Question

Find the smallest possible area of an isosceles triangle that is circumscribed about a circle of radius \(r\).

Short answer

Thus, the minimum area is \(3\sqrt 3 {r^2}\).

Step-by-step solution

Draw the figure of an isosceles triangle circumscribed about circle

Let radius of the circle \(OT = OS = r\) and angle \(\angle ABC = \theta \). Draw the angle bisector of angle ABC which intersects the center of the circle.

Thus, \(\tan \theta = \frac{h}{d}\;{\rm{and}}\;\tan \frac{\theta }{2} = \frac{r}{d}\) here, \(h\) is height of the triangle and \(d\) is the distance between B and T.

Area of the triangle

Obtain the area of the triangle as follows:

\(\begin{aligned}{c}A &= \frac{1}{2} \cdot BC \cdot AT\\ = \frac{1}{2}\left( {2d} \right)\left( h \right)\\ &= {d^2}\tan \theta \\ &= {d^2}\left( {\frac{{2\tan \frac{\theta }{2}}}{{1 - {{\tan }^2}\frac{\theta }{2}}}} \right)\end{aligned}\)

Substitute\(\;\tan \frac{\theta }{2} = \frac{r}{d}\)in the above equation gives,

\(\begin{aligned}{c}A = \frac{{2{d^2}\left( {\frac{r}{d}} \right)}}{{1 - \frac{{{r^2}}}{{{d^2}}}}}\\ = \frac{{2{d^3}r}}{{{d^2} - {r^2}}}\end{aligned}\)

Differentiate the area of triangle

Differentiate the area with respect to \(d\) gives:

\(\begin{aligned}{c}A' &= \frac{{\left( {6{d^4}r - 6{d^2}{r^3}} \right) - 4{d^4}r}}{{{{\left( {{d^2} - {r^2}} \right)}^2}}}\\ &= \frac{{2{d^4}r - 6{d^2}{r^3}}}{{{{\left( {{d^2} - {r^2}} \right)}^2}}}\\ &= \frac{{{d^2}r\left( {{d^2} - 3{r^2}} \right)}}{{{{\left( {{d^2} - {r^2}} \right)}^2}}}\end{aligned}\)

The derivative of the area is positive.

Obtain the minimum value of area of the triangle

Use the algebraic expression to expand as follows:

\(\left( {{d^2} - 3{r^2}} \right) = \left( {d + \sqrt 3 r} \right)\left( {d - \sqrt 3 r} \right)\)

Thus, the local minima occurs for \(\left( {d - \sqrt 3 r} \right)\) gives \(d = \sqrt 3 r\).

Use the formula for area as follows:

\(\begin{aligned}{c}A &= \frac{{2{d^3}r}}{{{d^2} - {r^2}}}\\ &= \frac{{2{{\left( {\sqrt 3 r} \right)}^3}r}}{{3{r^2} - {r^2}}}\\ &= \frac{{6\sqrt 3 {r^4}}}{{2{r^2}}}\\ &= 3\sqrt 3 {r^2}\end{aligned}\)

Hence, the minimum area is \(3\sqrt 3 {r^2}\).