Bulgarian Mathematical Competitions

$$\sin \angle C_1AH = \frac{C_1H}{2R_1} = \frac{A_1H}{2R_2} = \sin \angle A_1BH$$
and analogously
$$\sin \angle C_1AH = \sin \angle A_1BH = \sin \angle A_1CH = \sin \angle C_1BH = \sin \angle B_1AH.$$
Let $\angle C_1AH = \angle A_1BH = \angle B_1CH = \varphi$. Since $\angle AHB = \alpha + \beta = 180^\circ - \gamma$, then $\angle AC_1B = \gamma$, and hence $\angle AH_3B = 180^\circ - \gamma$. Analogously, $\angle BA_1C = \alpha$, $\angle BH_1C = 180^\circ - \alpha$, $\angle CB_1A = \beta$ and $\angle CH_2A = 180^\circ - \beta$. It follows that the points $H_1$, $H_2$ and $H_3$ belong to the circumcircle of $\triangle ABC$.

Now we shall use now vectors. Denote by $[\vec{a}]$ the image of the vector $\vec{a}$ under rotation through $360^\circ - 2\varphi$. We get from the proved above that $\angle C_1AH_3 = 90^\circ - \gamma = \angle CAH$. Hence $\angle CAH_3 = \varphi$ and then $\angle COH_3 = 2\varphi$. Analogously, $\angle AOH_1 = \angle BOH_2 = 2\varphi$.

Let $S$ and $T$ be the orthocenters of $\triangle A_1B_1C_1$ and $\triangle H_1H_2H_3$, respectively. We shall prove that $\overrightarrow{HS} = \overrightarrow{HT}$.

One has that
$$\begin{aligned} & \overrightarrow{HS} = \overrightarrow{HA_1} + \overrightarrow{HB_1} + \overrightarrow{HC_1} = \overrightarrow{HO} + \overrightarrow{OA_1} + \overrightarrow{HO} + \overrightarrow{OB_1} + \overrightarrow{HO} + \overrightarrow{OC_1} \\ & = 3\overrightarrow{HO} + \overrightarrow{OA_1} + \overrightarrow{OB_1} + \overrightarrow{OC_1} \\ & = \overrightarrow{HO} + [\overrightarrow{OA_1} + \overrightarrow{OB_1} + \overrightarrow{OC_1} + 2\overrightarrow{HO}] \\ & = \overrightarrow{HO} + [\overrightarrow{OH_1} + \overrightarrow{OH_2} + \overrightarrow{OH_3}] \\ & = \overrightarrow{HO} + [\overrightarrow{OH}] \end{aligned}$$

Similarly,
$$\overrightarrow{HT} = \overrightarrow{HO} + \overrightarrow{OT} = \overrightarrow{HO} + \overrightarrow{OH_1} + \overrightarrow{OH_2} + \overrightarrow{OH_3} = \overrightarrow{HO} + [\overrightarrow{OH}]$$
which means that $S \equiv T$.

The case $\angle C_1AH = \angle A_1CH = \angle B_1CH = \varphi$ can be considered in a similar way and we omit the details.