This week’s Riddler Classic is a geometry problem involving inscribed hexagons.

The larger regular hexagon in the diagram below has a side length of 1. What is the side length of the smaller regular hexagon?

If you look very closely, there are two more, even smaller hexagons on top. What are their side lengths?

Here is my solution:

[Show Solution]

Let $x_n$ denote the side length of the $n^\text{th}$ hexagon, and we will use the convention that $x_0=1$ refers to the largest one. We can compute the side lengths recursively using the Pythagorean theorem. Here is the picture:

Here, $O$ is the center of the circle, and $AB$ is the $n^\text{th}$ side length. So $|OA|=1$ and $|AB|=x_n$. The triangle $AOC$ is a right-angle triangle, which gives us $|OC| = \sqrt{1-\tfrac{1}{4}x_n^2}$. The $(n+1)^\text{st}$ hexagon has the property that point $F$ lies on the circle. Therefore, $|OF|=1$. By symmetry, the point $G$ is the midpoint of the top side, so we have $|FG| = \tfrac{1}{2}x_{n+1}$. Since a regular hexagon is made up of six equilateral triangles, the height is $|GC| = \sqrt{3} x_{n+1}$. Finally, $FOG$ is a right-angle triangle, so by the Pythagorean theorem, we have $|OF|^2 = |FG|^2 + |OG|^2$. Substituting in the lengths we found, we obtain

\[

1 = \frac{1}{4}x_{n+1}^2 + \left( \sqrt{3} x_{n+1} + \sqrt{1-\frac{1}{4}x_{n}^2}\right)^2

\]The recursion looks a little tidier if we express it in terms of the angles $\angle AOG$ and $\angle FOG$, but since we are after the side lengths, we will keep things in terms of $x_n$. Expanding the square above and simplifying, we obtain a quadratic equation for $x_{n+1}$:

\[

13 x_{n+1}^2 + 8\sqrt{3}\sqrt{1-\tfrac{1}{4}x_n^2}\, x_{n+1}-x_n^2=0

\]Solving for $x_{n+1}$, we see that there will be two roots, one positive and one negative. We can ignore the negative root, and we obtain

$\displaystyle

x_0=1,\quad

x_{n+1} = \frac{1}{13}\left( \sqrt{48+x_n^2}-\sqrt{48-12x_n^2}\right)

$

To find successive side lengths, simply start with $x_0=1$, and evaluate the expressions above in a recursive fashion. The first few lengths are:

\[

\left\{1,\frac{1}{13},\frac{\sqrt{8113}-90}{169},\frac{\sqrt{1387141-180 \sqrt{8113}}-12 \sqrt{8113}-90}{2197},\dots\right\}

\]Numerically, these evaluate to:

\[

\{1,\, 7.6923\times 10^{-2},\, 4.2718\times 10^{-4},\, 1.3170\times 10^{-8},\,\dots \}

\]The sequence decreases *very rapidly* to zero. We can see this if we plot the sequence of side lengths on a log scale:

In fact, $x_n\to 0$ doubly exponentially! To see why we can make a few approximations:

\begin{align}

x_{n+1} &= \frac{1}{13}\left( \sqrt{48+x_n^2}-\sqrt{48-12x_n^2}\right)\\

&= \frac{1}{13} \frac{(\sqrt{48+x_n^2}-\sqrt{48-12x_n^2})(\sqrt{48+x_n^2}+\sqrt{48-12x_n^2})}{\sqrt{48+x_n^2}+\sqrt{48-12x_n^2}}\\

&= \frac{x_n^2}{\sqrt{48+x_n^2}+\sqrt{48-12x_n^2}} \\

&\approx \frac{1}{2\sqrt{48}} x_n^2 \\

&=\frac{1}{8\sqrt{3}} x_n^2

\end{align}Using this approximation together with $x_0=1$, we obtain the approximate formula:

\[

x_n \approx \left(8 \sqrt{3}\right)^{-(2^n-1)}

\]Plotting the true side lengths together with the approximation, we see that the approximation is a bit smaller, but still very close: