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Let the triangle $A_1{B}_1{C}_1$ 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 $A_1{B}_1{C}_1$. Then $\angle O{B}_1{C}_1=90^{\circ }-\gamma$, $\angle O{C}_1{B}_1=90^{\circ }-\beta$, so $\angle B_{1}O{C}_1=180^{\circ }-(90^{\circ }-\gamma )-(90^{\circ }-\beta )=\beta +\gamma$. Since $\angle B_{1}A{C}_1+\angle B_{1}O{C}_1=\alpha +\beta +\gamma =180^{\circ }$, the quadrilateral $A{C}_{1}O{B}_1$ is cyclic. Hence $\angle OA{B}_1=O{C}_1{B}_1=90^{\circ }-\beta$ and $\angle OA{C}_1=\angle O{B}_1{C}_1=90^{\circ }-\gamma$.

Analogously, because quadrilaterals $B{A}_{1}O{C}_1$ and $C{B}_{1}O{A}_1$ are cyclic, we get $\angle OB{C}_1=90^{\circ }-\gamma$, $\angle OB{A}_1=90^{\circ }-\alpha$ and $\angle OC{A}_1=90^{\circ }-\alpha$, $\angle OC{B}_1=90^{\circ }-\beta$, so $O$ is the circumcentre of the triangle $ABC$.



Now assume that the point $O$ is the orthocentre of the triangle $A_1{B}_1{C}_1$ 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 $B^{\prime }$ on $CA$ and $C^{\prime }$ on $AB$ be such that the quadrilaterals $C{B}^{\prime }O{A}_1$ and $B{A}_{1}O{C}^{\prime }$ are cyclic. Then the quadrilateral $A{C}^{\prime }O{B}^{\prime }$ is also cyclic. Hence $\angle O{C}^{\prime }{B}^{\prime }=\angle OA{B}^{\prime }=\angle OAC=90^{\circ }-\beta$. Since the supplementary angle of the angle $\angle A_{1}O{C}^{\prime }$ is $\beta$, the line $A_{1}O$ is perpendicular to $B^{\prime }{C}^{\prime }$. This implies $B^{\prime }{C}^{\prime }\parallel B_1{C}_1$.

Since $O$ is the orthocentre of the triangle $A_1{B}_1{C}_1$, the line $B_1{C}_1$ is between $A$ and $O$, and $\angle B_{1}O{C}_1=180^{\circ }-{\alpha }_1$.

Since $A_{1}O{B}^{\prime }C$ is cyclic, $\angle A_{1}O{B}^{\prime }=180^{\circ }-\gamma$, and similarly $\angle A_{1}O{C}^{\prime }=180^{\circ }-\beta$. The sum of these two angles is $180^{\circ }+\alpha$ and therefore the line $B^{\prime }{C}^{\prime }$ lies between $A$ and $O$.

We may assume that the line $B^{\prime }{C}^{\prime }$ is closer to the point $A$ than the line $B_1{C}_1$ (the other case is similar). That implies $\angle B^{\prime }O{C}^{\prime }\le B_{1}O{C}_1$ and $\angle B^{\prime }{A}_1{C}^{\prime }\le B_1{A}_1{C}_1$. Adding these two inequalities gives $$\angle B^{\prime }O{C}^{\prime }+\angle B^{\prime }{A}_1{C}^{\prime }\le \angle B_{1}O{C}_1+\angle B_1{A}_1{C}_1.(‡)$$ Since $\angle B^{\prime }{A}_1{C}^{\prime }=\angle B^{\prime }{A}_{1}O+\angle O{A}_1{C}^{\prime }=\angle B^{\prime }CO+\angle OB{C}^{\prime }=\angle ACO+\angle OBA=90^{\circ }-\beta +90^{\circ }-\gamma =\alpha$, we have $\angle B^{\prime }O{C}^{\prime }+\angle B^{\prime }{A}_1{C}^{\prime }=180^{\circ }-\alpha +\alpha =180^{\circ }$.

Also, $\angle B_{1}O{C}_1+\angle B_1{A}_1{C}_1=180^{\circ }-{\alpha }_1+{\alpha }_1=180^{\circ }$.

Therefore in $(‡)$ we actually have equality, which can be attained only if $\angle B^{\prime }O{C}^{\prime }=\angle B_{1}O{C}_1$ and $\angle B^{\prime }{A}_1{C}^{\prime }=\angle B_1{A}_1{C}_1$, i.e. if ${\alpha }_1=\alpha$ and the lines $B^{\prime }{C}^{\prime }$ and $B_1{C}_1$ coincide. Then also ${\beta }_1=\beta$ and ${\gamma }_1=\gamma$, so the triangles $A_1{B}_1{C}_1$ and $ABC$ are similar.