I'm looking for better solutions to the following optimization problem, which I've solved so far in two ways. The problem: given a circle of radius $r$, what height $x$ yields the minimal area of an isosceles triangle circumscribed about the circle? The picture illustrates the problem, and the basis of my two solutions.
Solution 1: Brute force. Set up a coordinate system with the circle centered on the origin $(0,0)$, denote the apex as $(0,h)$, the point of base tangency as $(0,-r)$, and a base vertex as $(c,-r)$. Use implicit differentiation on the circle $x^2+y^2=r^2$ to identify the slope $\frac{dy}{dx}=-\frac{x}{y}$ and the tangent line for a side and solve where $y=-r$ to obtain $c$ in terms of $h$, which finally lets me define $A=\frac{1}{2}(2c)(x) = cx = c(h+r)$. Once differentiated and set to zero, you end with a cubic: $0=(h+r)(h-r)(h-2r)$ with the only sensible solution of $h=2r$ for a total $x=3r$ as the solution.
Solution 2: Similar triangles. Denote the length from a base vertex to the point of base tangency by $c_1$, and the length of the side tangency to the apex as $c_2$, so that the whole side is $c_1+c_2$ and the base is $2c_1$. By similar triangles, $\frac{r}{c_2}=\frac{c_1}{x}$, and $c_1=\frac{xr}{c_2}$. The $A=cx$ in Solution 1 is now $A=c_1x$, and $A=c_1x=\frac{xr}{c_2}x=\frac{x^2r}{c_2}$. With the Pythagorean theorem we have $c_2^2=(x-r)^2-r^2$, and since maximizing $A\ge0$ also maximizes $A^2$, we can set $A^2(x)=\frac{x^4r^2}{x^2-2rx}$ as the function to maximize, which yields $A^2(x)' = \frac{2r^2x^4(x-3r)}{x^2(x-2r)^2}$, preventing solutions $x=0, x=2r$ and again yielding $x=3r$ as the solution. This approach is nicer.
My question: Are there more direct solutions? I suspect that there are nice optimizations based on the inradius and semiperimeter, e.g. $A=r(2c_1+c_2)$, or using the vertex or base angles, but, without using the triangle similarity of Solution 2, I've not found solutions with those approaches. And sorry if this is a repost, I searched but didn't see this particular problem.
Note: edited to correct typo in title/question: seeking MIN not MAX of the triangle, thanks @MvG
Best Answer
Another aproach:
It can be shown that in an isosceles triangle if the altitude is $h$ and the radius of the inscribed circle is $r$ then the measures of the sides are:
$$AB=BC=\frac{h(h-r)}{\sqrt{ h(h-2r)}}$$
$$AC=\frac{2rh}{\sqrt {h(h-2r)}}$$
So the area is:
$$S=\frac{rh^2}{\sqrt {h(h-2r)}}$$
Taking derivative and equating to zero we finally get:
$h^2-3rh=0\rightarrow h=3r$