The Journal of Online Mathematics and Its Applications, Volume 8 (2008)
The Most Marvelous Theorem in Mathematics, Dan Kalman

## The Proof of Marden's Theorem

The proof I will give of Marden's Theorem is a hybrid of arguments by Bôcher and Marden. In overview, here is the organization. First, following Marden, we will see that if an ellipse has foci at the roots of p' and if that ellipse passes through the midpoint of one side of the triangle T, then it is actually tangent to that side at the midpoint. This permits us to construct three ellipses, each of which is tangent to one side of the triangle at its midpoint. Now consider one of these ellipses. Following Bôcher, we will see that it is actually tangent to all three sides of the triangle. It remains only to show that this one ellipse coincides with the other two ellipses. For this part of the argument, we use the fact that only one ellipse can exist with given foci and tangent to a given line (click here for details).

In accord with this outline we will consider two lemmas, one for Marden's part of the proof, and one for Bôcher's. But first, we need the following preliminary idea: with no loss of generality, we may translate, rotate, and scale our triangle in any convenient manner. Here, we use the fact that any complex number α can be expressed in polar form reiθ, and that multiplying any complex number by α has the effect of scaling by r and rotating through the angle θ. Similarly, we take advantage of the fact that complex numbers add like vectors in the plane. (For an explanation of these properties, click here.)  Together, these observations show that any combination of scaling, rotating, and translating in the complex plane can be imposed by a linear function M: CC. Let M(z) = αz + β, where α ≠ 0 and β are fixed complex numbers, and where α = reiθ. Then transforming z to M(z) is the same as the following geometric steps: scale z by r, rotate about the origin through angle θ, and translate by β. Likewise, if M is applied to every point of a figure, the resulting figure is a scaled, rotated, translated version of the original.

We can now justify the claim that imposing these transformations represents no loss of generality in our proofs. That requires showing that Marden's theorem holds for a triple {z1, z2, z3} if and only if it also holds for the transformed triple {M(z1), M(z2), M(z3)}. Actually, the invertibility of non-constant linear functions implies that we need only show one direction of the double implication. So, if we know that Marden's Theorem holds for {z1, z2, z3}, let us see that it must also hold for {M(z1), M(z2), M(z3)}.

In Figure 1, the triangle with vertices at the zj is shown, with the inscribed ellipse tangent at the midpoints of the sides. The foci of the ellipse, which must be the roots of p, are also shown. Looking at M geometrically, we observe that scaling, rotating, and translating this figure preserve all of the ingredients. That is, the image of the triangle is a similar triangle, and the image of the ellipse is still an inscribed ellipse tangent at the midpoints of the sides. We can also observe that the images of the foci of the original ellipse are the foci of the transformed ellipse. This follows from the two focus definition of an ellipse, and the fact that M expands or contracts all distances by an equal factor r. In particular, if in the original ellipse, the sum of the distances from any point to the two foci is D, then in the transformed ellipse, the sum of the distances from any point to the images of the two foci is rD. This proves that the transformed curve is in fact an ellipse, and that its foci are at the images of the foci of the original ellipse.

Figure 1.  Configuration for Marden's Theorem.

Now we know that the transformed triangle has vertices at the points M(zj), and that the inscribed ellipse tangent at the midpoints of the sides has foci at the images of the foci of the original ellipse. These are the points M(ξ) where ξ is a root of p. We want to show that these same M(ξ) are the roots of the derivative of a new polynomial, this time with its roots at the transformed vertices M(zj). Accordingly, define the new polynomial by pM(z) = (zM(z1))(zM(z2))(zM(z3)). The derivative is given by (pM)′(z). Our goal is to show that M carries the roots of p to roots of (pM)′.

To that end, let us substitute M(z) for z in the definition of pM. That gives us

 (1) pM(M(z)) = (M(z) − M(z1))(M(z) − M(z2))(M(z) − M(z3)).

Next, notice that since M(z) = αz + β, M(z) − M(zj) = α(zzj). Therefore, equation (1) simplifies to

pM(M(z)) = α3 p(z).

Now differentiate both sides of this equation, bearing in mind that M′(z) = α. We obtain

[pM(M(z))]′ = pM′(M(z))M′(z) = pM ′(M(z)) α

and

[α3 p(z)]′ = α3 p′(z)

Hence, equating these expressions leads to

pM′(M(z)) = α2 p'(z)

This shows that if z is a root of p, then M(z) is a root of pM. That is what we wished to show.

At last we are ready to prove Marden's Theorem. Following the outline discussed earlier, we will consider two lemmas, one based on the proof of Marden, and the other on the proof of Bôcher. We begin with Marden's contribution.

#### Lemma 1

Let the polynomial p(z), its roots z1, z2, z3, and the triangle T be as in the statement of Marden's Theorem. Let Q be the midpoint of a side of T. Then there exists a unique ellipse with foci at the roots of p and passing through Q, and this ellipse is actually tangent to the side of T containing Q.

#### Proof

As noted earlier, with no loss of generality we can rotate, translate, and scale the triangle in any way we choose. Accordingly, let us arrange things so that the side of T containing Q lies along the x axis centered at the origin, and with length 2, while the opposite vertex sits in the upper half plane. Thus, the vertices of the triangle (and the roots of p) are at 1, − 1, and w = a + bi where b > 0, and the origin takes the place of Q. See Figure 2.

