Question

If $O$ is the circumcentre 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)

$\frac{abc}{R^3}$
$\frac{4\Delta }{R^2}$

If O is the circumcentre of $\Delta ABC$, then $OA=OB=OC$ Let $R_1,R_2$ and $R_3$ be circumradii of $\Delta OBC,\Delta OCA$ and $\Delta OAB$ respectively.
In $\Delta OBC,2R_1=\frac{a}{sin2A}\Rightarrow \frac{a}{R_1}=2sin2A$
Similarly$\frac{b}{R_2}=2sin2B$ and $\frac{c}{R_3}=2sin2C$
$\Rightarrow \frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=2(sin2A+sin2B+sin2C)=8sinAsinBsinC$

$=8\frac{a}{2R}\frac{b}{2R}\frac{C}{2R}=\frac{abc}{R^3}$

This is also equal to $\frac{4\Delta }{R^2}$