside $△DEF$, and thus $H$ must be the incenter of $△DEF$. We have $$\begin{array}{rl}\angle (H{C}_2,C_{2}C)& =\angle (H{C}_2,C_{2}B)\\ & =\angle (HA,AB)\\ & =\angle (DA,AF)\\ & =\angle (DC,CF)\\ & =\angle (C_{2}C,CH)\end{array}$$ Thus, $\overrightarrow{C_{2}D}=\overrightarrow{DC}$, so $\overrightarrow{B_{1}D}=\overrightarrow{DB}$, $\overrightarrow{A_{2}E}=\overrightarrow{EA}$, $\overrightarrow{C_{1}E}=\overrightarrow{EC}$, $\overrightarrow{B_{2}F}=\overrightarrow{FB}$, and $\overrightarrow{A_{1}F}=\overrightarrow{FA}$ analogously (where $\overrightarrow{XY}$ denotes the vector $XY$). We have $$\begin{array}{rl}\angle (K{C}_2,C_2{B}_1)& =\angle (ED,DB)\\ & =\angle (EA,AB)\\ & =\angle (CA,AF)\\ & =\angle (CD,DF)\\ & =\angle (C_2{B}_1,B_{1}K)\end{array}$$ Thus, the internal angle bisector of $\angle MKL$ is the perpendicular bisector of $B_1{C}_2$. Since $B_1{C}_2$ is the image of $BC$ reflected about $HD$ and the perpendicular bisector of $BC$ passes through $O$, the perpendicular bisector of $B_1{C}_2$ passes through the reflection of $O$ about $H$, which we denote $O^{\prime }$. Similarly, the internal angle bisectors of $\angle KLM$ and $\angle LMK$ also pass through $O^{\prime }$. Therefore, $O^{\prime }$ is the incenter of $△KLM$. The incenters of $△DEF$ and $△KLM$ both are on $\ell$, the Euler line of $△ABC$, so $X$ lies on $\ell$. Since the circumcenter of $△DEF$ (the nine point center of $△ABC$) lies on $\ell$, and $X$ lies on $\ell$, the circumcenter of $△KLM$ also lies on $\ell$, as required.