Fraction of area covered by three circles Take a square with edges of size $10$. Now take take three circles of radius $5$.


*

*Prove that you can't cover the square with these three circles.

*Find the maximum proportion of the area of the square that you can cover.


The first question comes from a high-school contest. My question is more about the second one and if it is possible, I would like to know if one can find a solution to the second one using only basic tools.
 A: I can give an illustration of the second... the answer seems to be north of 99%, but an exact value would be a grind. If you assume that the bottom two circles are going to touch the corners and middle of the bottom edge, you could determine the optimum position of the top circle, but there's no guarantee that's actually the best configuration.

A rough calculation gives 0.04 units uncovered (out of 100) using this configuration, with the top circle centred midway between the top corner and the side intersection of the bottom circles.
A: I will give an answer to first question : 
(1) Three circles $C_i$ must cover vertices $a_i$ in a square. So
there exists a circle $C_1$ s.t. it contains two vertices $a_1,\
a_2$. So the center $C_1$ is in a side in a square.
(1.1) If $C_2$ contains $a_3,\ a_4$, then there exist two points
$b_2\in \overline{a_2a_3},\ b_4 \in \overline{a_1a_4}$ s.t. $$
|a_2b_2|=|a_4b_4|=\varepsilon $$
Note that $|b_2b_4| > 10$ if $\varepsilon$ is small. Then $b_2,\
b_4$ can not be covered by $C_3$.
(1.2) If $C_2$ contains $a_3$ and if $C_3$ contains $a_4$, then
since $\varepsilon$ is small, $C_2$ must contain $b_2$ and $C_3$
must contain $b_4$. If $\varepsilon$ is small, then a midpoint in
$\overline{a_3a_4}$ can not be covered by any circles.
