>Given a finite set of unit circles in the plane such that the area of their union $U$ is $S$, what is the largest possible bound $kS$ for some constant $k$ such that there exists a subset of mutually disjoint circles such that the area of their union is greater than or equal to $kS$?

It can be proven that a bound $k' = \frac{2}{9}$ satisfies the conditions (even though it isn't maximal) and that any number greater than $\frac{1}{4}$ for $k'$ doesn't work.

To prove that $k' = \frac{2}{9}$ is a sufficient bound, partition the plane $\pi$ into regular hexagons each having innards $2$. Then fix one of these hexagons, denoted by $\gamma$. Then for any other hexagon $x$ in the partition there exists a unique translation $\tau_x$ taking it onto $\gamma$. Define the mapping $\phi: \pi \to \gamma$ as follows: If $A$ belongs to the interior of a hexagon $x$, then $\phi(A) = \tau_x(A)$. Then we also have that the total area of the images of the union of the given circles equals $S$ and that the area of the hexagon $\gamma$ is $8\sqrt{3}$.

>**Lemma 1.** *If $U = u_1 \cup u_2 \cup u_3 \cup \cdots $ where $u_i = \{A \in U | |f^{-1}(\{f(A)\})| = i\}$, then $\text{area}(\phi(u_n)) = \frac{1}{n} \cdot \text{area}(u_n)$.*

*Proof.* We have $u_i = \{A \in U |   |f^{-1}(\{f(A)\})| = i\} = \{A \in U | \nu_{\phi}(A) = i\}$ where $\nu_{f}: A \to \mathbb{N}$ and $\nu_f(A) = |\{a' \in A| f(a') = f(a)\}|$. Then given our piecewise isometry $\phi: \mathbb{R}^2 \to \mathbb{R}^2$ we have that $\text{area}(\phi(u_n)) = \frac{1}{n} \cdot \text{area}(u_n)$ since each point in the output is counted $n$ times.

>**Lemma 2.** Using the definition for the piecewise isometry $\phi$, we have $\phi(u_i) \cap \phi(u_j) = \emptyset$ when $i \neq j$. 

*Proof.* Let $y \in \phi(u_i)$. Then $|\phi^{-1}(y)| = i$, but if $y \in \phi(u_j)$ then $|\phi^{-1}(y)| = j \neq i$, contradiction.

Now we have $\phi(U) = \phi(u_1) \cup \phi(u_2) \cup \cdots \phi(u_n)$. Then $\text{area}(\phi(u)) = \text{area}(u_1)+\frac{1}{2}\text{area}(u_2)+\cdots+\frac{1}{k}\text{area}(u_k)$. Now let $k$ be the maximum value such that $u_k \neq \emptyset$ and therefore $$8\sqrt{3} \geq \text{area}(\phi(U)) \geq \frac{1}{k}(\text{area}(u_1)+\text{area}(u_2)+\cdots+(u_k)) = \frac{1}{k} \text{area}(U) = \frac{S}{k} \implies k \geq \frac{S}{8\sqrt{3}}.$$

So take $B$ to be any $B \in \phi(u_k)$ and therefore $|\phi^{-1}(\{B\})| = k \geq \frac{S}{8\sqrt{3}}$

>**Question.** I am wondering how much more difficult it would be to achieve the best possible bound for this question. Does the same isometry help to yield a better possible bound?