Question

If in a triangle, $R$ and r are the circumradius and inradius respectively and $r_1,r_2$ and $r_3$ are in $H.P$. then $H.M$ of the exradii of the triangle is?

• A. $3R$
• B. $2R$
• C. $R+r$
• D. $\frac{1}{2R}$

Answer 1

Answer: (A)

$3R$

let $r_1$ ,$r_2$,$r_3$ be the three exradii

and we know exradii $=\frac{\Delta }{s-a}$

So $r_1=\frac{\Delta }{s-a}$, $r_2=\frac{\Delta }{s-b}$, $r_3=\frac{\Delta }{s-a}$ --------------$\left(1\right)$

where $\Delta$ is the area of triangle and $s=\frac{(a+b+c)}{2}$

the harmonic means of the three exradii are

H.M $=\frac{3}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}}$ -------------$\left(2\right)$

putting the value of $r_1,r_2,r_3$ from equ(1) to equ($2$) H. M. $=\frac{3\Delta }{s-a+s-b+s-c}$

H. M. $=\frac{3\Delta }{3s-(a+b+c)}$

H. M. $=\frac{3\Delta }{3s-\left(2s\right)}$ ...... Because $2s=a+b+c$

So, H. M. $=\frac{3\Delta }{s}$ - - - - - - - - $\left(3\right)$

and we know that

$r\left(inradius\right)=\frac{\Delta }{s}$

Put it on equation($3$)

so H.M of exradii = $3r$