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.
For a 3-topping pizza, determine the probability that some region of the pizza with non-zero area is covered by all 3 toppings. (The diagram above shows an example where no region is covered by all 3 toppings.)

Solution:
Suppose that the first topping is placed on the top half of the pizza. (Again, we can rotate the pizza so that this is the case.)

Assume that the second topping is placed on the half of the pizza that is above the diameter that makes an angle of $\theta$ clockwise with the horizontal as shown. In other words, the topping covers the pizza from $\theta$ to $\theta+180^{\circ}$.

We may assume that $0^{\circ} \leq \theta \leq 180^{\circ}$. If $180^{\circ} \leq \theta \leq 360^{\circ}$, the resulting pizza can be seen as a reflection of the one shown.
<img_4026>

Consider the third diameter added, shown dotted in the diagram above. Suppose that its angle with the horizontal is $\alpha$. (In the diagram, $\alpha<90^{\circ}$.) We assume that the topping is added on the half pizza clockwise beginning at the angle of $\alpha$, and that this topping stays in the same relative position as the diameter sweeps around the circle.

For what angles $\alpha$ will there be a portion of the pizza covered with all three toppings?

If $0^{\circ} \leq \alpha<180^{\circ}$, there will be a portion covered with three toppings; this portion is above the right half of the horizontal diameter.

If $180^{\circ} \leq \alpha<180^{\circ}+\theta$, the third diameter will pass through the two regions with angle $\theta$ and the third topping will be below this diameter, so there will not be a region covered



with three toppings.

If $180^{\circ}+\theta \leq \alpha \leq 360^{\circ}$, the third topping starts to cover the leftmost part of the region currently covered with two toppings, and so a region is covered with three toppings.

Therefore, for an angle $\theta$ with $0^{\circ} \leq \theta \leq 180^{\circ}$, a region of the pizza is covered with three toppings when $0^{\circ} \leq \alpha<180^{\circ}$ and when $180^{\circ}+\theta \leq \alpha \leq 360^{\circ}$.

To determine the desired probability, we graph points $(\theta, \alpha)$. A particular choice of diameters corresponds to a choice of angles $\theta$ and $\alpha$ with $0^{\circ} \leq \theta \leq 180^{\circ}$ and $0^{\circ} \leq \alpha \leq 360^{\circ}$, which corresponds to a point on the graph below.

The probability that we are looking for then equals the area of the region of this graph where three toppings are in a portion of the pizza divided by the total allowable area of the graph.

The shaded region of the graph corresponds to instances where a portion of the pizza will be covered by three toppings.

<img_3744>

This shaded region consists of the entire portion of the graph where $0^{\circ} \leq \alpha \leq 180^{\circ}$ (regardless of $\theta$ ) as well as the region above the line with equation $\alpha=\theta+180^{\circ}$ (that is, the region with $\theta+180^{\circ} \leq \alpha \leq 360^{\circ}$ ).

Since the slope of the line is 1 , it divides the upper half of the region, which is a square, into two pieces of equal area.

Therefore, $\frac{3}{4}$ of the graph is shaded, which means that the probability that a region of the pizza is covered by all three toppings is $\frac{3}{4}$.

Final answer: $\frac{3}{4}$