4

The question was inspired by this answer.

Condition for perfect packing of ellipse with circles: $n$ circles can be perfectly packed along the major axis of the ellipse with semi-axes $a,b\quad$ iif

\begin{align} \frac ba&=\sin\frac{\pi}{2\,n} \tag{1}\label{1} . \end{align}

Here "perfectly packed" means that the chain of circles with the centers on the major axis is inscribed into the ellipse, and the radii of the circles at both ends agree with the curvature of the ellipse at the ends of the major axis.

Condition \eqref{1} works for both odd (ex. $n=7$)

enter image description here

and even $n$ (ex. $n=8$)

enter image description here

Question:

Is this a well-known condition? Any reference?

Addendum

General formula for the radius of $k$-th circle in a chain of $n$ perfectly packed circles along the major axis of the ellipse \begin{align} r_k&=a\,\sin\frac{\pi}{2\,n}\,\sin\frac{(2\,k+1)\,\pi}{2\,n} ,\quad k=0,\dots,n-1 , \end{align} where $a$ is the major semi-axis, and the location of the center of $k$-th circle is found as \begin{align} O_k&= F_1\cdot\cos^2 \frac{(2\,k+1)\,\pi}{4\,n} +F_2\cdot\sin^2 \frac{(2\,k+1)\,\pi}{4\,n} , \end{align} where $F_1,F_2$ are the focal points of the ellipse.

Distances from $F_1,F_2$ to the tangent point of the $k$-th circle with the ellipse are

\begin{align} |F_1T_k| &= 2\,a\,\sin^2\frac{(2\,k+1)\,\pi}{4\,n} ,\\ |F_2T_k| &= 2\,a\,\cos^2\frac{(2\,k+1)\,\pi}{4\,n} . \end{align}

\begin{align} \cos\angle T_kO_kF_2&= \frac{\sin\frac{k\,\pi}n-\sin\frac{(k+1)\,\pi}n} {\sin\frac{k\,\pi}n+\sin\frac{(k+1)\,\pi}n} . \end{align}

enter image description here

g.kov
  • 13,911
  • It's neater to write the condition in terms of cosine: $$e = \cos\frac{\pi}{2n}$$ where $e$ is, of course, the eccentricity. :) Upon realizing this, I re-wrote my answer to the circle-chain question, replacing the cumbersome $\arctan(b/c)$ with $\arccos e$. I also wrote an Addendum to explicitly derive your perfectly-packable condition. – Blue Mar 25 '20 at 22:28
  • @Blue: Nice, +1 – g.kov Mar 26 '20 at 01:44
  • Regarding your Addendum: From my answer, we can deduce your formula for the $k$-th radius, and also that the $x$-coordinate of the center of the $k$-th circle is $x_k=a\cos\frac{\pi}{2n}\cos\frac{(2k+1)\pi}{2n}$. Note that (literally) together, these give $$x_k+r_k = a\cos\frac{k\pi}{n}$$ These values are the $x$-coordinates points of tangency of the chained circles. Coincidentally, they're also the $x$-coordinates of the vertices of a regular $2n$-gon whose circumdiameter is the ellipse's major axis. Nifty! ... I wonder if this connection leads to a more-direct proof. – Blue Mar 27 '20 at 21:15
  • @Blue: Yes, the connection with regular polygon is fascinating. Btw, your comments already weight the answer. – g.kov Mar 29 '20 at 02:27

0 Answers0