The Journal of Online Mathematics and Its Applications, Volume 8 (2008)

The Most Marvelous Theorem in Mathematics, Dan Kalman

*How can we be sure that every triangle has an inscribed ellipse of the kind stipulated in Marden's theorem?* The basic idea of a proof can be illustrated quite dramatically using dynamic geometry, as shown below. Center the square frame in your browser window, and click the green arrow to start. After viewing the animation, scroll down to read an explanation.

In this demonstration, an arbitrary triangle is positioned so that its base is on the `x` axis, centered at the origin. The third point of the triangle may be taken as any point above the `x `axis. Clearly, any triangle can be positioned in this fashion. Once the triangle has been positioned, an equilateral triangle with an inscribed circle is constructed, sharing the same base as the original triangle. Note that the inscribed circle is tangent to the sides at their midpoints.

Two transformations are now performed: a vertical stretch and a horizontal shear. It is important to note that these may be formulated as linear transformations of the plane, so that they preserve points of tangency and midpoints, and carry any ellipse to another ellipse (see Linear transformations take ellipses to ellipses). In particular, they must carry a circle (which is a special kind of ellipse) to an ellipse. Therefore, the two transformations carry the inscribed circle in the equilateral triangle to an inscribed ellipse in the original triangle, and this inscribed ellipse is tangent to the sides at their midpoints.

The preceding argument demonstrates the *existence* of an inscribed ellipse in any triangle, tangent to the sides at the midpoints. We must also see that such an ellipse is unique. To that end, consider a triangle `T` with an inscribed midpoint ellipse `E`. Consider the transformation used in the demonstration above, taking an equilateral triangle to `T`. If we reverse the transformation, deforming `T` back to the equilateral triangle, `E` is transformed into some ellipse, inscribed in the equilateral triangle and tangent at the midpoints of the sides. To complete the uniqueness proof, it suffices to show that this new ellipse is actually a circle. That will show that the ellipse we began with is actually the one produced by the construction used in the existence proof, which must therefore be uniquely determined.

So we are left with a new goal: prove that an equilateral triangle has only one inscribed ellipse, tangent at the midpoints of the sides, namely a circle. As before, we begin with an arbitrary inscribed ellipse tangent at the midpoints of the sides, but this time in an equilateral triangle. Let us place that triangle so the inscribed ellipse is centered at the origin, and has its axes on the coordinate axes. This ellipse will then be given by the standard equation . See Figure 1.

Next, apply the linear transformation given by the matrix `M` = , with `r` = `a`/`b`. This takes the ellipse into a circle of radius `a`, now inscribed in a stretched triangle, but still tangent at the midpoints of the sides. See Figure 2.

Actually, this new triangle must still be equilateral. To see this, draw lines from the center of the circle to each vertex and each midpoint, creating six right triangles, as shown with six different colored triangles in Figure 3. Each right triangle has a radius of the circle for one leg, and half of a side of the original triangle for another. Any two right triangles sharing one of the radial lines must therefore be congruent, and that implies that the hypotenuses of the triangles are all equal. This in turn shows that the six right triangles are all congruent, and so the large triangle they combine to form is equilateral, as asserted.

This is enough to show that `a` = `b`, and in particular, that the original ellipse was actually a circle. If the vertices of the original equilateral triangle are at points `P`, `Q`, and `R`, then the (non-parallel) vectors `P``Q` and `P``R` have equal lengths. Applying matrix `M` to these vectors produces results that also have equal lengths. That implies that `M` always maps vectors of equal lengths to vectors of equal lengths. On the other hand, `M` takes the unit vectors along the `x` and `y` axes to vectors of length 1 and `r`. Thus, `r` = 1, showing that `a` = `b`. This completes the proof.