#### 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.

#### Solution

\[\text {Let ABC is an isosceles triangle with AB=AC=x and a circle with centre O and radius r is inscribed in the triangle.}\]

\[\text { O,A and O,E and O,D are joined.}\]

\[\text { From }\Delta ABF, \]

\[{AF}^2 + {BF}^2 = {AB}^2\]

\[\Rightarrow {(3r)}^2 + {(\frac{y}{2})}^2 = x^2 . . . . . (1)\]

\[\text { Again, From }\Delta ADO, {(2r)}^2 = r^2 + {AD}^2\]

\[\Rightarrow 3 r^2 = {AD}^2 \]

\[\Rightarrow AD=\sqrt{3}r \]

\[\text { Now, BD=BF and EC=FC(Since tangents drawn from an external point are equal })\]

\[\text { Now, AD+DB=x}\]

\[\Rightarrow (\sqrt{3}r) + (\frac{y}{2}) = x\]

\[\Rightarrow \frac{y}{2} = x -\sqrt{3} ............. (2)\]

\[\begin{array}{l}\therefore {(3r)}^2 + {(x - \sqrt{3}r)}^2 = x^2 \\ \Rightarrow 9 r^2 + x^2 - 2\sqrt{3}rx + 3 r^2 = x^2 \\ \Rightarrow 12 r^2 = 2\sqrt{3}rx \\ \Rightarrow 6r = \sqrt{3}x \\ \Rightarrow x = \frac{6r}{\sqrt{3}}\end{array}\]

\[\begin{array}{l}\text { Now, From }(2), \\ \frac{y}{2} = \frac{6}{\sqrt{3}}r - \sqrt{3}r \\ \Rightarrow \frac{y}{2} = \frac{6\sqrt{3}}{3}r - \sqrt{3}r \\ \Rightarrow \frac{y}{2} = \frac{(6\sqrt{3} - 3\sqrt{3})r}{3} \\ \Rightarrow \frac{y}{2} = \frac{3\sqrt{3}r}{3} \\ \Rightarrow y = 2\sqrt{3}r \\ \text { Perimeter } = 2x + y \\ = 2\left( \frac{6}{\sqrt{3}}r \right) + 2\sqrt{3}r \\ = \frac{12}{\sqrt{3}}r + 2\sqrt{3}r \\ = \frac{12r + 6r}{\sqrt{3}} \\ = \frac{18}{\sqrt{3}}r \\ = \frac{18 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}r \\ = 6\sqrt{3}r\end{array}\]