Elite Math Program

Tag: ellipse

Marden’s Theorem: How the Derivative of a Cubic Finds an Ellipse
What can the derivative of a polynomial tell us about geometry? For a cubic polynomial with three complex roots, the answer is surprisingly precise: the derivative locates the two foci of a particular ellipse.

Let

$p (z) = (z - z_{1}) (z - z_{2}) (z - z_{3}),$

where $z_{1}, z_{2}, z_{3}$ are three distinct, noncollinear complex numbers.

Every complex number can be represented as a point in the complex plane. Therefore the three roots of p become three points, and those three points form a triangle.

Now differentiate. Since p is cubic, $p^{'} (z)$ is quadratic and has two roots, counted with multiplicity. Call them $w_{1}$ and $w_{2} .$

Where are these two critical points located, and what do they have to do with the triangle formed by the original three roots?

Figure 1. Three noncollinear roots z₁, z₂, and z₃ of a cubic polynomial determine a triangle in the complex plane.

What does the derivative know about the triangle?

There is already a classical theorem that gives us some information. The Gauss–Lucas theorem states that the zeros of the derivative of a polynomial lie in the convex hull of the zeros of the polynomial.

For our cubic, the convex hull of the three roots is simply the triangle with vertices $z_{1}, z_{2}, z_{3} .$ Therefore the two zeros of $p^{'}$ must lie inside this triangle.

That is already an interesting connection between differentiation and geometry. But for a cubic, something much stronger is true.

Figure 2. By the Gauss–Lucas theorem, the two zeros w₁ and w₂ of the derivative lie inside the triangle determined by the three roots of the cubic.

Marden’s theorem

Recall the Steiner inellipse of a triangle: the unique ellipse tangent to the three sides at their midpoints.

We discussed this ellipse and an area characterization of it in The Steiner Inellipse and a Surprising Area Characterization .

Marden’s theorem reveals a completely different way in which the same ellipse appears.

Marden’s Theorem. Let
$p (z) = (z - z_{1}) (z - z_{2}) (z - z_{3}),$
where the three roots are noncollinear. Then the two zeros of $p^{'} (z)$ are exactly the two foci of the Steiner inellipse of the triangle whose vertices are $z_{1}, z_{2}, z_{3} .$

This is much stronger than Gauss–Lucas. Gauss–Lucas tells us that the critical points are somewhere inside the triangle. Marden’s theorem identifies their exact geometric meaning.

The derivative of the cubic has found the foci of an ellipse determined by the roots of the original polynomial.

Figure 3. Marden’s theorem: the two zeros w₁ and w₂ of the derivative are exactly the foci of the Steiner inellipse of the triangle formed by z₁, z₂, and z₃.

A concrete example

Let us see the theorem in action. Choose the three roots
z1=−2, z2=1+2i, z3=1−2i.
The corresponding cubic is
p(z) = (z+2) (z−1−2i) (z−1+2i).
Multiplying the conjugate factors first gives
(z−1−2i) (z−1+2i) = z2 −2z +5.
Hence
p(z) = (z+2) ( z2 −2z +5 ) = z3 +z +10.
Now differentiate:
p′(z) = 3z2 +1.
The critical points satisfy
3z2 +1 =0,
so

$z = \pm \frac{i}{\sqrt{3}} .$

Marden’s theorem now tells us something geometric that would be very difficult to guess merely by looking at the polynomial:

The two foci of the Steiner inellipse are
i3 and − i3.
Figure 4. For p(z) = z³ + z + 10, the roots form the displayed triangle, while the zeros ±i/√3 of p′(z) are exactly the foci of its Steiner inellipse.

Why is this so surprising?

The derivative is usually introduced as an analytic object: it measures instantaneous rate of change, gives the slope of a tangent line, and locates critical points.

Here it is doing something that looks completely different.

Start with three complex numbers. Use them as the roots of a cubic. Differentiate the polynomial. Solve one quadratic equation. The two answers are not merely points somewhere inside the triangle—they are the two foci of a distinguished ellipse.

In symbols, the chain of ideas is

$three roots \to triangle \to differentiate \to two critical points \to two foci .$

The same Steiner inellipse therefore admits two very different descriptions.

In our earlier article, it appeared from an area condition: a point lies on the ellipse precisely when three corner triangles have total area equal to half the area of the original triangle.

Here the ellipse appears from differentiation: its two foci are the zeros of the derivative of the cubic whose roots are the vertices of the triangle.

This is a beautiful example of a recurring theme in mathematics: algebra, calculus, and geometry can encode the same object in completely different ways.

References and further reading
1. D. Kalman, “An Elementary Proof of Marden’s Theorem” , The American Mathematical Monthly, Vol. 115, No. 4 (2008), pp. 330–338.
2. A. Eydelzon, “On a New Property of the Steiner Inellipse” , The American Mathematical Monthly, Vol. 127, No. 10 (2020), pp. 933–935.
October 7, 2026
Steiner Inellipse: A Surprising Area Characterization
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.

Figure 1. The Steiner inellipse of triangle ABC.
It is the unique ellipse tangent to the three sides at their midpoints,
and its center is the centroid G.

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

$T_{1}, T_{2}, T_{3} .$

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?

$T_{1} + T_{2} + T_{3} = \frac{T}{2} ?$

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

Figure 2. Through an arbitrary interior point M,
draw three lines parallel to the sides of triangle ABC.
The three corner triangles have areas T₁, T₂, and T₃.

The surprising answer

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

Theorem. Let ABC be a triangle with area $T$ , and let M be a point in its interior. Through M, draw three lines parallel to the sides of the triangle, cutting off three corner triangles with areas $T_{1}$ , $T_{2}$ , and $T_{3}$ . Then
$M \in Steiner inellipse \Leftrightarrow T_{1} + T_{2} + T_{3} = \frac{T}{2} .$

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.

Figure 3. A point M on the Steiner inellipse.
For this point, the three corner areas satisfy T₁ + T₂ + T₃ = T/2.

The equilateral case

Consider the equilateral triangle with vertices
A=(−1,0), B=(0,3), C=(1,0).
The base has length 2 and the height is $\sqrt{3}$ , so the area of the triangle is
T = 2·3 2 = 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
T1 = 34 ( 1 +a − b3 ) 2 , T2 = 34 ( 1 −a − b3 ) 2 ,
and
T3 = b2 3 .
Now impose our area condition:
T1 + T2 + T3 = T2 = 32.
Substituting the three expressions above and simplifying gives
a2 + b2 − 2b 3 = 0.
Completing the square transforms this into

$a^{2} + {(b - \frac{1}{\sqrt{3}})}^{2} = \frac{1}{3} .$

But this is the equation of the circle centered at
( 0, 13 )
with radius $\frac{1}{\sqrt{3}} .$ This is precisely the incircle of our equilateral triangle.

Therefore, in the equilateral case, the points satisfying
T1 + T2 + T3 = T2
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 .

Figure 4. In the equilateral case, the Steiner inellipse is the incircle. The area condition produces a circle centered at (0, 1/√3) with radius 1/√3.

Returning to the original triangle

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

$T_{1} + T_{2} + T_{3} = \frac{T}{2}$

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

$\sqrt{T_{1}} + \sqrt{T_{2}} + \sqrt{T_{3}} = \sqrt{T} .$

The contrast between the two identities is worth noticing.

For every interior point M,

$\sqrt{T_{1}} + \sqrt{T_{2}} + \sqrt{T_{3}} = \sqrt{T} .$

But without the square roots,

$T_{1} + T_{2} + T_{3} = \frac{T}{2}$

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 .
October 7, 2026