### Geometry question about ellipse packing

Posted:

**January 12th, 2022, 1:34 am**This is in Sandbox because it has nothing to do with cellular automata, but maybe I can find someone here with a better understanding of geometry (and ellipses) who can help. I was recently working on packing three congruent ellipses into a unit circle (for a decorative design) so each was tangent to its neighbors, and matched the curvature (radius 1) where it was tangent to the circle.

By setting up some equations and solving in a symbolic solver, I was able to determine that an ellipse with semi-major axis a=sqrt(3/7) and semi-minor axis b=3/7 and satisfies these conditions. It has curvature a^2/b = 1, and each matches the slope of some line at 120° increments around the circle to which it is tangent. I'll leave the algebraic solution as an exercise. Any ellipse with proportions sqrt(7/3) to 1 will pack into some circle this way. Some other things work out nicely, though. The ellipses are tangent at a distance of exactly 1/2 from the center. Also (because of the 3/7 semi-minor axis) there's just enough room to pack a circle of 1/7 radius inside that is tangent to all three.

In terms of getting what I need for my design, I'm done, but solving the equations leaves me with no intuition at all about the result. I keep thinking there must be a simpler way to understand this geometrically.

Also, this generalizes to different numbers of ellipses (trivially 1, which is a circle, then 2, as well as 4 and 5 shown below). For 4 ellipses, the inner tangent circle has radius 1/3. For 5, I don't have an exact solution since I started with tan 54° for the slope. I haven't tried 6. I wonder if there is a name for this kind of packing. I was originally trying to fill a circle with a design, and I think this is visually appealing. Note that the 5-ellipse case looks like it could be an orthographic projection of circles drawn on a dodecahedron, including the center circle. I'm going to wager that it isn't, though, since the analogy fails for a cube.

For completeness, the formula for the radius of the inner tangent circle is: 1 - 2/(tan(pi/2 - pi/n)^2 + 2)

By setting up some equations and solving in a symbolic solver, I was able to determine that an ellipse with semi-major axis a=sqrt(3/7) and semi-minor axis b=3/7 and satisfies these conditions. It has curvature a^2/b = 1, and each matches the slope of some line at 120° increments around the circle to which it is tangent. I'll leave the algebraic solution as an exercise. Any ellipse with proportions sqrt(7/3) to 1 will pack into some circle this way. Some other things work out nicely, though. The ellipses are tangent at a distance of exactly 1/2 from the center. Also (because of the 3/7 semi-minor axis) there's just enough room to pack a circle of 1/7 radius inside that is tangent to all three.

In terms of getting what I need for my design, I'm done, but solving the equations leaves me with no intuition at all about the result. I keep thinking there must be a simpler way to understand this geometrically.

**Added**: I didn't mention above, but my first thought was that if you draw circles on the faces of a cube and find an orthographic projection on the diagonal you should get these ellipses. Those have proportions sqrt(3) to 1 and thus too large a radius of curvature. In practice you could get away with it and people might not notice.Also, this generalizes to different numbers of ellipses (trivially 1, which is a circle, then 2, as well as 4 and 5 shown below). For 4 ellipses, the inner tangent circle has radius 1/3. For 5, I don't have an exact solution since I started with tan 54° for the slope. I haven't tried 6. I wonder if there is a name for this kind of packing. I was originally trying to fill a circle with a design, and I think this is visually appealing. Note that the 5-ellipse case looks like it could be an orthographic projection of circles drawn on a dodecahedron, including the center circle. I'm going to wager that it isn't, though, since the analogy fails for a cube.

**Added:**Going up to 6, the proportions of the ellipses are sqrt(5) to 1, and the center circle radius is 3/5 the outer circle. Maybe I should try to work out a formula by interpolation: 0, 1/7, 1/3, ???, 3/5. Anything nice to go in the blank? It is approximately 0.4864.**One more update:**After tweaking my system of equations a little more in Wolfram Alpha, I can conclude that the ratio of major to minor axis for the packing of n ellipses is sqrt(tan(pi/2 - pi/n)^2 + 2) (angles in radians, which was not my choice, but I needed it to plot in Desmos). This still doesn't help my intuition at all. It is interesting how nicely some of the numbers work out for special angles.For completeness, the formula for the radius of the inner tangent circle is: 1 - 2/(tan(pi/2 - pi/n)^2 + 2)