China Mathematical Olympiad

Firstly we claim that $A^{\prime },B,O,C$ are concyclic points. Otherwise, extend $AO$ to intersect the circumcircle of $△BOC$ at point $P$ which is different from $A^{\prime }$. We get $$\angle BPA=\angle BCO=\angle CBO=\angle CPA.$$ Then $△P{A}^{\prime }C\cong △P{A}^{\prime }B$, and $A^{\prime }B=A^{\prime }C$. So $AB=AC$ and that is a contradiction since $△ABC$ is not isosceles. So $$\angle BC{A}^{\prime }=\angle BO{A}^{\prime }=180^{\circ }-2\angle C,$$ and $\angle A^{\prime }C{A}_1=\angle C$. Further, we have $$\frac{H_A{A}_1}{AC}=\sin \angle C=\cos \angle A_{2}A{A}^{\prime }=\frac{A{A}_2}{A{A}^{\prime }}$$ and $$\angle A^{\prime }AC=\angle A_{2}A{H}_A=\frac{\pi }{2}-\angle B.$$ So $△A_{2}A{H}_A\sim △A^{\prime }AC$. In the same way, $△A_1{H}_{A}A\sim △A^{\prime }BA$. Then $\angle A_2{H}_{A}A=\angle AC{A}^{\prime }$ and $\angle A_1{H}_{A}A=\angle AB{A}^{\prime }$. Consequently, $$\begin{array}{rl}\angle A_1{H}_A{A}_2& =2\pi -\angle A_2{H}_{A}A-\angle A_1{H}_{A}A\\ & =2\pi -\angle AC{A}^{\prime }-\angle AB{A}^{\prime }\\ & =\angle A+2\left(\frac{\pi }{2}-\angle A\right)\\ & =\pi -\angle A.\end{array}$$ We get $$\frac{R}{R_A}=\frac{\frac{R}{A_1{A}_2}}{\frac{2\sin \angle A_1{H}_A{A}_2}{\sin \angle A}}=\frac{2R}{A_1{A}_2}=\frac{2R}{A{A}^{\prime }};\stackrel{◯}{1}$$ The last equality holds since $A,A_2,A^{\prime },A_1$ lie on the same circle with $A{A}^{\prime }$ as the diameter. Now, draw $A{A}^{\prime \prime }\perp A^{\prime }C$ with point $A^{\prime \prime }$ on line $A^{\prime }C$. Since $\angle AC{A}^{\prime \prime }=\angle A^{\prime }C{A}_1=\angle C$, we have $A{A}^{\prime \prime }=A{H}_A$. Then $$\begin{array}{rl}A{A}^{\prime }& =\frac{A{A}^{\prime \prime }}{\sin \angle A{A}^{\prime }C}=\frac{A{H}_A}{\sin (90^{\circ }-\angle A)}\\ & =\frac{A{H}_A}{\cos \angle A}=\frac{2S_{△ABC}}{BC\cos \angle A}.\stackrel{◯}{2}\end{array}$$ From ①, ② we get $$\begin{array}{rl}\frac{1}{R_A}& =\frac{BC\cos \angle A}{S_{△ABC}}=\frac{\cos \angle A}{R\sin \angle B\sin \angle C}\\ & =\frac{1}{R}(1-\cot \angle B\cot \angle C).\end{array}$$ In the same way $$\frac{1}{R_B}=\frac{1}{R}(1-\cot \angle C\cot \angle A)$$ and $$\frac{1}{R_C}=\frac{1}{R}(1-\cot \angle A\cot \angle B).$$ Notice that $$\cot \angle A\cot \angle B+\cot \angle B\cot \angle C+\cot \angle C\cot \angle A=1.$$ We then have $$\frac{1}{R_A}+\frac{1}{R_B}+\frac{1}{R_C}=\frac{2}{R}.$$