Question

Prove that the cubic equation whose roots are the radii $r_1,r_2,r_3$ of escribed circles of a triangle is $t^3-(r+4R)t+s^{2}t-s^{2}r=0$

Answer 1

$r_1,r_2,r_3{t}^3-(r+4R)t+s^{2}t-s^{2}r=0r_1+r_2+r_3-r=4R{r}_1+r_2+r_3=4R+r-----\left(i\right)r_1{r}_2+r_2{r}_3+r_3{r}_1=\frac{r_1{r}_2{r}_3}{r}=s^2⟹r_1{r}_2+r_2{r}_3+r_3{r}_1=s^2-----\left(ii\right)$

and, $r_1{r}_2{r}_3=s^{2}r-----\left(iii\right)$

For a cubic equation, $x^3+a{x}^2=bx+c$,

Sum of the roots=$\alpha +\beta +\gamma =-a$

Sum of the roots taken two at a time=$\alpha \beta +\beta \gamma +\gamma \alpha =b$

and product of the roots=$\alpha \beta \gamma =-c$

$\therefore$ Cubic equation with $r_1$, $r_2$, $r_3$ as roots is

$t^2-(r+4R)t^2+s^{2}t-s^{2}r=0$