# Find the ellipse inscribed in a triangle having the maximum area

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.

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Notice under any affine transform, the ratio of areas between any two figures are preserved. Furthermore, if a pair of curves is tangent at some point, so does their image under the affine transform. This means you can first use an affine transform to bring you triangle to an equilateral triangle, solve the simpler problem there (you get a circle) and then inverse transform to get the desired ellipse. – achille hui Jun 9 '14 at 9:16
@achillehui: I think this should be an answer. – Ted Shifrin Jun 9 '14 at 13:06
@achillehui: if you expand your comment and put it in the answer I shall accept it . – Riccardo.Alestra Jun 9 '14 at 13:08

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.

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I'm really impressed by your answers. +1. – Edwin Jun 9 '14 at 18:28