Subfield: Combinatorics


Question: At Pizza by Alex, toppings are put on circular pizzas in a random way. Every topping is placed on a randomly chosen semicircular half of the pizza and each topping's semi-circle is chosen independently. For each topping, Alex starts by drawing a diameter whose angle with the horizonal is selected

<img_3522>
uniformly at random. This divides the pizza into two semi-circles. One of the two halves is then chosen at random to be covered by the topping.
Suppose that $N$ is a positive integer. For an $N$-topping pizza, determine the probability, in terms of $N$, that some region of the pizza with non-zero area is covered by all $N$ toppings.

Solution:
The main idea of this solution is that the toppings all overlap exactly when there is one topping with the property that all other toppings "begin" somewhere in that toppings semi-circle. In the rest of this solution, we determine the probability using this fact and then justify this fact.

Suppose that, for $1 \leq j \leq N$, topping $j$ is put on the semi-circle that starts at an angle of $\theta_{j}$ clockwise from the horizontal left-hand radius and continues to an angle of $\theta_{j}+180^{\circ}$, where $0^{\circ} \leq \theta_{j}<360^{\circ}$. By establishing these variables and this convention, we are fixing both the angle of the diameter and the semi-circle defined by this diameter on which the topping is placed.

Suppose that there is some region of the pizza with non-zero area that is covered by all $N$ toppings.

This region will be a sector with two bounding radii, each of which must be half of a diameter that defines one of the toppings.

Suppose that the radius at the clockwise "end" of the sector is the end of the semi-circle where topping $X$ is placed, and that the radius at the counter-clockwise "beginning" of the sector is the start of the semi-circle where topping $Y$ is placed.



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This means that each of the other $N-2$ toppings begins between (in the clockwise sense) the points where topping $X$ begins and where topping $Y$ begins.

Consider the beginning angle for topping $X, \theta_{X}$.

To say that the other $N-1$ toppings begin at some point before topping $X$ ends is the same as saying that each $\theta_{j}$ with $j \neq X$ is between $\theta_{X}$ and $\theta_{X}+180^{\circ}$.

Here, we can allow for the possibility that $\theta_{X}+180^{\circ}$ is greater than $360^{\circ}$ by saying that an angle equivalent to $\theta_{j}$ (which is either $\theta_{j}$ or $\theta_{j}+360^{\circ}$ ) is between $\theta_{X}$ and $\theta_{X}+180^{\circ}$. For each $j \neq X$, the angle $\theta_{j}$ is randomly, uniformly and independently chosen on the circle, so there is a probability of $\frac{1}{2}$ that this angle (or its equivalent) will be in the semicircle between $\theta_{X}$ and $\theta_{X}+180^{\circ}$.

Since there are $N-1$ such angles, the probability that all are between $\theta_{X}$ and $\theta_{X}+180^{\circ}$ is $\frac{1}{2^{N-1}}$.

Since there are $N$ possible selections for the first topping that can end the common sector, then the desired probability will be $\frac{N}{2^{N-1}}$ as long as we can show that no set of angles can give two different sectors that are both covered with all toppings.

To show this last fact, we suppose without loss of generality that

$$
0^{\circ}=\theta_{1}<\theta_{2}<\theta_{3}<\cdots<\theta_{N-1}<\theta_{N}<180^{\circ}
$$

(We can relabel the toppings if necessary to obtain this order and rotate the pizza so that topping 1 begins at $0^{\circ}$.)

We need to show that it is not possible to have a $Z$ for which $\theta_{Z}, \theta_{Z+1}, \ldots, \theta_{N}, \theta_{1}, \theta_{2}, \ldots, \theta_{Z-1}$ all lie in a semi-circle starting with $\theta_{Z}$.

Since $\theta_{Z}<180^{\circ}$ and $\theta_{1}$ can be thought of as $360^{\circ}$, then this is not possible as $\theta_{1}$ and the angles after it are all not within $180^{\circ}$ of $\theta_{Z}$.

Therefore, it is not possible to have two such regions with the same set of angles, and so the desired probability is $\frac{N}{2^{N-1}}$.

Final answer: $\frac{N}{2^{N-1}}$