Problem 1233

Solution 1. Let $M$ be the midpoint of $BC$. We denote by $\mathcal{S}$ the central symmetry with respect to the point $M$. If $A'=\mathcal{S}(A)$, then $BA'CH$ is cyclic because

$$\angle BA'C + \angle BHC = \angle BAC + \angle BHC = 180^\circ$$

since $\angle BA'C = \angle BAC$ and $\angle BHC = 180^\circ - \angle BAC$. We also observe that $\mathcal{S}(A) = A'$, $\mathcal{S}(B) = C$, and $\mathcal{S}(C) = B$, which implies that $\mathcal{S}$ maps the circumcircle of triangle $ABC$ to the circumcircle of $BHC$. Specifically, $\mathcal{S}$ maps the center $O$ of the circumcircle of $ABC$ to the center $S$ of the circumcircle of $BHC$.

Furthermore, from $OM \perp BC$ it follows that $OS \perp BC$ and $OM = MS$. It is known that $AH = 2 \cdot OM$, which can be proven by noting that $AO$ and $HM$ intersect the circumcircle of $ABC$ at point $H'$ and that $OM$ is the midline of triangle $AHH'$.

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Now we can say that $AH = 2 \cdot OM = OS$. Moreover, $AH \perp BC$ and $OS \perp BC$, which means that $AH$ is parallel to $OS$. This implies that $AHSO$ is a parallelogram, from which we conclude that $AS$ passes through the midpoint $E$ of $OH$. We have that $AH = AT_a$ by the definition of $T_a$ and $SH = ST_a$ because $S$ is the center of the circumcircle of $BHC$. We conclude that $AS$ is the perpendicular bisector of $HT_a$, but $E$ lies on the line $AS$, which means that $ET_a = EH$.

Similarly, we obtain that $ET_b = ET_c = EH$. This means that $ET_a = ET_b = ET_c = EH$, from which it follows that $HT_aT_bT_c$ is cyclic with center at $E$.