Tag: equilateral triangle

Given a triangle, there are many ellipses that can be drawn inside it. Among them, one has a particularly beautiful and distinguished place in geometry: the Steiner inellipse.

What is the Steiner inellipse?

Let ABC be any triangle. The Steiner inellipse is the unique ellipse contained in the triangle that is tangent to the three sides at their midpoints.

Thus, if D, E, and F are the midpoints of BC, CA, and AB, respectively, the Steiner inellipse passes through all three points and is tangent to the corresponding sides there.

Its center is the centroid G of the triangle—the point where the three medians intersect.

The Steiner inellipse also has an extremal property: among all ellipses contained in a given triangle, it has the largest possible area.

Why does the Steiner inellipse naturally appear?

One way to understand the Steiner inellipse is through affine geometry. Every triangle can be obtained from an equilateral triangle by an invertible affine transformation.

For an equilateral triangle, the Steiner inellipse is simply its incircle. The incircle touches the three sides at their midpoints and is centered at the common centroid, incenter, and circumcenter.

Under an affine transformation, a circle generally becomes an ellipse, midpoints remain midpoints, and tangency is preserved. Therefore the incircle of an equilateral triangle is transformed into an ellipse tangent to the three sides of the new triangle at their midpoints. That ellipse is precisely the Steiner inellipse.

So the Steiner inellipse may be viewed as the affine image of the incircle of an equilateral triangle.

A different way to recognize the same ellipse

The definition above characterizes the Steiner inellipse by tangency: it touches the sides of the triangle at three special points.

But there is another, rather unexpected way to detect whether a point lies on this ellipse—one that does not initially mention an ellipse, tangency, or even distances.

It uses only parallel lines and areas.

Choose an arbitrary point M inside triangle ABC. Through M, draw three lines, each parallel to one side of the triangle.

These three lines cut off three smaller triangles at the vertices A, B, and C. Let their areas be

Let $T$ denote the area of the original triangle.

Now ask a simple question:

For which points M is the sum of the three corner areas exactly one-half of the area of the original triangle?

At first glance, there is no obvious reason that the answer should involve an ellipse at all.

The surprising answer

The answer is remarkably simple: the points satisfying this area condition are exactly the points on the Steiner inellipse.

In other words, a point M lies on the Steiner inellipse if and only if the three corner triangles together have exactly half the area of the original triangle.

This is unexpected because the construction itself contains no ellipse. We choose a point, draw three parallel lines, and measure three areas. Yet the condition that their sum equals one-half of the total area traces out precisely the Steiner inellipse.

Why should this be an ellipse?

The key is affine geometry. An invertible affine transformation sends triangles to triangles, preserves parallelism and ratios of areas, and sends ellipses to ellipses.

We can therefore transform our original triangle into an equilateral triangle without changing the essential area condition. In an equilateral triangle, the Steiner inellipse becomes something much more familiar: the incircle.

So it is enough to determine which points satisfy the area condition in the equilateral case.

The equilateral case

Consider the equilateral triangle with vertices

$$A=(-1,0),B=(0,\sqrt{3}),C=(1,0).$$

The base has length 2 and the height is $\sqrt{3}$, so the area of the triangle is

$$T=\frac{2·\sqrt{3}}{2}=\sqrt{3}.$$

Let $M=(a,b)$ be an interior point. The three lines through M parallel to the sides cut off three smaller triangles. Because each corner triangle is similar to the original equilateral triangle, their areas can be written in terms of a and b.

A direct calculation gives

$$T_1=\frac{\sqrt{3}}{4}{(1+a-\frac{b}{\sqrt{3}})}^2,$$ $$T_2=\frac{\sqrt{3}}{4}{(1-a-\frac{b}{\sqrt{3}})}^2,$$

and

$$T_3=\frac{b^2}{\sqrt{3}}.$$

Now impose our area condition:

$$T_1+T_2+T_3=\frac{T}{2}=\frac{\sqrt{3}}{2}.$$

Substituting the three expressions above and simplifying gives

$$a^2+b^2-\frac{2b}{\sqrt{3}}=0.$$

Completing the square transforms this into

But this is the equation of the circle centered at

$$(0,\frac{1}{\sqrt{3}})$$

with radius $\frac{1}{\sqrt{3}}.$ This is precisely the incircle of our equilateral triangle.

Therefore, in the equilateral case, the points satisfying

$$T_1+T_2+T_3=\frac{T}{2}$$

are exactly the points on the incircle.

Equilateral triangles often turn geometric questions into especially elegant problems. For another example, see Equilateral Triangle Maximum Area .

Returning to the original triangle

We have proved that, for an equilateral triangle, the condition

describes exactly the incircle.

Now apply the inverse affine transformation that carries the equilateral triangle back to the original triangle. Parallel lines remain parallel, and all areas are multiplied by the same factor, so the area condition is preserved. The incircle is transformed into the Steiner inellipse.

Therefore, for any triangle, a point M satisfies the area condition if and only if M lies on the Steiner inellipse.

A related area identity

There is another elegant relation involving the same three corner triangles. Unlike the characterization above, this identity holds for every interior point M, not only for points on the Steiner inellipse.

Since each corner triangle is similar to the original triangle, the ratio of corresponding side lengths is the square root of the ratio of the corresponding areas. The three relevant length ratios add to 1, which gives

The contrast between the two identities is worth noticing.

For every interior point M,

But without the square roots,

holds precisely when M lies on the Steiner inellipse. Thus the same three corner areas give both a universal identity and a geometric characterization of a special ellipse.

There is another remarkable way in which the Steiner inellipse appears. If the vertices of the triangle are regarded as the three complex roots of a cubic polynomial, the zeros of its derivative are exactly the two foci of the Steiner inellipse. See Marden’s Theorem: How the Derivative of a Cubic Finds an Ellipse .

References and further reading

1. A. Eydelzon, “On a New Property of the Steiner Inellipse” , The American Mathematical Monthly, Vol. 127, No. 10 (2020), pp. 933–935.
2. NCS/MAA Team Contest, Thirteenth Annual Contest (2009), Problem 9 , “Square roots of area ratios.”
3. D. Kalman, “An Elementary Proof of Marden’s Theorem” , The American Mathematical Monthly, Vol. 115, No. 4 (2008), pp. 330–338.

The first reference contains the area characterization of the Steiner inellipse discussed in this article. The second gives an earlier appearance of the classical square-root area problem. The third provides additional background on the Steiner inellipse and its connection with Marden’s theorem.

A triangle is a two-dimensional simplex, while a tetrahedron is its three-dimensional counterpart. For a related exploration of simplex geometry, see The Geometry of a Tetrahedron: From Pythagoras to Vector Identities .