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O is the circumcenter of $Δ A B C$
and $R_{1}, R_{2}, R_{3}$ are respectively the radii of the circumcircles of the triangles OBA,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 circumcenter of $Δ A B C$ then
$O A = O B = O C = R$
Let $R_{1}, R_{2}$ and
$R_{3}$ be the circumradii of $Δ O B C$ $Δ O C A$ and $Δ O A B$
respectively.
In $Δ O B C$ $2 R_{1} = \frac{a}{sin 2 A}$ or $\frac{a}{R_{1}} = 2 sin 2 A$
Similarly, $\frac{b}{R_{2}} = 2 sin 2 B$ and $\frac{c}{R_{3}} = 2 sin 2 C$
$∴ \frac{a}{R_{1}} + \frac{b}{R_{2}} + \frac{c}{R_{3}} = 2 (sin 2 A + sin 2 B + sin 2 C)$ $= 8 sin A sin B sin C = 8 \frac{a}{2 R} \frac{b}{2 R} \frac{c}{2 R} = \frac{a b c}{R^{3}}$

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