Question 123
Let $\triangle ABC$ be a non-isosceles acute triangle, and point $O$ is the circumcenter. Let $A'$ be a point on the line $AO$ such that $\angle BA'A = \angle CA'A$. Construct $A'A_1 \perp AC$, $A'A_2 \perp AB$ with $A_1$ on $AC$, $A_2$ on $AB$ respectively. $AH_A$ is perpendicular to $BC$ at $H_A$. Write $R_A$ as the circumradius of $\triangle H_AA_1A_2$. Similarly we have $R_B, R_C$. Prove that
$$
\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C} = \frac{2}{R},
$$
where $R$ is the circumradius of $\triangle ABC$.
$$
\frac{1}{R_A} + \frac{1}{R_B} + \frac{1}{R_C} = \frac{2}{R},
$$
where $R$ is the circumradius of $\triangle ABC$.