Proof of the Pizza Theorem

Question

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

Answer 1

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.

Answer 2

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 $2\theta$.

Answer 3

Here is my informal proof of the basic form of the pizza theorem.

Draw a circle and pick a point anywhere in it for the grid center---the point where the four chords cross. Rotate the grid to achieve another orientation if desired.

If any chord of the grid lies on top of a diagonal of the circle, the total light and dark areas are equal, by symmetry. We assume a starting point where all the chords are off-diagonal.

Turn the drawing so that one chord is horizontal and the center of the circle lies above it. This does not change relative areas, since we're just turning the whole drawing.

In the key step, move the grid center parallel to the horizontal chord until the vertical chord passes through the center of the circle. The final position has equal light and dark areas, by symmetry.

We will show that horizontal movement does not change relative areas. Since a position with equal light and dark areas is achieved without changing the proportion of each, the starting position must also have equal light and dark areas.

Proof.

First, horizontal movement of each of three chords changes generates changes in curved areas at opposite ends of each chord. But these are exactly the same shape (for a given chord), and opposing tips have different shading, so the changes cancel.

What remains are six rectangles, and a rather busy center, which we ignore for now.

In the figure, the grid center is sliding to the left, and the rectangles are colored according to which type of area is increasing during the movement.

Sliding to the left a distance $\sqrt{2} \cdot \Delta x$ increases the dark area by $$ \Delta x \cdot (\sqrt{2} a + d + f) $$

and decreases it by $$ \Delta x \cdot (\sqrt{2} b + c + e) $$

The net change is $$ \Delta A = \Delta x \ [ \ \sqrt{2} (a - b) + (d - c) + (f - e) \ ] $$

We can show that $\Delta A = 0$.

The distance from any chord center to the grid center is equal to one-half of the difference of the long and short arms of the chord.

For the two chords at $45^{\circ}$, consider the positions of the chord centers with respect to the circle center and the grid center. These four points form a rectangle with all vertices lying on an isosceles right triangle whose sides are the perpendicular bisectors of the chords. The grid center lies on the hypotenuse. The rectangle's perimeter is easily shown to be constant, using similar triangles.

The total of the two distances together is invariant and equal to $\sqrt{2}$ times the distance from the vertical chord's center to the grid center. The latter distance doesn't change during horizontal movement either.

That is: $$ \sqrt{2} (a - b)/2 = (c - d)/2 + (e - f)/2 \\ \sqrt{2} (a - b)/2 + (d - c)/2 + (f - e)/2 = 0 $$

$\Delta A = 0$ follows easily.

For the center, symmetry demands that light and dark have equal claims on its area, regardless of the exact shape required.

I've never seen this approach discussed, but it is pretty simple, so I assume it may have been given before. Much of the literature is behind a paywall.

There is a longer discussion in my geometry book (38 MB pdf).