Team Selection Test for IMO 2007

Let $H$ be the orthocenter of the triangle $A_1{B}_1{C}_1$. By the similarity of the triangles $ABC$ and $A_1{B}_1{C}_1$, we have $\angle A_1{B}_1{C}_1=\angle ABC$. From this, $\angle A_{1}H{C}_1=180^{\circ }-\angle A_1{B}_1{C}_1=180^{\circ }-\angle ABC=\angle A_{1}B{C}_1$ is obtained. Therefore the points $A_1$, $B$, $H$, $C_1$ are concyclic. It follows that $\angle HBA=\angle H{A}_1{C}_1$. Similarly, $\angle HAC=\angle H{C}_1{B}_1$.



Also since $A_1$, $B$, $H$, $C_1$ are concyclic, the point $H$ does not lie in the triangle $A_1{C}_{1}B$. Similarly, $H$ does not lie in the triangles $B_1{A}_{1}C$ and $C_1{B}_{1}A$. So $H$ lies inside the triangle $ABC$.

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Then, $\angle HAB=\angle CAB-\angle HAC=\angle C_1{A}_1{B}_1-\angle H{C}_1{B}_1=\angle C_1{A}_1{B}_1-\angle H{A}_1{B}_1=\angle H{A}_1{C}_1=\angle HBA$, and consequently, $HA=HB$. Similarly, $HA=HC$. Therefore the point $H$ is the circumcenter of the triangle $ABC$.