Find the ellipse inscribed in a triangle having the maximum area

Question

If we have a triangle of sides $a,b,c$, there are infinite ellipsis inscribed in the triangle. How can I find that having the maximum area? Is this ellipse the circle, or in what cases the maximum area ellipse is the circle? Thanks.

Answer 1

One way to solve this problem is using affine transform, i.e transformation of the form:

$$\mathbb{R}^2 \ni (u,v)\quad\mapsto\quad (x,y) \in\mathbb{R}^2 \;\;\text{ with}\;\; \begin{cases}x &= a u + b v + c\\y &= du + ev + f\end{cases}$$ where $a,b,c,d,e,f \in \mathbb{R}$ subject to the constraint $\Delta = ae - bd \ne 0$.

The key properties that we need to use are

1. under an affine transform, the area of all geometric figure get scaled by same factor $|\Delta|$. As a result, the ratio of areas of any two geometric figures is invariant under such a transform.
2. Given any two triangles or two circles/ellipses, there is an affine transform that send one to another.
3. Under an affine transform, triangles map to triangles, circles/ellipses map to circles/ellipses.

4. Several geometric relationships are invariant under affine transform. e.g.

   • if two curves are tangent to each other at some point, so does their images.
   • if a point is the mid-point of another two points, so does their images.

This means given any triangle $T$, the problem of finding the largest ellipse (in sense of area) inside $T$ can be solved as follows:

• By 2. there is an affine transform $\phi$ which maps $T$ to an equilateral triangle $T'$.
• By 3. if $E$ is any circle or ellipse inside $T$, $E' = \phi(E)$ is a circle or ellipse inside $T'$.
• By 1. $$\frac{\text{Area}(E)}{\text{Area}(T)} = \frac{\text{Area}(E')}{\text{Area}(T')}$$ As a result, the problem of finding the largest ellipse $E$ in $T$ is equivalent to one finding the largest ellipse $E'$ in $T'$.
• Intuitively, the largest circle/ellipse inside $T'$ is its incircle $C_{in}$. Assume this is the case, the largest circle/ellipse inside $T$ will be $\phi^{-1}(C_{in})$.

To identify $E_{max} = \phi^{-1}(C_{in})$ geometrically, we use following fact:

The in-circle of an equilateral triangle is a circle tangent to the mid-points of its three sides.

By 4. we know $E_{max}$ will be an ellipse tangent to the mid points of the three sides. Since each point tangent to a side is equivalent to specify two points on a curve and 5 points is enough to completely specify a conic. We have more than enough to fully specify $E_{max}$.

Such an ellipse is called the Steiner inellipse of triangle $T$. For other interesting geometric properties of it, please consult the wiki pages and the references there.

There is one loose end to close, that is the assertion that

The in-circle $C_{in}$ is the largest ellipse inside $T'$.

To show this, we use 2. again. For any ellipse $E'$ in $T'$, there is always an affine transform $\psi$ which send $E'$ to the unit circle $C$. Let $T'' = \psi(T')$. Since $$\frac{\text{Area}(E')}{\text{Area}(T')} = \frac{\text{Area}(C)}{\text{Area}(T'')}$$ The problem of finding the largest $E'$ inside $T'$ is equivalent to finding the smallest triangle $T''$ that encloses the unit circle $C$.

If $T''$ is the smallest triangle, it is obvious its three sides are tangent to the unit circle $C$. Let $2\alpha, 2\beta, 2\gamma$ be the three angles of $T''$. It is clear they satisfy following constraint

$$0 < \alpha, \beta, \gamma \quad\text{ and }\quad \alpha+\beta+\gamma = \frac{\pi}{2}\tag{*1}$$ and the area of $T''$ is given by

$$\text{Area}(T'') = \cot(\alpha) + \cot(\beta) + \cot(\gamma) \tag{*2}$$

Notice $\cot(\theta)$ is a strictly convex function in $\theta$. By Jensen's inequality, the configuration that minimize $(*2)$ subject to constraint $(*1)$ are those where $\alpha = \beta = \gamma$. This implies $T''$ is an equilateral triangle and hence $\psi$ is simply a scaling transform.

As a corollary, the largest ellipse $E'$ inside $T'$ is the in-circle.