Let’s imagine that we go 2 persons to an Italian and ask for a pizza. When the bartender brings it, we prepare to divide it in 8 pieces. For it we do first 2 perpendicular courts and then other 2 for the bisectrices of the previous cuts. I joust when we are going to begin eating it, we realize that we have done the cuts… but they do not happen for the center, and skylight, now there are pieces bigger than others. Will it be possible to distribute the pizza so that each of 2 messmates eat exactly the same pizza quantity (without cutting again, skylight)?
This situation is already not a problem thanks to the mathematicians Rick Mabry and Paul Deiermann, for whom that thing about to calculate the area of every piece and to add up was too boring and they decided to complete a problem as common as this.
First of all we are going to establish the rules of the problem. We have a pizza perfectly circular to that we do to him straight courts that go from rim to rim of the pizza (ropes of the circumference), so that all the cuts happen for a point (that we are going to suppose that it is not the geometric center of the pizza), and, also, we are going to make sure that the angles that form the cuts are all be equal.

Although all this could look like a real leg-pull (mathematically speaking), the reality is that the problem comes of ancient, since the first time that appears is in 1967 when L. J. Upton (Mathematics Magazine 40 (5), p. 163) proposes the problem for 4 courts forming angles of 45th.
But let’s explore a little the problem. The simplest case is when one of the cuts happens exactly for the center of the pizza, since this way it is very easy to verify that in this case, there remains the pizza divided of symmetrical form with regard to the above mentioned cut, therefore if we begin to eat each of 2 messmates of alternative form adjacent pieces, in the end each one will take exactly half of the pizza, since this way if the first messmate takes a piece, the second messmate, he will eat (in some moment) the symmetrical piece.
Now then: what happens if no cut happens for the center of the pizza? (it would be the case of the previous drawing): this begins becoming slightly serious.
The case of 1 only cut is the simplest and it is not even necessary to explain it. For 2 courts, we can see a very very simple graphic demonstration of that the one that eats the piece that contains the center, will eat more:
In the drawing is seen how the pieces with númetros equal have the same area, then the one that eats the gray pieces (the biggest of which contains the center of the pizza), will eat more.
For the case of of 4 courts, which it was the problem raised by Upton in 1967, the challenge did not last too much, since in 1968 Michael Goldberg (and Robert Brennan and Hussein Demir, of independent form) they managed to solve the problem, and even to generalize it for the case in which there is an even number of courts (cf. Mathematics Magazine 41 p.46).
The Goldberg result says to us that if a times even number cuts the pizza (major strict that 2) so that all the cuts should happen for a point, which is not the center of the pizza and which, also, no cut spends for the center, then if we are taking adjacent pieces of alternative form, cda messmate will eat exactly half of the pizza.
But…: what happens if we do an odd number of courts? Will the same happen? In this case the things are complicated enough. The first result on this matter dates of 1994 when, again in Mathematics Magazine (vol 67, p.304), Larry Carter proposes the problem for the case of 3 courts, indicating that the solution must be that the messmate who chooses the part with the center will eat less.
So complicated has been the problem, which has not been completely solved until May, 2009, when the mathematicians Mabry and Deiremann (quoted at the beginning of the article) published the following article: Of Cheese and Crust: To proof of the Pizza Conjecture and other tasty results, American Mathematical Monthly, 116 (5), pp.423-438 (this magazine is not any, in the field of mathematics has an impact index in 2008 of 0.361 and 176 occupies the position of between 215 magazines index-linked in the Journal of Citations Report).
In the above mentioned article, using big talent of calculation and skills of algebraic series and trigonometrical functions, they manage to complete the result. This way, they manage to demonstrate that, in case of an odd number of courts, the following thing happens:

• If 3, 7, 11, 15 are realized…, 4n-1… courts, and we distribute adjacent sectors of alternative form, the messmate who eats the piece that contains the center, he will eat more than other (this also is valid if 1 is realized exactly ó 2 courts).
• On the contrary, if courts are cut in 5, 9, 13, 17…, 4n+1…, and we make the habitual share-out, of that time the messmate who takes the portion that contains the center will eat less than other.

In short, that the saying from which it departs and distributes, takes the best part I could be done realid, whenever you know the mathematics necessary to understand the Theorem of the equitable share-out of a pizza.
Nevertheless, this is not the only problem that Marbry and Deierman treat in his article, but they study some related porblemas like: Who will eat more crust? or: who will eat more cheese? or even what would happen if instead of a pizza we had a calzone.
I hope that, from now on, when you should go to an Italian with your couple and should ask for a pizza, you should remember that once, a mathematical bloguero told you the method to be able to eat more than she and look like the whole gentleman.