Cutting a pizza – it’s easy as pi

A bunch of mathematicians (no doubt uni students) have attempted to solve the dilemma of distributing pizza slices evenly to people who have made equal contributions to the pizza buying cause. This article explains.

The problem that bothered them was this. Suppose the harried waiter cuts the pizza off-centre, but with all the edge-to-edge cuts crossing at a single point, and with the same angle between adjacent cuts. The off-centre cuts mean the slices will not all be the same size, so if two people take turns to take neighbouring slices, will they get equal shares by the time they have gone right round the pizza – and if not, who will get more?

It’s complex. Apparently. If you have two diners, and the pizza is cut an even number of times, the trick is to take alternate pieces.

It has been known since the 1960s that when N is even and greater than 2, an answer to the first question is for Gray and White to choose alternate slices about the point P of concurrency.

The conclusion – from the paper that’ll cost you $20 to buy – was this:

It was conjectured by Stan Wagon and others, that for N=3,7,11,15,…, whoever gets the center gets the most pizza, while for N=5,9,13,17,…, whoever gets the center gets the least. We prove this Pizza Conjecture by first showing its equivalence to a (pretty wild) trigonometric inequality. This inequality is proved with the aid of a theorem that counts lattice paths. Our main theorem is sufficiently general that, as a bonus, results concerning the equiangular slicing of other dishes are obtained.

One can only assume all this would be easier with one of these plates.