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!

“fence post” problems

Saw this conversation on Twitter last week:

Once of the most surprising lessons I learned teaching my kids math came from my older son struggling with this type of problem. He has struggled with them for *years* and seemingly no amount of discussion / practice / reading / and etc has made this problems easier.

I’m baffled, but one thing for sure is that I understand that this type of problem can be difficult for kids.

I decided to try out the problem with my kids this morning. My younger son went first and didn’t have too much trouble:


My older son, on the other hand, stumbled a little. In fact, you’ll see that his initial reaction is to label the dots with 1/7ths:


So, I assume that lots of kids will have little trouble with problems like this one, but some kids will struggle. For those kids, this type of problem is far more difficult than you can imagine.