67th NMO Selection Tests for JBMO

a) The angles $\angle ABC$ and $\angle A{B}_1{C}_1$ are equal, therefore so are their complementary angles, $\angle BA{A}_1$ and $\angle A_{2}AC$. It follows that the rays ($AH$ and ($A{A}_2$ are isogonal, hence $A_2\in (AO$). As $AO=CO$, we have $\angle ACO\equiv \angle OAC\equiv \angle A{A}_2{B}_1$, therefore points $O,A_2,B_1,C$ are cocyclic. (The arguments above hold in the case $A_2\in (AO)$ as well as in the case $O\in (A{A}_2)$.)

b) From the power of the point $A$ with respect to the circles through $O,A_2,B_1,C$ and $H,A_1,C,B_1$ respectively, it follows that $A{B}_1\cdot AC=AO\cdot A{A}_2$ and $A{B}_1\cdot AC=AH\cdot A{A}_1$. Since $AO\cdot A{A}_2=AH\cdot A{A}_1$, from the converse of the power of the point theorem, it follows that the points $O,A_2,H,A_1$ are cocyclic.



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Alternative solution.

Alternative Solution. For b). Consider $\{A_3\}=(A{A}_2\cap B_1{C}_1$ and let $O_1$ be the midpoint of $[AH]$. Then $O_1$ is the circumcenter of triangle $A{B}_1{C}_1$. As $[O_1{A}_3]$ is a midsegment in triangle $AH{A}_2$, we have $\angle H{A}_{2}A\equiv \angle O_1{A}_{3}A$. But the similarity of triangles $ABC$ and $A{B}_1{C}_1$ leads to the equality of the corresponding angles $\angle A{A}_{1}O$ and $\angle A{A}_3{O}_1$, which means that $\angle A{A}_{2}H\equiv \angle A{A}_{1}O$ and the conclusion.