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This is the theorem : Theorem. If a circular pizza is cut into $4n$ slices by $2n$ concurrent cuts (which run right across the pizza) at equal angles to each other, and n people share the pizza by taking every n’th slice (thus receiving four slices each) then they receive equal shares.

The proof is : Let $\alpha =\frac{\pi}{2n}$ and let $r(\theta)$ be the distance from $P$, the point of concurrency, to the edge of the pizza. Then the $k$’th person’s share is enter image description here

I can understand to solve the integral, but I can not understand the history of that integral can show up? and what is $\theta$ ?

Please help me, thanks.

abdmahin
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2 Answers2

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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 $\theta$.

Figure

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.

MvG
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This theorem says that;
when a circular pizza with radius R is radially cut into $4 n$ slices with the same wedge angle $\theta=\pi/(2 n)$, a total area of 4 slices cross-shaped is the same area $S=\pi R^2/n$.
So we see that as $n \rightarrow \infty$, $S(\theta)=2 R^2 \theta~\rightarrow~dS =2 R^2 d \theta$ and 4 slices cross-shaped converge to 4 line segments cross-shaped.
Assuming the lengths of these 4 line segments are $(x_1, x_2, y_1, y_2)$, we see that
$d S = (1/2)({x_1}^2+{x_2}^2+{y_1}^2+{y_2}^2)d \theta=2 R^2 d \theta$
Therefore, we only have to prove that
${x_1}^2+{x_2}^2+{y_1}^2+{y_2}^2=4 R^2=$ const.
where $(x_1, x_2, y_1, y_2)$ are arm lengths of a cross-shape made by two chords of a circle that intersects at right angles. Assuming these crossed chords are on two coordinate axes,
4 end points of the chords are $(x_1,0),(x_2,0),(0,y_1),(0,y_2)$ and the center position of the circle is $(a,b)$, the formula of the circle is
$(x-a)^2+(y-b)^2=R^2~\rightarrow~ (x^2-2ax)+(y^2-2by)+a^2+b^2-R^2 = 0$
Then, we see that
$y=0$ $ \rightarrow~ x_1+x_2=2a,~x_1 x_2=a^2+b^2-R^2$
$x=0$ $ \rightarrow~ y_1+y_2=2b,~y_1 y_2=a^2+b^2-R^2$
Thus we have
${x_1}^2+{x_2}^2+{y_1}^2+{y_2}^2 = (x_1+x_2)^2- 2 x_1 x_2+(y_1+y_2)^2-2 y_1 y_2= 4 R^2$

tossy
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