fair-divisiongeometryoptimization
Two friends wants to share a pizza. One of them loves the edge of the pizza and the other one hates it. Both consider the pizza to get tastier the closer to the center you get. What is the fairest way to cut the pizza if you are only allowed four straight cuts, such that the two sets of slices sum up to the same area and one set contains all edges?
I have attached an intuitive sketch, which seems somewhat fair, but I don't know how to approach a problem like this one.
The approach uses polar coordinates. The pair $(\theta,r)$ define a point in that coordinate system, with $P$ in the origin. So for each angle $\theta$ you can compute the corresponding distance $r(\theta)$ of a point on the circle in the direction denoted by $\theta$.
The area element in polar coordinates is
$$\mathrm dA = \frac12 r^2\mathrm d\theta$$
which explains the area computation formula
$$A=\int_{\theta_1}^{\theta_2}\frac12r(\theta)^2\mathrm d\theta$$
Now in your specific case, you integrate over four areas at the same time. These areas are parametrized by angles $\theta$ which differ by multiples of $\frac\pi2=90°$. The boundaries denote one single slice.
So what is $r(\theta)$? If the pizza has its center at $(C_x,C_y)$ relative to $P$, and has a radius of $R$, then $r$ has to satisfy the equation
\begin{align*} (r\cos\theta - C_x)^2 + (r\sin\theta - C_y)^2 &= R^2 \\ r^2 - 2(C_x\cos\theta+C_y\sin\theta)r + (C_x^2+C_y^2-R^2) &= 0 \\ C_x\cos\theta+C_y\sin\theta+\sqrt{(C_x\cos\theta+C_y\sin\theta)^2-C_x^2-C_y^2+R^2} &= r(\theta) \end{align*}
I'm choosing the greater of the two solutions of the quadratic equation since the other would lead to a negative $r$, the other point where the line for angle $\theta$ intersects the circle. Polar coordinates use $r>0$.
We now want to obtain a relationship between the integrand at angles $\theta$, $\theta+\tfrac12\pi$, $\theta+\pi$ and $\theta+\tfrac32\pi$, i.e. corresponding parts from each of the four slices. In order to compute these, let's use some abbreviations.
\begin{align*} P(\theta)&:=C_x\cos\theta+C_y\sin\theta\\ D(\theta)&:=P(\theta)^2-C_x^2-C_y^2+R^2\\ r(\theta)&=P(\theta)+\sqrt{D(\theta)}\\ r(\theta)^2&=P(\theta)^2+2P(\theta)\sqrt{D(\theta)}+D(\theta)\\ P(\theta+\pi)&=-P(\theta)\\ r(\theta)^2+r(\theta+\pi)^2&=2P(\theta)^2+2D(\theta)\\ r(\theta)^2+r(\theta+\pi)^2&= 4(C_x\cos\theta+C_y\sin\theta)^2+2(R^2-C_x^2-C_y^2)\\ r(\theta+\tfrac12\pi)^2+r(\theta-\tfrac12\pi)^2&= 4(C_x\sin\theta-C_y\cos\theta)^2+2(R^2-C_x^2-C_y^2)\\ P(\theta)^2=(C_x\cos\theta+C_y\sin\theta)^2&= C_x^2\cos^2\theta+2C_xC_y\sin\theta\cos\theta+C_y^2\sin^2\theta\\ P(\theta+\tfrac12\pi)^2=(C_x\sin\theta-C_y\cos\theta)^2&= C_x^2\sin^2\theta-2C_xC_y\sin\theta\cos\theta+C_y^2\cos^2\theta\\ P(\theta)^2+P(\theta+\tfrac12\pi)^2&=(C_x^2+C_y^2)(\sin^2\theta+\cos^2\theta) =C_x^2+C_y^2\\ \sum_{k=0}^3r(\theta+\tfrac k2\pi)^2&=4(C_x^2+C_y^2)+4(R^2-C_x^2-C_y^2)=4R^2 \end{align*}
So the sum of the area elements will always be $2R^2\mathrm d\theta$, which explains why all $n$ people get equal shares of this pizza.
I don't believe the claim. Look at the figure below, in which each "thin" area is smaller than its two neighbors. Clearly the sum of the thin areas is smaller than the sum of the fat areas.
It's possible that there's some arrangement of areas (i.e., I get two large and two small, and you get two large and two small) in which the pizza area comes out equal...but it's not "alternating odd and even".
And if you move the "cut point" even closer to the edge, it's possible to get a situation in which one slice is more than half of the pizza, in which case there cannot possibly be a fair division. 
I think that your friend must have forgotten something. (And as Claude's comment shows...the missing thing is "the angles at the slice-center must all be equal", which the ones in your diagram are not.)
Here's a possible solution: pick a point $C$ to be the center, and let $P$ be the point of the circle near $C$. Draw the line $PC$, which will be a diameter, and then draw three more lines. Give one person all the slices on one side of $PC$, and the other one all the slices on the other side. They then each get half the pizza. This answers the original question, although not your friend's claim. A perhaps cleaner way:
Draw any diameter of the circle as your first line, and then draw three more lines that meet at some off-center point of that diameter. Then give each person the slices on one side of the diameter.
Best Answer
Since I raised the possibility in one of my comments, here is a possible solution that gives ~42% of the center to the edge lover. It does however rely on the freedom to displace the pieces between the cuts. Given that this is a problem of practical nature, I don't see any harm in allowing displacement. Folding the pizza before or between cuts would probably result in undesired side effects, but it is possible that could yield an even fairer distribution.