Question

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

Answer 1

$$\begin{gathered} \begin{array}{l}\text{Let ABC is an isosceles triangle with AB=AC=}x\text{and a circle with centre O and radius}r\text{is inscribed in the triangle.}\\ \text{O,A and O,E and O,D are joined.}\\ \text{From}\Delta \text{ABF,}\\ {\text{AF}}^2+{\text{BF}}^2={\text{AB}}^2\\ \Rightarrow {\left(3r\right)}^2+{\left(\frac{y}{2}\right)}^2=x^2.....\left(1\right)\\ \\ \text{Again,From}\Delta \text{ADO,}{\left(2r\right)}^2=r^2+{\text{AD}}^2\\ \Rightarrow 3r^2={\text{AD}}^2\\ \Rightarrow \text{AD=}\sqrt{3}r\\ \text{Now, BD=BF and EC=FC}\left(\text{Since tangents drawn from an external point are equal}\right)\\ \text{Now, AD+DB=}x\\ \Rightarrow \left(\sqrt{3}r\right)+\left(\frac{y}{2}\right)=x\\ \Rightarrow \frac{y}{2}=x-\sqrt{3}.....\left(2\right)\end{array} \\ \begin{array}{l}\therefore {\left(3r\right)}^2+{(x-\sqrt{3}r)}^2=x^2\\ \Rightarrow 9r^2+x^2-2\sqrt{3}rx+3r^2=x^2\\ \Rightarrow 12r^2=2\sqrt{3}rx\\ \Rightarrow 6r=\sqrt{3}x\\ \Rightarrow x=\frac{6r}{\sqrt{3}}\end{array} \\ \begin{array}{l}\text{Now, From}\left(\text{2}\right),\\ \frac{y}{2}=\frac{6}{\sqrt{3}}r-\sqrt{3}r\\ \Rightarrow \frac{y}{2}=\frac{6\sqrt{3}}{3}r-\sqrt{3}r\\ \Rightarrow \frac{y}{2}=\frac{(6\sqrt{3}-3\sqrt{3})r}{3}\\ \Rightarrow \frac{y}{2}=\frac{3\sqrt{3}r}{3}\\ \Rightarrow y=2\sqrt{3}r\\ \text{Perimeter}=2x+y\\ =2\left(\frac{6}{\sqrt{3}}r\right)+2\sqrt{3}r\\ =\frac{12}{\sqrt{3}}r+2\sqrt{3}r\\ =\frac{12r+6r}{\sqrt{3}}\\ =\frac{18}{\sqrt{3}}r\\ =\frac{18\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}r\\ =6\sqrt{3}r\end{array} \end{gathered}$$