Question

If O is the circumcenter of $\Delta ABC$ and $R_1,R_2,$ and $R_3$ are the radii of the circumcircles of triangles OBC, OCA, and OAB, respectively, then $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}$ has the value equal to

• A. $\frac{abc}{R^3}$
• B. $\frac{R^3}{abc}$
• C. $\frac{4\Delta }{R^2}$
• D. $\frac{\Delta }{4R^2}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{abc}{R^3}$
C $\frac{4\Delta }{R^2}$

In $△OBC$
$\angle BOC=2A$
From sine rule: $\frac{BC}{\sin \angle BOC}=2R_1$
$\Rightarrow \frac{a}{R_1}=2\sin 2A$
similarly $\frac{b}{R_2}=2\sin 2B$ and $\frac{c}{R_3}=2\sin 2C$
now, $y=\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=2(\sin 2A+\sin 2B+\sin 2C)$
$\Rightarrow y=4\sin (A+B)\cos (A-B)+4\sin C\cos C$
$\Rightarrow y=4\sin C(\cos (A-B)-\cos (A+B))=8\sin A\sin B\sin C$
$\Rightarrow y=\frac{abc}{R^3}$ [ Sine rule ]