Croatian Mathematical Competitions

Let the triangle $A_1B_1C_1$ be similar to the triangle $ABC$ ($\angle A = \angle A_1 = \alpha$, $\angle B = \angle B_1 = \beta$, $\angle C = \angle C_1 = \gamma$), and let the point $O$ be the orthocentre of the triangle $A_1B_1C_1$. Then $\angle OB_1C_1 = 90^\circ - \gamma$, $\angle OC_1B_1 = 90^\circ - \beta$, so $\angle B_1OC_1 = 180^\circ - (90^\circ - \gamma) - (90^\circ - \beta) = \beta + \gamma$. Since $\angle B_1AC_1 + \angle B_1OC_1 = \alpha + \beta + \gamma = 180^\circ$, the quadrilateral $AC_1OB_1$ is cyclic. Hence $\angle OAB_1 = OC_1B_1 = 90^\circ - \beta$ and $\angle OAC_1 = \angle OB_1C_1 = 90^\circ - \gamma$.

Analogously, because quadrilaterals $BA_1OC_1$ and $CB_1OA_1$ are cyclic, we get $\angle OBC_1 = 90^\circ - \gamma$, $\angle OBA_1 = 90^\circ - \alpha$ and $\angle OCA_1 = 90^\circ - \alpha$, $\angle OCB_1 = 90^\circ - \beta$, so $O$ is the circumcentre of the triangle $ABC$.

Now assume that the point $O$ is the orthocentre of the triangle $A_1B_1C_1$ and the circumcentre of the triangle $ABC$. Let $\angle A = \alpha$, $\angle B = \beta$, $\angle C = \gamma$ and $\angle A_1 = \alpha_1$, $\angle B_1 = \beta_1$, $\angle C_1 = \gamma_1$. Let the points $B'$ on $CA$ and $C'$ on $AB$ be such that the quadrilaterals $CB'OA_1$ and $BA_1OC'$ are cyclic. Then the quadrilateral $AC'OB'$ is also cyclic. Hence $\angle OC'B' = \angle OAB' = \angle OAC = 90^\circ - \beta$. Since the supplementary angle of the angle $\angle A_1OC'$ is $\beta$, the line $A_1O$ is perpendicular to $B'C'$. This implies $B'C' \parallel B_1C_1$.

Since $O$ is the orthocentre of the triangle $A_1B_1C_1$, the line $B_1C_1$ is between $A$ and $O$, and $\angle B_1OC_1 = 180^\circ - \alpha_1$.

Since $A_1OB'C$ is cyclic, $\angle A_1OB' = 180^\circ - \gamma$, and similarly $\angle A_1OC' = 180^\circ - \beta$. The sum of these two angles is $180^\circ + \alpha$ and therefore the line $B'C'$ lies between $A$ and $O$.

We may assume that the line $B'C'$ is closer to the point $A$ than the line $B_1C_1$ (the other case is similar). That implies $\angle B'OC' \leq B_1OC_1$ and $\angle B'A_1C' \leq B_1A_1C_1$. Adding these two inequalities gives
$$\angle B'OC' + \angle B'A_1C' \leq \angle B_1OC_1 + \angle B_1A_1C_1. \qquad (\ddagger)$$
Since $\angle B'A_1C' = \angle B'A_1O + \angle OA_1C' = \angle B'CO + \angle OBC' = \angle ACO + \angle OBA = 90^\circ - \beta + 90^\circ - \gamma = \alpha$, we have $\angle B'OC' + \angle B'A_1C' = 180^\circ - \alpha + \alpha = 180^\circ$.

Also, $\angle B_1OC_1 + \angle B_1A_1C_1 = 180^\circ - \alpha_1 + \alpha_1 = 180^\circ$.

Therefore in $(\ddagger)$ we actually have equality, which can be attained only if $\angle B'OC' = \angle B_1OC_1$ and $\angle B'A_1C' = \angle B_1A_1C_1$, i.e. if $\alpha_1 = \alpha$ and the lines $B'C'$ and $B_1C_1$ coincide. Then also $\beta_1 = \beta$ and $\gamma_1 = \gamma$, so the triangles $A_1B_1C_1$ and $ABC$ are similar.