help: isosceles triangle circumscribing a circle of radius r Please help me show that: the equilateral triangle of altitute $3r$ is the isosceles triangle of least area circumscribing a circle of radius $r$.
Iassumed the following:
base = $2a$
height = $h$
radius of circle = $r$
Area = $\frac{1}{2}(2a)h = ah$
$tan(2\theta) = \frac{h}{a}$ and  $tan\theta = \frac{r}{a} $
Ii tried using double angle identities to represent h in terms of a and \theta but I have failed miserably.
This is what I ended up with:
$\displaystyle{%
h={2r\cos^{2}\left(\theta\right) \over 2\cos^{2}\left(\theta\right) - 1}}$
I know i need to differentiate the area and set to zero, but I can't get it. please direct me how to solve this problem. Thank you;

First, two corrections in the question 
1) The height will be $3r/2$
2) This will maximize the area. Since the minimum area of an isosceles triangle will be of height $2r$ and zero base, hence zero area.

From the figure, $h=r+\sqrt{r^2-a^2}$, area $A=ah=a(r+\sqrt{r^2-a^2})$. Now equate $\frac{\partial A}{\partial a}=0$ to get the $a$ which maximizes the area. For your info, 
$$
\frac{\partial A}{\partial a}= \frac{r\sqrt{r^2-a^2} + r^2-2a^2}{\sqrt{r^2-a^2}}; \\
\frac{\partial A}{\partial a}=0 \Rightarrow r\sqrt{r^2-a^2} + r^2-2a^2 = 0 \\
(2a^2-r^2)^2 = r^2(r^2-a^2) \\
a = \frac{\sqrt{3}}{2}r
$$
Hope this helps.
