Question

If $O$ is the circumcentre of $\Delta ABC$ and $R_1,R_2,R_3$ are the radii of the circum-circle of triangle $OBC,OCA$ and $OAB$ respectively, then the value of $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=\frac{abc}{R^n}$ , then $n$ equals to

• A. $1$
• B. $-1$
• C. $-3$
• D. $3$

Answer 1

Answer: (D)

$3$

$\because R=\frac{abc}{4\Delta }$
If ${\Delta }_1,{\Delta }_2,{\Delta }_3$ be the area of triangle $OBC,OCA$ and $OAB$ respectively, then $R_1=\frac{OB.OC.BC}{4{\Delta }_1}=\frac{R^{2}a}{4{\Delta }_1}\Rightarrow \frac{a}{R_1}=\frac{4{\Delta }_1}{r^2}$
Similar $\frac{b}{R_2}=\frac{4{\Delta }_2}{R^2},\frac{c}{R_3}=\frac{4{\Delta }_3}{R^2}$
$\therefore \frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=\frac{4}{R^2}({\Delta }_1+{\Delta }_2+{\Delta }_3)$
$=\frac{4\Delta }{R^2}=\frac{4}{R^2}\times \frac{abc}{4R}=\frac{abc}{R^3}$