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?



Is always possible to slice a pizza in a fair way
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.


