, 2017

Let $D, E$, and $F$ be the feet of the altitudes from $A, B$, and $C$, respectively with respect to $\triangle ABC$. Let $B_1B_2$ intersect $C_1C_2$ at $K$, $C_1C_2$ intersects $A_1A_2$ at $L$, and $A_1A_2$ intersects $B_1B_2$ at $M$. Let $O$ be the circumcenter of $\triangle ABC$.
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We have
$$\begin{align*} \angle(KL, DE) &= \angle(KL, BC) + \angle(BC, DE) \\ &= \angle(C_1C_2, C_2B) + \angle(BD, DE) \\ &= \angle(C_1A, AB) + \angle(BA, AE) \\ &= \angle(CA, AB) + \angle(AB, CA) \\ &= 0 \end{align*}$$
Thus, $KL \parallel DE$, so $LM \parallel EF$ and $MK \parallel FD$ analogously.
Therefore, there is a homothety sending $\triangle DEF$ to $\triangle KLM$. Call its center $X$.
Consider
$$\begin{align*} \angle(FD, DH) &= \angle(FB, BH) \\ &= \angle(FB, BE) \\ &= \angle(FC, CE) \\ &= \angle(HC, CE) \\ &= \angle(HD, DE) \end{align*}$$
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 $\triangle DEF$. Since $\triangle ABC$ is acute, we observe that $H$ is

inside $\triangle DEF$, and thus $H$ must be the incenter of $\triangle DEF$. We have
$$\begin{align*} \angle(HC_2, C_2C) &= \angle(HC_2, C_2B) \\ &= \angle(HA, AB) \\ &= \angle(DA, AF) \\ &= \angle(DC, CF) \\ &= \angle(C_2C, CH) \end{align*}$$
Thus, $\overrightarrow{C_2D} = \overrightarrow{DC}$, so $\overrightarrow{B_1D} = \overrightarrow{DB}$, $\overrightarrow{A_2E} = \overrightarrow{EA}$, $\overrightarrow{C_1E} = \overrightarrow{EC}$, $\overrightarrow{B_2F} = \overrightarrow{FB}$, and $\overrightarrow{A_1F} = \overrightarrow{FA}$ analogously (where $\overrightarrow{XY}$ denotes the vector $XY$).
We have
$$\begin{align*} \angle(KC_2, C_2B_1) &= \angle(ED, DB) \\ &= \angle(EA, AB) \\ &= \angle(CA, AF) \\ &= \angle(CD, DF) \\ &= \angle(C_2B_1, B_1K) \end{align*}$$
Thus, the internal angle bisector of $\angle MKL$ is the perpendicular bisector of $B_1C_2$. Since $B_1C_2$ is the image of $BC$ reflected about $HD$ and the perpendicular bisector of $BC$ passes through $O$, the perpendicular bisector of $B_1C_2$ passes through the reflection of $O$ about $H$, which we denote $O'$. Similarly, the internal angle bisectors of $\angle KLM$ and $\angle LMK$ also pass through $O'$. Therefore, $O'$ is the incenter of $\triangle KLM$.
The incenters of $\triangle DEF$ and $\triangle KLM$ both are on $\ell$, the Euler line of $\triangle ABC$, so $X$ lies on $\ell$. Since the circumcenter of $\triangle DEF$ (the nine point center of $\triangle ABC$) lies on $\ell$, and $X$ lies on $\ell$, the circumcenter of $\triangle KLM$ also lies on $\ell$, as required.