Proof of the Pizza Theorem

How do we prove the Pizza Theorem?

I tried a coordinate bash (I also involved the concept of finding areas through definite integration)... But was too complicated.

I read about it at the following link: http://en.wikipedia.org/wiki/Pizza_theorem

Lemma. Consider $n\geq2$ equally spaced chords through the point $(a,0)$ of the unit disk, having slopes $\phi_k:=\phi_0+{k\pi\over n}$ $\>(1\leq k\leq n)$. Then the sum of the squares of the $2n$ resulting chord pieces is $2n$, independently of $\phi_0$.

Proof. Intersecting the line $s\mapsto (a+s\cos\phi, s\sin\phi)$ with the unit circle we get the equation $s^2+2as\cos\phi+a^2-1=0$. If $s_1$ and $s_2$ are its two solutions we can say that $$s_1^2+s_2^2=(s_1+s_2)^2-2s_1s_2=4a^2 \cos^2\phi-2(a^2-1)=2+2a^2\cos(2\phi).$$ Noting that $\sum_{k=1}^n\cos(2\phi_k)=0$ we conclude that the sum of the $2n$ considered squares is $2n$.

Proof of the Theorem. Assume that we have $4n\geq8$ equally spaced blades, and let $S(\phi)$ denote the shaded pizza area when one blade points in direction $\phi$. When $\phi\mapsto r(\phi)$ is the polar representation of the circular pizza boundary with respect to the center of the cutter then it is easy to see that $$S'(\phi)=\pm{1\over2}\sum_{k=1}^{4n} (-1)^k r^2\left(\phi+{k\pi\over2n}\right)\ .$$ Here the right side is $\equiv0$, since the alternating sum of the $4n$ blade-length-squares vanishes according to the Lemma. It is then easy to see that $S(\phi)\equiv$ half the area of the pizza.

Applying the Theorem to two concentric pizzas of radii $1$ and $1+\epsilon$ one concludes that the "crust area" is halved as well, and in the limit $\epsilon\to0+$ it follows that the shaded areas together cover half the circumference of the pizza.

Draw the unit circle and fix a point $A(x,0)$ on the $x$-axis inside the unit disk with $x>0$

Given an angle $\theta \in [0; \pi/2]$ we do the cutting in $8$ parts and color the pieces alternating red and blue. In order to prove the theorem it is enough to show that the function $\theta \mapsto \text{red area}$ is constant :

if the red area is constant then so is the blue area, and by moving $\theta$ from $0$ to $\pi/8$ we switch the red and blue areas, so both area must be equal.

To show that it is constant we differentiate the area with respect to $\theta$. What we get is the difference between $4$ very thin slices of angle $d\theta$ placed at angles $\theta + k\pi/2$, with $4$ equally thin slices placed at angles $\theta + \pi/4 + k\pi/2$.

The area of one thin slice placed at $\theta$ and with angle $d\theta$ is $l^2d\theta/2$ where $l$ is the length of the slice. By the law of cosines, we get that the $4$ lengths are the positive solutions to the equations $1 = x^2 + l^2 \pm 2xl \cos\theta$ and $1 = x^2 + l^2 \pm 2xl \sin \theta$, which are, if $c = x\cos \theta$ and $s = x\sin \theta$, $\sqrt{1-c^2} \pm s$ and $\sqrt{1-s^2} \pm c$.

The sum of the $l^2$ is $((1-c^2)+s^2+2\sqrt{1-c^2}s) + ((1-c^2)+s^2-2\sqrt{1-c^2}s) + ((1-s^2)+c^2+2\sqrt{1-s^2}c) + ((1-s^2)+c^2+2\sqrt{1-s^2}c) = 4$ is actually independant of both $x$ and $\theta$.

This proves something even stronger : if you look at the area of the unit disk covered by a cross that rotates around its center by an angle of $\theta$, then this area is $2\theta$.