Rugs in a Room

Consider a point $Q$ , which is distributed uniformly at random in the room. Also, let us denote the event that the $i$ th rug contains the point $Q$ by $E_i$ . We have $\mathbb{P}(E_i)=\frac{1}{5}, \forall i=1,2,\ldots, 9$ . Now, using Bonferroni's Inequality , we have $1\geq \mathbb{P}(\cup_{i=1}^{9}E_i)\geq \sum_{i=1}^{9}\mathbb{P}(E_i)-\sum_{1=i<j=9} \mathbb{P}(E_i\cap E_j)\geq \frac{9}{5}-\binom{9}{2}\max \mathbb{P}(E_i\cap E_j).$ This implies, $\max \mathbb{P}(E_i\cap E_j) \geq \frac{1}{45}.$ Since the total area of the room is $5 m^2$ , it follows that the least amount of overlap area of two rugs is at least $5 \times \frac{1}{45}=\frac{1}{9}$ . Now, it is easy to come up with symmetrical rugs (see, e.g., the annular room layout in the other solution attempt) which meets this bound.

Note: This is a nice example of using Probabilistic (or more generally, Measure theoretic) ideas to solve a seemingly Geometrical problem.

we will focus on numbers (areas), and ignore the shapes, since they can be arbitrary.

this is the sequence of my thought process:

1. Assumption : the minimizing layout will not involve 3 overlapping rugs. (this start with pure intuition, proven later)

2. Layout : assuming a ring shaped room, the layout is shown below.

3. Find the overlapping area y. $\begin{aligned} \text{room area} = 9x + 9y &= 5 \\ \text{rug area} = 2x + y &= 1 \\ \\ \therefore x = \frac{4}{9} , y = \frac{1}{9} \\ \therefore a + b = 1 + 9 = 10 \end{aligned}$

4. Proving assumption in step 1. We will focus on 3 rugs. Making another assumption from intuition, minimizing layout will be symmetrical (left). It can be redistributed (right) such that (1) the total covered area remain unchanged, and (2) 3 overlapping area is 0. $\begin{aligned} \text{covered area} = x + 3y +3z &= 3a + 3b\\ \text{rug area} = x + 2y + z &= a + 2b \\ \therefore a = -\frac{1}{3} x + z , b = \frac{2}{3} x + y \\ \end{aligned}$ Hence, two rugs overlapping area decreases from $(x + y)$ to $(\frac{2}{3} x + y)$ , (partially) proving the assumption.

5. Loose ends :

   (a) 4 overlapping rugs and above?

   (b) Is there a minimizing layout which is not symmetrical?

   (c) What if $a = -\frac{1}{3} x + z < 0$ in step 4?

The solution is by no means rigorous, but my patience is limited.