# What is the maximum number of pieces that a pizza can be cut into by 7 knife cuts? (NBHM 2005)

I am seeing this question very first time and do not know any formal way to solve it. Which part of mathematics it is related to?

What is the maximum number of pieces that a pizza can be cut into by 7 knife cuts?

A pizza is usually round shaped. So I was cutting a circle into pieces by seven straight lines. Minimum number is eight. But no maximum I am getting.

It is better to add complete answer with references. Thank you for your help.

• This sounds like something belonging to Combinatorial Geometry. – Austin Mohr Jan 21 '14 at 17:12
• math.stackexchange.com/questions/169432/… would be of some help to you i believe... – user87543 Jan 22 '14 at 1:42
• This link is helpful. – Supriyo Jan 22 '14 at 2:23

## 5 Answers

This is a known problem: it is appropriately called the Circle Cutting Problem.

I suggest to you to read the linked page --- it is quite informative. However, the important part is that with $n$ cuts, you can divide a circle into at most $f(n)$ parts, where:

$$f(n) = \frac12\left(n^2 + n + 2\right)$$

In your case, $f(7) = \frac12(49+7+2) = \frac12(58) = 29$.

• Thank you for the link and answer. – Supriyo Jan 22 '14 at 2:23
• @Newb How do we prove the above formula? – user464147 Oct 18 '17 at 16:44

Let's apply Euler's polyhedron formula$$1-V+E-F+1=0$$ to the plane graph formed by the $n$ chords (straight knife cuts), assumed to be pairwise intersecting with no three concurrent, and the rim of the pizza. The number we want is $$f(n)=F-1=E-V+1$$ since we don't want to count the outer face which has no pizza. The number of vertices is$$V=\binom n2+2n,$$i.e., $\binom n2$ points where two chords cross, and $2n$ points where chords meet the rim. The number of edges is$$E=n^2+2n,$$i.e., each of the $n$ chords is cut into $n$ segments, and the rim in cut into $2n$ arcs. Our final answer is $$f(n)=E-V+1=n^2+2n-\binom n2-2n+1=\frac{n^2+n+2}2=\binom{n+1}2+1$$and so, for $7$ cuts,$$f(7)=\binom82+1=29.$$

• Ooh, very nice. – Newb Jan 21 '14 at 20:07
• Thank you for this nice construction. But I do not know algebraic topology. – Supriyo Jan 22 '14 at 2:24
• Neither do I. I think Euler's polyhedron formula was mentioned in the solid geometry class I took in 12th grade. – bof Jan 22 '14 at 6:30

When you make a straight cut through a whole or already-sliced pizza, you divide each piece that you slice through into two pieces; that is, you create $p$ new pieces, where $p$ is the number of pieces you slice through. Your cut starts in some piece, and then crosses into a new piece for each line it traverses; so $p=c+1$, where $c$ is the number of cuts you slice through. And for the $n$-th cut, you can slice through at most all of the preceding $n-1$ cuts, so $p=c+1\le n$.

Putting this together, the maximum number of pieces after $n$ cuts is at most $$1+1+2+3+\ldots+n=1+n(n+1)/2,$$ or $29$ pieces after $7$ cuts.

• Thank you for your answer. – Supriyo Jan 22 '14 at 2:22

Let us try to cut the pizza into as many pieces as possible. Put $p_n$ as the number of pieces after $n$ cuts. Then $p_0 = 1$ and $p_1 = 2$. Next $p_3 = 4$ if we cut across the first cut. For a general $n$, we cut across all $n$ lines and get $p_n = n + p_{n-1}$. Hence $$p_n = 1 + \sum_{k=1}^n p_k = 1 + {n(n+1)\over 2}.$$

• Thank you for your answer. – Supriyo Jan 22 '14 at 2:22

For 'n' number of cuts,

i) Mininum Number of pieces = (n+1)

ii) Maximum number of pieces = Sum of the values of n

Eg : For 3 cuts, Min pieces = 3+1 = 4 Max pieces = 1+2+3 = 6

• What's the reasoning behind this answer? – HSN Jan 25 '17 at 12:50
• Max pieces is 7 for 3 cuts – RobAu Oct 1 '17 at 15:56