Question

If $O$ is the circum-centre of $\Delta ABC$ and $R_1{R}_2,R_3$ are the radii of the circum-circles of triangles $OBC,OCA$ and $OAB$ respectively.

Then $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}$ is equal to?

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

Answer 1

The correct option is B $\frac{abc}{R^3}$

Using Sine law in $\Delta ABC;$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Using Sine law in $\Delta OCB;$

$\frac{a}{\sin 2A}=\frac{R}{\sin B_1}=\frac{R}{\sin C_1}=2R_1$

$⟹\frac{a}{R_1}=2\sin 2A$

[$\because \angle BOC=2A;$ as the angle form by a chord at the center is double than formed at surface of the circle]

Similarly in $\Delta OCA\text{and}\Delta OAB$;

$\frac{b}{R_2}=2\sin 2B$

$\frac{c}{R_3}=2\sin 2C$

Now;

$\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=2\left(\sin 2A+\sin 2B+\sin 2C\right)$

$=2\times 4\sin A.\sin B.\sin C$

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

Hence;

$\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=\frac{abc}{R^3}$