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
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.
Best Answer
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.