Question

Let f,g,h be the length of the perpendiculars from the circumcenter of the $△ABC$ in the sides a,b and c respectively. If $\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\lambda \frac{abc}{fgh}$, then the value of $\lambda$ is -

• A. $1/4$
• B. $1/2$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

In $△ABC,\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$f=R\cos A,g=R\cos B,h=R\cos C$

$\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=\frac{2R\sin A}{R\cos A}+\frac{2R\sin B}{R\cos B}+\frac{2R\sin C}{R\cos C}=2(\tan A+\tan B+\tan C)=2\tan A\tan B\tan C$

$\frac{abc}{fgh}=\frac{\left(2R\sin A\right)\left(2R\sin B\right)\left(2R\sin C\right)}{\left(R\cos A\right)\left(R\cos B\right)\left(R\cos C\right)}=\tan A\tan B\tan C$
$⟹\frac{a}{f}+\frac{b}{g}+\frac{c}{h}=2\frac{abc}{fgh}⟹\lambda =2$