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The Orthocentre

O is the circ...

Question

O is the circumcentre of $△ A B C$ and $R_{1}, R_{2}, R_{3}$ are respectively, the radii of the circumcircles of the triangles OBC, OCA, and OAB. Prove that $\frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} = \frac{a b c}{R^{3}}$

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Solution

If $O$ is the circumcentre of $Δ A B C$ , then
$O A = O B = O C = R$
Let $R_{1}, R_{2}$ and $R_{3}$ be circumradii of $Δ O B C, Δ O C A$ and $Δ O A B,$ respectively.
$\begin{matrix} In Δ O B C, 2 R_{1} = \frac{a}{sin 2 A} \Rightarrow \frac{a}{R_{1}} = 2 sin 2 A Similarly, \frac{a}{R_{2}} = 2 sin 2 B and \frac{a}{R_{3}} = 2 sin 2 C \Rightarrow \frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} = 2 (sin 2 A + sin 2 B + sin 2 C) = 8 sin A sin B sin C = 8 \frac{a}{2 R} \frac{b}{2 R} \frac{c}{2 R} = \frac{a b c}{R^{3}} \end{matrix}$

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