Taiwan IMO Selection Camp

We first prove two lemmas.

Lemma 1. Let $P, Q$ be a pair of isogonal conjugate points of $\triangle ABC$. If $O, O_a, O_b, O_c, T$ are respectively the circumcenters of $\triangle ABC, \triangle BPC, \triangle CPA, \triangle APB$ and $\triangle O_aO_bO_c$, then $PQ$ is parallel to $OT$.

Proof of Lemma 1. Let $\triangle Q_aQ_bQ_c$ be the pedal triangle of $Q$ with respect to $\triangle ABC$, and let $V$ be the circumcenter of $\triangle Q_aQ_bQ_c$. Since $Q_bQ_c, Q_cQ_a, Q_aQ_b$ are respectively perpendicular to $AP, BP, CP$, the configuration $\triangle Q_aQ_bQ_c \cup Q \cup V$ is homothetic to $\triangle O_aO_bO_c \cup O \cup T$. Since $V$ is the midpoint of $PQ$, it follows that $OT$ is parallel to $PQ$. This completes the proof of Lemma 1.

Lemma 2. Given $\triangle ABC$ and a point $U$ such that $\triangle BUC$ is equilateral. Let $V$ be the isogonal conjugate of $U$ with respect to $\triangle ABC$. Then $UV$ is parallel to the Euler line of $\triangle ABC$.

Proof of Lemma 2. Let $BV, CV$ meet the circumcircle of $\triangle ABC$ at points $E, F$ respectively, and let $N$ be the midpoint of $EF$. Let $G, O$ be respectively the centroid and circumcenter of $\triangle ABC$, and let $M$ be the midpoint of $BC$. By an angle computation, $\triangle AEF$ is equilateral, so $A, O, N$ are collinear, and
$$\frac{ON}{AO} = \frac{1}{2} = \frac{GM}{AG},$$
hence $MN$ is parallel to the Euler line $GO$ of $\triangle ABC$.

On the other hand, by an angle computation we have $\angle^*VBU = \angle^*FAC$, $\angle^*VCU = \angle^*EAB$ (here $\angle^*$ denotes a directed angle), so by using the Law of Sines to compute directed areas, we get
$$\begin{align*} [\triangle UMV] - [\triangle UNV] \\ &= \frac{1}{2}([\triangle UBV] + [\triangle UCV]) - \frac{1}{2}([\triangle UEV] + [\triangle UFV]) \\ &= \frac{1}{2}([\triangle UBE] + [\triangle UCF]) = 0. \end{align*}$$
This means that $MN$ is parallel to $UV$, and hence $UV$ is also parallel to the Euler line of $\triangle ABC$. This completes the proof of Lemma 2.

Returning to the original problem. Let $O$ be the circumcenter of $\triangle ABC$, and let $V$ be the reflection of $O$ across $BC$.
Since the midpoint of $A'V$ is the nine-point center of $\triangle A'BC$ and $N$ is the midpoint of $AV$, $AN$ is parallel to the Euler line of $\triangle A'BC$. Hence it suffices to further prove that $AT$ is parallel to the Euler line of $\triangle A'BC$.

Let $X, Y, Z, K$ be respectively the circumcenters of $\triangle BCD, \triangle A'BD, \triangle A'CD$ and $\triangle XYZ$.
Since $XY, XZ$ are respectively perpendicular to $BD, CD$, we have $\angle^*YOZ = 120^\circ$. Hence $\triangle KYZ$ is an isosceles triangle with base $YZ$ and base angle $30^\circ$. Also note that $YZ, ZO, OY$ are respectively perpendicular to $A'D, A'C, A'B$, so $\triangle AEF$ is homothetic to $\triangle OZY$. Hence $\triangle AEF \cup T$ is homothetic to $\triangle OZY \cup K$, from which it follows that $AT$ is parallel to $OK$.

On the other hand, let $D'$ be the isogonal conjugate of $D$ with respect to $\triangle A'BC$. By Lemma 1, $DD'$ is parallel to $OK$. Also, by Lemma 2, $DD'$ is parallel to the Euler line of $\triangle A'BC$. Combining the above, we obtain that $AT$ is parallel to the Euler line of $\triangle A'BC$. This completes the proof.