Max Chunks from 5 Cuts in a Sphere

By cutting a sphere with five planes, what is the maximum number of resulting chunks you can obtain?

Introduction:

Most of the time, with this type of problem, solving a simpler or particular case is the key to understanding the context. So, let us look at the 2-dimensional case of cutting a circle.

0 lines will generate one piece, the original circle.
1 line will generate exactly two pieces (if the line is not tangent to the circle)
2 lines will create a maximum of 4
a third line will subsequently add a maximum of three new pieces, for a total of 7.

Changes in number of pieces when adding one new cut

Solution:

To maximize the areas created, the (k + 1)th line will have to not pass through any existing intersection point and intersect with all the previous lines inside the given circle.

Right now, we will not prove that this is indeed possible, but you should try and understand why this is the case. We formalize the set of assumptions we described before.

If we focus solely on the latest added line, we see that it has two intersection points with the circle, and between them, k intersection points with the k lines. This means that the segment inside the circle splits into other k+1 distinct segments.

Each of them is inside one of the slices we had prior and generates a new border. So, k+1 of the previous pieces are split in two.

Formally, $A(k+1)$ is $A(k) + k + 1$ .

We have the desired recursion for A, with the general formula $\dfrac{n(n+1)}{2} + 1$ .

Now, we extrapolate to the $3D[math] case, the one we were interested in from the beginning. We again formalize the number of chunks the sphere can be cut into by n planes, by denoting it with [math]S(n)$ . It’s easy to see that $S(1)$ is two and $S(2)$ is four. Again we look at the situation when we add a new plane to an existing configuration. We know from before that this circle will be split into a maximum of $A(k)$ pieces, and, in a similar manner to the $2D$ case, each piece will split one of the existing chunks in two.

Setting a similar set of assumptions for the addition of a new plane, we can conclude that, when moving from k to k+1 planes, we can add a maximum of $A(k)$ chunks.

Expanding this recurrence, we get that $S(n) = \dfrac{n^3 + 5n}{6 + 1}$ .

Note:

This problem is just asking for a particular “Cake Number”; this topic is also related to the Lazy Caterer’s Sequence.

Max Chunks from 5 Cuts in a Sphere: Solving the Puzzle

By cutting a sphere with five planes, what is the maximum number of resulting chunks you can obtain?

Introduction:

Solution:

Note:

Video Solution

Feel free to share your thoughts and ideas in the Latex-enabled comment section below!

Leave a Reply Cancel reply