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Given a pizza, represented by the unit disk $D_1(0,0)=\{(x,y)\in\mathbb{R}^2\mid \|(x,y)\|\leqslant 1\}$, and given $N$ slices of $r$-pepperoni, represented by disks $D_r(a_i,b_i)=\{(x,y)\in\mathbb{R}^2\mid \|(x,y)-(a_i,b_i)\|\leqslant r\}$, for $i\in[1..N]$.

Assume that the $N$ slices are randomly distributed on the pizza and that:

  1. for each $i\in[1..N]$, $D_r(a_i,b_i)\subseteq D_1(0,0)$; and
  2. for each $i,j\in[1..N]$, if $i\neq j$ then $D_r(a_i,b_i)\cap D_r(a_j,b_j)=\emptyset$.

Suppose that the pizza is shared by $2$ people. Is it always possible to slice the pizza along a single straight line such that the two parts have the same area (same amount of pizza) and the same total area of pepperoni?

What about generalizations to $K$ people and cuts along line segments?

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The 2-person version is simply the 2-dimensional version of the Ham Sandwich Theorem: . – Robin Chapman Oct 6 '10 at 15:50
Thanks for the references. A recursive application of the Borsuk-Ulam Theorem seems to provide a solution for $K$ people. First slice the pizza in two, then for each of the parts $P_1$ and $P_2$, define a function $f_i$ from the unit sphere to the real numbers, such that $f_i(p)$ is the amount of pepperoni on the left side of the hyperplane $H_i(p)$ with normal vector $(0,p)$ and such that $H_i(p)$ slices $P_i$ in two parts of $1/3$rd and $2/3$rd of the area of $P_i$, respectively. The antipodal points then provide a way of cutting $P_i$. This can be repeated for the 4rd person, and so on. – Sirolf Oct 7 '10 at 11:20
up vote 2 down vote accepted

Intermediate Value Theorem (for two people).

We draw a $x$-axis through the centre of the disk. For an $\theta \in R$ we draw an axix which makes an angle of $\theta$-degrees with the x-axis (note that the axix for $\theta+\pi$ gives exactly the opposite direction).

Lets look now at the two half disks defined by this $\theta$-axis.

Let $f(\theta)$ denote the quantity of peperoni on the left half disk minus he quantity of peperoni on the rigth half disk (since the axis is oriented, left-rigth make sense, one looks in the direction of the axis).

$f$ is continuous in $\theta$, and since $f(\theta)+f(\theta+\pi)=0$, either $f \equiv 0$ or $f$ takes both positive and negative values. Now the IVT completes the proof.

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Thanks, this is helpful. I would still like to see a method for more than two people. – Sirolf Oct 7 '10 at 9:39
There should be some generalization of the ham sandwich theorem that takes care of more than two people (I think). It is trivial to generalize to a power of two people, for example (by iterating ham sandwich thm). However, a related, and mostly unsolved problem, is to divide a polygon into n parts with equal area and perimeter (some special cases are known only). – Per Alexandersson May 13 '13 at 21:19

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