Figure 2.  Configuration for Lemma 1.

Now we know the roots of p (which we may assume to be monic), so we have

p(z) = (z − 1)(z + 1)(zw) = z3wz2z + w

Differentiating we find

Thus, if the roots of p are z4 = r4eiθ4 and z5 = r5eiθ5, we conclude (using properties of quadratics) that z4 + z5 = 2w/3 and z4z5 = − 1/3. The first of these shows that at least one of the roots of p must be in the upper half-plane, and the second shows that θ4 + θ5 = π (details here). This in turn also tells us that both roots must be in the upper half-plane. Moreover, considering these roots as vectors drawn from the origin, the angles the vectors make with the positive x axis are supplementary. Thus, either both roots are on the y axis, or one root makes an acute angle with the positive x axis and the other roots makes an equal angle with the negative x axis.

Now it is time to construct our ellipse. Note that the origin is not on the closed line segment joining the roots of p, because those roots are in the upper half-plane. Therefore, there is a unique ellipse with foci at the roots of p and passing through the origin (details here), which is the midpoint of the side that lies on the x axis. We have already seen that the lines from the origin to the foci of this ellipse make equal acute (or possibly right) angles with the x axis. Therefore, using the optical property of the ellipse, it is tangent to the x axis. Thus, we have shown that the ellipse with foci at the roots of p and passing through the midpoint of one side of the triangle T is actually tangent to that side.

On to Bôcher's contribution. Before proceeding, one notational convention needs explanation. If z and w are nonzero complex numbers, we can think of them as points in the plane. A vector can be drawn from the origin to each of these points, creating an angle θ with 0 ≤ θ ≤ π. As a convenience, we shall refer to this as the angle between the complex numbers z and w, thus avoiding a more accurate but also more long-winded reference to the angle between the vectors from the origin to the complex numbers z and w. Keep this convention in mind as you read the next Lemma and its proof.

#### Lemma 2

Let the polynomial p(z), its roots z1, z2, z3, and the triangle T be as in the statement of Marden's Theorem. Consider the ellipse with foci at the roots of p and which is tangent to one side of T . Then this same ellipse is tangent to the other two sides of T.

#### Proof

Once again, we are free to position the triangle any way we wish. The ellipse E is tangent to one side of the triangle, and we again place this side along the x axis. However, this time we put one vertex at the origin, and the other at 1. The remaining vertex will again be placed at w = a + bi where b > 0. This configuration will permit us to see that the ellipse is tangent to the side of the triangle from the origin to w.

Figure 3: Configuration for Bôcher's Result.

With roots at 0, 1, and w, p(z) can be taken as z(z − 1)(zw). Multiplying out these factors gives

p(z) = z3 − (1+w)z2 + wz

p'(z) = 3z2 − 2(1 + w)z + w.

Arguing as before, the sum of the roots of p equals (2/3)(1 + w). This shows that at least one of the roots of p must be in the upper half-plane. But here we also know that these roots are the foci of an ellipse tangent to the x-axis. Therefore, both are in the upper half plane, allowing us to express the roots of p as z4 = r4eiθ4 and z5 = r5eiθ5, where 0 < θ4θ5 < π. Also observe that the product of the roots of p equals a positive multiple of w. This implies that the angles for z4 and z5 must add up to the angle for w. These features are illustrated in Figure 3.

As indicated in the figure, the angle between z5 and w is θ4 + θ5 - θ5 = θ4. On the other hand, we know from the extended optical property that the two tangent lines from 0 to the ellipse E make equal angles with the lines from 0 to the foci. One of the tangent lines is the x axis, and that makes an angle of θ4 with z4. The other tangent line must make an equal angle with z5. But we already know that the line from 0 to w is just such a line. This implies that the line from 0 to w is tangent to the ellipse, which is what we wished to show. (See Figure 4).

Figure 4: The ellipse is tangent to 0w.

It remains to show that the side from 1 to w is also tangent to the ellipse. This can be established by essentially the same proof, but with the triangle translated horizontally so that it has vertices at −1 and 0, rather than at 0 and 1, as in Figure 5. Carrying out this plan completes the proof of the lemma.

Figure 5: Shifted Configuration.

And now, finally, we arrive at the

#### Proof of Marden's Theorem

As usual, we assume that polynomial p, its roots zj, and triangle T are as in the statement of the theorem. Using the roots of p as foci, draw an ellipse E that passes through the midpoint of one side of the triangle. By Lemma 1, E is actually tangent to that side of T. By Lemma 2, E is also tangent to the other two sides of T. Now we claim that the points of tangency with these other two sides must be the midpoints. If not, repeat the construction above with a new side, producing an ellipse E. Since E and E have the same foci, and are both tangent to the same line (in fact to three lines), they must actually coincide (details here). But this would show that E and E both contact the new side in the same point, namely, the midpoint. By symmetry, the same conclusion holds for the remaining side of the triangle. Thus, the original ellipse E is tangent to all three sides at their midpoints. That completes the proof of Marden's Theorem.