Finding the Maximum Number of Areas

Hi all,

I originally posted this in the pre-university geometry forum without much luck, and since it's a puzzle I'll repost it here and hope that someone will take up the challenge!

Basically, I wanted to see what the maximum number of areas I could divide a circle into with N number of straight cuts -- basically, drawing a circle on a sheet of paper and then using a ruler to draw straight lines through it. After that I worked out the same thing but for cutting up a sphere with N 2D planes -- or in other words cutting up a cake with a large, straight knife, where you may not move the pieces between cuts.

I managed to get these right and have confirmed it with other sources (see links below). These two work as great puzzles in their own right, if anyone wants to give them a try before looking at the answers.

For cutting circles, see: Circle Division by Lines -- from Wolfram MathWorld

And for cutting spheres (or in this case cubes), see: Cube Division by Planes -- from Wolfram MathWorld

**Lastly (and the one I'm currently stuck on), I want to work out the maximum number of areas N three-dimensional planes can divide a 4D hypersphere into. **

I have an answer, but it doesn't make much sense and I'm already out of my depth, so I was hoping someone could help me out. The answer I've come up with is as follows:

I hope you have fun with these puzzles and are able to help me out with the more difficult 4 dimensional question. Thanks!

Re: Finding the Maximum Number of Areas

As a general remark: Circle, Cube, whatever does not matter - the unbounded problem gives the same numbers.

4 3-dimensional cuts can produce 16 4-dimensional volumes. The easiest way to see this is to use the "coordinate cubes" as cuts: x_1=0, x_2=0, x_3=0, x_4=0

Defining the numbers for the n-dimensional problem with i cuts as f(n,i), I think that the formula f(n,i)=f(n,i-1)+f(n-1,i-1) (with the special case n=3 explained at OEIS) should hold.

f(n,0)=1

f(1,i)=i+1 (these are i dots on a line)

f(2,i)=1+i(i+1)/2 and the formula fits: f(2,i) = f(2,i-1)+f(1,i-1)=1+(i-1)(i)/2+(i-1)+1=1+i(i+1)/2

Another interesting thing: Formulas like these are usually polynomials, here with the dimension of the cutted volume as order. As the value for i<=n cuts is always 2^i (use the coordinates as cuts), the n+1 trivial cases are sufficient to determine all n+1 parameters of the polynomial.