International Mathematical Olympiad

Proof I Let $B_0$, $C_0$ be the midpoints of $CA$, $AB$ respectively. Denote $A^{\prime }$ as the other intersection of the circle centered at $B_0$ which passes through $H$ and the circle centered at $C_0$ which passes through $H$. We know that $A^{\prime }H\perp C_0{B}_0$. Since $B_0$, $C_0$ are the midpoints of $CA$, $AB$ respectively, $B_0{C}_0\parallel BC$. Therefore $A^{\prime }H\perp BC$. This yields that $A^{\prime }$ lies on the segment $AH$.

By the Secant-Secant theorem, it follows that $$A{C}_1\cdot A{C}_2=A{A}^{\prime }\cdot AH=A{B}_1\cdot A{B}_2,$$ So $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.

Let the intersection of the perpendicular bisectors of $B_1{B}_2$, $C_1{C}_2$ be $O$. Then $O$ is the circumcenter of quadrilateral $B_1{B}_2{C}_1{C}_2$, as well as the circumcenter of $△ABC$. So $$O{B}_1=O{B}_2=O{C}_1=O{C}_2.$$ Similarly, $$O{A}_1=O{A}_2=O{B}_1=O{B}_2.$$ Therefore, six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$, and radius $O{A}_1$.

Proof II Let $O$ be the circumcenter of triangle $ABC$, and $D$, $E$, $F$ the midpoints of $BC$, $CA$, $AB$ respectively. $BH$ intersects $DF$ at point $P$, then $BH\perp DF$. By Pythagoras' theorem, we have $$B{F}^2-F{H}^2=B{P}^2-P{H}^2=B{D}^2-D{H}^2,\stackrel{◯}{1}$$ $$B{O}^2-A_1{O}^2=B{D}^2-A_1{D}^2=B{D}^2-D{H}^2.\stackrel{◯}{2}$$ Similarly, $$B{O}^2-C_2{O}^2=B{F}^2-F{H}^2,\stackrel{◯}{3}$$ By ①, ②, ③, $A_{1}O=C_{2}O$. It is obvious that $A_{1}O=A_{2}O$, $C_{1}O=C_{2}O$, thus $$A_{1}O=A_{2}O=C_{1}O=C_{2}O.$$ Similarly, $$A_{1}O=A_{2}O=B_{1}O=B_{2}O.$$ Therefore, six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$.