# A fun surprise in Bjorn Poonen’s n-dimensional sphere problem

I’ve written two posts about a problem I heard Bjorn Poonen discussing last week:

A Strange Problem I overheard Bjorn Poonen discussing

Bjorn Poonen’s n-dimensional sphere problem with kids

There’s additional one really cool part of the problem that I wanted to write about – it doesn’t take much of a modification to change the answer to the last question.

Here’s a quick summary of the problem (though click on the first link above if you don’t want spoilers):

Start in two dimensions. Suppose that you have a 2×2 square and that you chop it into 4 1×1 squares. Inside each of those 4 squares draw the inscribed circle, and then draw the circle at the center of the square that is tangent to each of those 4 circles. The situation looks like this picture:

Now extend the idea to 3 dimensions. So, start with a 2x2x2 cube, chop it into 8 cubes, inscribe spheres in each of those cubes, and finally draw the sphere in the middle of the cube that is tangent to each of those 8 spheres. I don’t know how much it helps, but here’s basically half of the picture with our Zometool set and some tennis balls:

Here’s the problem:

(A) As you extend this idea into higher and higher dimensions, is there a dimension in which the sphere in the middle is actually larger than each of spheres inscribed in the 1x1x1 . . . x1 n-dimensional boxes?

(B) Is there a dimension in which the sphere in the middle actually extends outside of the 2x2x2 . . . x2 n-dimensional box?

(C) Is there a dimension in which the n-dimensional volume of the center sphere is larger than the 2x2x2 . . . x2 n-dimensional box?

The answer to each of the three questions is yes, but a small change to the problem changes the answer to part (c).

In the project I did with my kids, we arrived at this formula for the ratio of the volume of the sphere to the volume of the cube in dimension 2k:

Ratio = $( \frac{\pi}{16} )^k (\frac{1}{k!} ) (\sqrt{2k} - 1)^{2k}$

Again, just for clarity, this is the ratio in dimension 2k.

To make life a little easier, we’ll ignore the -1 in the last term and look at the slightly larger quantity:

$( \frac{\pi}{16} )^k (\frac{1}{k!} ) (\sqrt{2k})^{2k}$ = $( \frac{\pi}{8} )^k (\frac{1}{k!} ) (k^k)$

Stirling’s formula tells us that a good approximation for $k!$ when $k$ is larger is:

$k! = \sqrt{2 \pi k}(k^k)(e^{-k})$

Substituting this formula in to our ratio formula, we obtain this approximation to the ratio of the area of the sphere to the area of the cube in dimension $2k$:

Ratio = $(\frac{\pi * e}{8} )^k (\frac{1}{\sqrt{2 \pi k}} )$

In otherwords, the reason that this ratio gets above 1, and actually tends to infinity (!!), is that $\pi * e$ is greater than 8. How’s that for a fun fact 🙂

Also, that product isn’t that much greater than 8. In fact, if we shrink the radius of the sphere by 4%, the volume of the sphere would not exceed the volume of the cube in any dimension – amazing!