# How to divide a pizza into $n$ parts?

Let's say you have invited $(n-1)$ people for dinner. You decide that the main course consists of one pizza for each guest, so you order $n$ pizzas. Unfortunately, the pizza guy on the scooter trips on his way to your house and loses all but one pizza.

As you don't want to disappoint your guests too much, you try to divide the (round) pizza up into $n$ equal pieces, so everyone gets an equal share. You are, however, not in the company of someone with a compass. You do have an unmarked ruler and a pen that can leave a colored, eatable solution on the pizza. You can also fold the pizza as much as you want. You can use the creases on the pizza that are left one the surface of the pizza after one has unfolded it.

We assume that the pizza is perfectly round.

Question: Is there a method by which we can divide the pizza into any $n$ equal parts using the prescribed materials and rules?

Thanks,

Max

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By the way, the pizza guy on the scooter is all right after he trips... – Max Muller Apr 24 '11 at 20:41
Should the parts be equal in area or also equal in shape? – joriki Apr 24 '11 at 20:42
Here's how I interpret the question; @Max, please correct me if I'm wrong: We don't have a compass, so we can't draw arbitrary circles. The ruler is marked and stays unmarked. The pen is for marking the pizza, not for marking the ruler. This is a mathematical problem and not an exercise in lateral thinking, so the idea is not to figure out how to use the given instruments to draw arbitrary circles without a compass, but to decide whether $n$ equal shares can be constructed using only an unmarked ruler and folding. (Note that folding allows us to draw certain circle arcs.) Correct? – joriki Apr 24 '11 at 21:03
This is why you should always invite Godel, he won't eat anything anyway. :-) – Asaf Karagila Apr 24 '11 at 21:28
There's always $k$ people that are crust hogs, $j$ people that are crust tolerant and $n-k-j$ people that don't eat the crust. Very important! – Ryan Budney Apr 25 '11 at 0:33

Quick trip to google brought this up.

Here is evidence that it can be done:

It is possible to do all compass and straightedge constructions without the straightedge. That is, it is possible, using only a compass, to find the intersection of two lines given two points on each, and to find the tangent points to circles. It is not, however, possible to do all constructions using only a straightedge. It is possible to do them with straightedge alone given one circle and its center.

The Circle is our pizza.

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It is not the answer to the question, but thanks for making me lookt at this beautiful solution. – Phira Apr 24 '11 at 20:52
@user9325, the solution is edited. It MAY be a proper solution now. – picakhu Apr 24 '11 at 21:31
@ picakhu thank you for your answer. It's getting pretty late in the Netherlands at the moment so I'm going to sleep now and read through everything carefully tomorrow. – Max Muller Apr 24 '11 at 21:39
@picakhu: a) That's a passage in a Wikipedia article, not a proof. b) It only means that you can do all constructions of points that way; it doesn't mean you can construct circles with a radius different from the given circle. So you can't use this to construct the nice semicircle solution that you linked to. – joriki Apr 24 '11 at 23:03
@picakhu I'm sorry but I've read all the links and I agree with joriki and user9325, as you need to construct the semicircles in the solution you need to, something that (apparently) can't be done by means of straightedge alone (I can understand your link though, I think the wikipedia article is a bit misleading...) – Max Muller Apr 25 '11 at 12:12