Question

Prove that the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6\sqrt{3r}$.

Answer 1

Step 1: Draw the isosceles triangle.

Let $ABC$ be isosceles is $AB=AC$

$r$ is radius of circle

${AF}^{\mathit{2}}\mathit{+}{BF}^{\mathit{2}}\mathit{=}{AB}^{\mathit{2}}$

$\mathit{(}\mathit{3}r{\mathit{)}}^{\mathit{2}}\mathit{+}\mathit{\left(}{y}^{\mathit{2}}\mathit{\right)}\mathit{=}{x}^{\mathit{2}}.............(1)$

From $\Delta AOD$

$\begin{array}{r}(2r{)}^2=r^2+(AD{)}^2\\ =3r^2=A{D}^2\\ =AD=\sqrt{3}r\end{array}$

Step 2: Now $BD=BF\&EC=FC$ (as tangents are equal from an external point)

$AD+DB=x$

$\begin{array}{c}=\sqrt{3}r+y^2=x\\ y^2=x=\sqrt{3}r-\left(2\right)\end{array}$

$\text{From}\left(1\right)\mathrm{\&}\left(2\right)$

$\begin{array}{c}\\ \therefore (3r{)}^2+(x-\sqrt{3}r)=x^2\\ 9r^2+x^2+3r^2-2\sqrt{3}rx=x^2\\ =9r^2+3r^2-2\sqrt{3}rx=0\\ =12r^2=2\sqrt{3}rx\\ =6r=\sqrt{3}x\\ \therefore x=\frac{6}{\sqrt{3}}r\end{array}$

From $\left(2\right)$

$\begin{array}{c}y/2=(6/\sqrt{3})r-\sqrt{3}r\\ y/2=(3\sqrt{3}/3)r\\ y=2\sqrt{3}r\end{array}$

Step 3: Find the perimeter

perimeter $=2x+y$

$\begin{array}{c}=2(6/\sqrt{3})\mathrm{r}+2\sqrt{3}\mathrm{r}\\ =(12\mathrm{r}+6\mathrm{r})/\sqrt{3}\\ =(18/\sqrt{3})\mathrm{r}\\ =6\sqrt{3}\mathrm{r}\end{array}$

Hence, proved.