Here’s an idea I got from Albert Chern on Twitter. He did all the hard work, and I think he also drew the picture I’m going to use. I’ll just express the idea in a different way.
Here’s a strange fact about Newtonian gravity.
Consider three parallel ‘line masses’ that have a constant mass per length—the same constant for each one. Choose a plane orthogonal to these lines. There will typically be two points on this plane, say a and b, where a mass can sit in equilibrium, with the gravitational pull from all three lines masses cancelling out. This will be an unstable equilibrium.
Put a mass at point a. Remove the three line masses—but keep in mind the triangle they formed where they pierced your plane!
You can now orbit a test particle in an elliptical orbit around the mass at a in such a way that:
• one focus of this ellipse is a,
• the other focus is b, and
• the ellipse fits inside the triangle, just touching the midpoint of each side of the triangle.
Even better, this ellipse has the largest possible area of any ellipse contained in the triangle!
Here is Chern’s picture:
The triangle’s corners are the three points where the line masses pierce your chosen plane. These line masses create a gravitational potential, and the contour lines are level curves of this potential.
You can see that the points a and b are at saddle points of the potential. Thus, a mass placed at either a and b will be in an unstable equilibrium.
You can see the ellipse with a and b as its foci, snugly fitting into the triangle.
You can sort of see that the ellipse touches the midpoints of the triangle’s edges.
What you can’t see is that this ellipse has the largest possible area for any ellipse fitting into the triangle!
Now let me explain the math. While the gravitational potential of a point mass in 3d space is proportional to , the gravitational potential of a line mass in 3d space is proportional to which is also the gravitational potential of a point mass in 2d space.
So, if we have three equal line masses, which are parallel and pierce an orthogonal plane at points and then their gravitational potential, as a function on this plane, will be proportional to
Here I’m using as our name for an arbitrary point on this plane, because the next trick is to think of this plane as the complex plane!
Where are the critical points (in fact saddle points) of this potential? They are just points where the gradient of vanishes. To find these points, we can just take the exponential of and see where the gradient of that vanishes. This is a nice idea because
The gradient of this function will vanish whenever
Since is a cubic polynomial, is a quadratic, hence proportional to
for some a and b. Now we use
Marden’s theorem. Suppose the zeros of a cubic polynomial are non-collinear. Then there is a unique ellipse inscribed in the triangle with vertices and tangent to the sides at their midpoints. The foci of this ellipse are the zeroes of the derivative of
For a short proof of this theorem go here:
This ellipse is called the Steiner inellipse of the triangle:
• Wikipedia, Steiner inellipse.
The proof that it has the largest area of any ellipse inscribed in the triangle goes like this. Using a linear transformation of the plane you can map any triangle to an equilateral triangle. It’s obvious that there’s a circle inscribed in any equilateral triangle, touching each of the triangle’s midpoints. It’s at least very plausible that that this circle is the ellipse of largest area contained in the triangle. If we can prove this we’re done.
Why? Because linear transformations map circles to ellipses, and map midpoints of line segments to midpoints of line segments, and simply rescale areas by a constant fact. So applying the inverse linear transformation to the circle inscribed in the equilateral triangle, we get an ellipse inscribed in our original triangle, which will touch this triangle’s midpoints, and have the maximum possible area of any ellipse contained in this triangle!