Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, then

$r_1^2+r_2^2+r_3^3+r^2=?$
(where $r$ $=$ in-radius, R $=$ circumradius, $r_1,r_2,r_3$ are ex-radii)

• A. $16R^2-a^2-b^2-c^2$
• B. $8R^2-a^2-b^2-c^2$
• C. $16R^2+a^2+b^2+c^2$
• D. $8R^2+a^2+b^2+c^2$

Answer 1

The correct option is A $16R^2-a^2-b^2-c^2$

$(r_1+r_2+r_3{)}^2=r_1^2+r_2^2+r_3^2-2r(r_1+r_2+r_3)+2(r_1{r}_2+r_2{r}_3+r_3{r}_1)$

We know that $r_1+r_2+r_3=4R+r$

Put this in above eqnation,

$(4R{)}^2=r_1^2+r_2^2+r_3^2-2r(r_1+r_2+r_3)+2s^2$ ------ $r_1{r}_2+r_2{r}_3+r_3{r}_1=s^2$

$r_1^2+r_2^2+r_3^2=16R^2+2r(r_1+r_2+r_3)-2s^2$ ------$1$

we know $r=\frac{\Delta }{s}$ and $r_1=\frac{\Delta }{s-a}$ similarly for $r_2$ and $r_3$

Hence, $r{r}_1+r{r}_2+r{r}_3=[\frac{s^2}{s(s-a)}+\frac{s^2}{s(s-b)}+\frac{s^2}{s(s-c)}]$

On solving above equation, we will get

$r{r}_1+r{r}_2+r{r}_3=(ab+bc+ca)-s^2$ ------eqn($2$)

Put eqn($2$) in eqn($1$) we will get, $r_1^2+r_2^2+r_3^2=16R^2+2(ab+bc+ca)-2s^2-2s^2$

$r_1^2+r_2^2+r_3^2=16R^2+2(ab+bc+ca)-4s^2$ -----$s=\frac{(a+b+c)}{2}$

On solving above eqnation, we will get

$r_1^2+r_2^2+r_3^2=16R^2-a^2-b^2-c^2$