Question

If perimeters of $△LMN$ and $△ABC$ are $\lambda$ and $\mu$, then the value of $\frac{\lambda }{\mu }$ is:

• A. $\frac{r}{R}$
• B. $\frac{R}{r}$
• C. $\frac{rR}{△}$
• D. $\frac{△}{rR}$

Answer 1

The correct option is A $\frac{r}{R}$

Here, $△MNL$ is pedal triangle, then
$MN=a\cos A,NL=b\cos B$ and $ML=c\cos C$
and $\angle MLN=\pi -2A,\angle LMN=\pi -2B,$ and $\angle MNL=\pi -2C$
$\lambda =MN+NL+LM$
$=a\cos A+b\cos B+c\cos C$
Using sine rule,for $a,b,c$ we have
$=2R\sin A\cos A+2R\sin B\cos B+2R\sin C\cos C$
Using multiple angle formula $2\sin \theta \cos \theta =\sin 2\theta$ we get
$=R(\sin 2A+\sin 2B+\sin 2C)$
Consider, $\sin 2A+\sin 2B+\sin 2C$
$=2\sin (A+B)\cos (A-B)+2\sin C\cos C$
$=2\sin (\pi -C)\cos (A-B)+2\sin C\cos C$
$=2\sin C\cos (A-B)+2\sin C\cos \pi -(A+B)$
$=2\sin C[\cos (A-B)-\cos (A+B)]$
On applying transformation formula above we get
$=2\sin C\left[2\sin A\sin B\right]$
$=4\sin A\sin B\sin C$
Substituting for $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$ we get
$\lambda =R(\sin 2A+\sin 2B+\sin 2C)$
$=4R\sin A\sin B\sin C$
Using sine rule, we have
$=4R\frac{a}{2R}\times \frac{b}{2R}\times \frac{c}{2R}$
$=\frac{abc}{2R^2}$
$=\frac{4R△}{2R^2}$
$=\frac{2△}{R}$
and $\mu =2s=\frac{2△}{r}$
$\therefore \frac{\lambda }{\mu }=\frac{\frac{2△}{R}}{\frac{2△}{r}}=\frac{r}{R}$