# Patrick Honner’s Pi day exercise in 4d [This is a quick post I wrote while my younger son was at a little math enrichment activity. Sorry it looks like it was written in a hurry and not proof read . . . ]

Earlier today my older son and I played around with Patrick Honner’s Pi Day exercise:

That project is here:

Patrick Honner’s Pi Day Exercise

After we finished my son wondered about extending the exercise to 4 dimensions!

But, extending to 4 dimensions isn’t as easy as it seems. For one thing, the “volume” and “surface area” of a 4 dimensional sphere involve $\latex \pi^2$ not $\pi$:

“Volume” = $(1/2) \pi^2 R^4$

“Surface Area” = $2 \pi^2 R^3$

So, we’ll modify Honner’s 3d $\pi$ formula to be $\pi^2$ = (1/128) (Surface Area^4) / (Volume^3). That’ll give us a value for $\pi^2$ and then we can compute $\pi$.

So, I found the “volume” and “surface area” of the 4 dimensional regular Polytopes here:

Polytopes

Calculating $"\pi"$ for the regular 4 dimensional polytopes gave values of approximately:

5-Cell: 8.63

8-Cell: 5.66

16-Cell: 4.62

24-Cell: 4.00

120-Cell: 3.38

600-Cell: 3.24

We’ve actually made a 3D version of the 120-cell with our Zometool set: That project is here, and maybe helps see that the shape is getting sort of spherical.

A Stellated 120-Cell made from our Zometool set

Another way to see some of these 4-dimensional shapes is to check out the game Hypernom:

Using Hypernom to get kids talking about math

Anyway, thanks for Patrick Honner for a fun Pi day!

# Patrick Honner’s Pi Day Exercise Saw this tweet from Patrick Honner yesterday:

I liked the activity for a lot of reasons – it was a great way to review a little geometry and arithmetic, a nice opportunity to discuss more 3D geometry, and finally a lucky coincidence with our project from yesterday talking about the truncated cube and the truncated dodecahedron:

What Will It Look Like? The unlucky part is that my younger son slept in because of daylight savings time – boo 😦 Oh well, here’s what I did with my older son:

Part 1: We first talked about the surface area and volume formulas for a sphere and then discussed how you could use the ratio in Honner’s post to define $\pi$ in an unusual way.

Once we had the idea that $\pi = (1/36) \frac{SA^3}{V^2}$ we extended that idea to other 3 dimensional shapes. The first shape was a cube.

Also, sorry for the bad camera shot initially on this one – didn’t realize I’d cut off the bottom of our whiteboard until mid way through the video – oops.

After the Cube discussion we moved on to a Tetrahedron. Since I wasn’t looking to study the tetrahedron too much today, we pulled the volume formula from Wikipedia.

Next we took a look at the two shapes from yesterday and tried to guess which one was more like a sphere:

Finally, we did the calculations for these to shapes to find out what “ $\pi$ was for the truncated cube and the truncated dodecahedron according to Patrick Honner’s formula:

I left some of the calculations at the end of the video as an exercise for my son. After he finished the calculations he wanted to try to extend the idea to 4 dimensions – yes!! Sadly this idea proved to be a little harder than he expected, but maybe we’ll look at it tonight.

So, a great exercise – thanks to Patrick Honner for posting it!