Minimum area of isosceles triangle circumscribed about a circle

Question

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

Answer 1

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:

$h^2-3rh=0\rightarrow h=3r$

Answer 2

This answer is for the original question, which was searching for the maximal area.

You have a fundamental problem in your approach: you assume that the value is bounded so that the maximum is a finite number. With that assumption it makes sense to look at zeros of the derivative to indicate local extremal (i.e. maximal or minimal) values.

But the assumption is wrong. Imagine what happens as you increase the height, pulling the top point higher and higher. The base edge will remain larger than the diameter of the circle, approaching it in the limit. The height on the other hand can become arbitrarily large. Taken together that means the area can become arbitrarily large.

Your $x=3r$ is indeed extremal, but it is the minimum, not the maximum. If you draw that, or compute some more distances, you will find that you got the condition for an equilateral triangle.

You might be able to come up with some clever arguments why the equilateral has to be smallest. Probably by showing that every deviation from that adds more area to the tip than it removes at the base. But whether any of that cleverness is more convincing than your brute force coordinate computation in the end I don't know. There is something to be said for not requiring any knowledge besides some very basic formulas for computing the area. This elegance question may be more a matter of taste than an objective criterion, though.

Answer 3

The OP requested solutions that minimize the area of an isosceles triangle circumscribed about a fixed radius, with special interest in solutions based on the semiperimeter, inradius, and/or apex angle. Those solutions and one other follow:

Solution 3: minimizing the semiperimeter

For a triangle with inradius $r$ and semiperimeter $s$ we have $A=rs$. Since $r$ is given, to minimize $A$ we need only to minimize $s=s_1+s_2+s_3$ in the image below.

Drafting on the solution from @sirious, we have the segments $s_1,s_2,s_3$ in terms of $h$ as:

$s_1$: $(h-r)^2=r^2+s_1^2 \ \Rightarrow \ s_1 = \sqrt{h^2-2hr}$

Then by similar triangles:

$s_2$: $\frac{h-r}{s_1}=\frac{s_1+s_2}{h}$

$h^2-hr = s_1^2+s_1 s_2$

$h^2-hr = h^2-2hr +s_1 s_2$

$\frac{hr}{s_1} = s_2 = \frac{hr}{\sqrt{h^2-2hr}}$

$s_3$: We immediately have $s_3=s_2$ from known properties of tangents to a circle from a point.

This can also be derived algebraically from $\frac{s_1}{r}=\frac{h}{base}$.

The semiperimeter $s$ is $s=s_1+s_2+s_3 = s_1+2s_2$

$s=\sqrt{h^2-2hr}+2\frac{hr}{\sqrt{h^2-2hr}} = \frac{h^2}{\sqrt{h^2-2hr}}$

Taking the derivative of $s$ with respect to $h$ leads to:

$ s'(h) = \frac{2h}{\sqrt{h^2-2hr}} -\frac{h^2(2h-2r)}{2(h^2-2hr)^{\frac{3}{2}}} = \frac{2h(h^2-2hr)-h^2(h-r)}{2(h^2-2hr)^{\frac{3}{2}}} = \frac{h^3-3h^2r}{2(h^2-2hr)^{\frac{3}{2}}} = \frac{h^2(h-3r)}{2(h^2-2hr)^{\frac{3}{2}}}$

Setting $s'=0$ yields $h=3r$, a minimum.

One could then solve for the sides and base and find that the triangle is equilateral.

Solution 4: minimize via apex angles

Using trig and similar triangles we have:

$ sin(\theta)=\frac{r}{y} \Rightarrow y=\frac{r}{sin(\theta)}$

$s_1 = \frac{rcos(\theta)}{sin(\theta)}$

$\frac{s_1}{r} = \frac{y+r}{s_3} \Rightarrow s_1s_3=r^2+ry \Rightarrow s_3=\frac{r(sin(\theta)+1)}{cos(\theta)}$

$A=s_3(y+r) = \frac{r^2(sin(\theta)+1)^2}{sin(\theta)cos(\theta)}$

Taking $\frac{dA}{d\theta}$ and with a few steps of simplifying we obtain a numerator of $(sin(\theta)+1)^2(2sin(\theta)-1)$ which gives $\theta=\frac{\pi}{6}$ as the solution. This solution immediately implies the triangle is equilateral.

Solution 5: optimization via double angles

This post is getting long, this will be terse.

$\frac{r}{y}=sin(\theta) \Rightarrow y = \frac{r}{sin(\theta)}$

$\frac{r}{z}=sin(\beta) \Rightarrow z = \frac{r}{sin(\beta)}$

Note that $sin(\theta)=cos(2\beta)$.

$ A=(y+r)(zcos(\beta)) = r^2\frac{cos(\beta)}{sin(\beta)}\big(\frac{1}{sin(\theta)}+1\big) = r^2\frac{cos(\beta)}{sin(\beta)}\big(\frac{1}{cos(2\beta)}+1\big) = r^2cot(\beta)(sec(2\beta)+1)$

That benign-looking equation is a mess to clean up after differentiating, but has a minimum in the interval $[0,\frac{\pi}{2}]$ at $\beta = \frac{\pi}{6}$, again indicating an equilateral triangle.

I had originally set this up with $\theta = \frac{\pi}{4}+x$ and $\beta = \frac{\frac{pi}{4}-x}{2}$ with the hope that $x=\frac{\pi}{12}$ would pop out but the algebra got pretty tangled and never seemed to work out nicely.