We covered the unit square with three congruent circular disks. What is the minimum radius of the disks?
We will show that an unit square can be covered by three circles of radius $\frac{\sqrt{65}}{16} \approx 0.5039$, but not by smaller circles.

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Let $E$ be the midpoint of the side $CD$ of the unit square $ABCD$, and let $F$ and $G$ be the points on the adjacent sides $AD$ and $BC$, respectively, such that $AG = GE = EF = FB = 2r$ (see figure). Simple calculation shows that in this case $AF = BG = \frac{1}{8}, r = \frac{\sqrt{65}}{16}$. If we now place three circles of radius $r$ such that their centers are the midpoints of the segments $AG$, $GE$, and $EF$, respectively, then the first circle passes through points $A, B, F, G$; the second through points $C, E, G$; and the third through points $D, E, F$, and the three circles clearly cover the entire unit square.

Suppose that the unit square can be covered by three circles of radius $r' < r$. Then the points $A$ and $B$ must be covered by the same circle, which we will call the first circle, because if $A$ were covered by the first and $B$ by the second circle, then $AB = 1 > 2r'$. The points $G$ and $F$ must also be covered by the same circle, which we will call the second circle, because if $G$ were covered by the second and $F$ by the third circle, then $EF = EG = 2r > 2r'$ and $AE = \frac{\sqrt{80}}{8} > 2r > 2r'$, so the point $E$ could not be covered by any circle. The second circle cannot cover the points $C$ and $D$ because $CF = DG = \frac{\sqrt{113}}{8} > \frac{\sqrt{65}}{8} = 2r > 2r'$, so these points must be covered by the third circle.

In summary: the first circle covers points $A$ and $B$, the second circle covers points $F$ and $G$, and the third circle covers points $C$ and $D$. Now consider the midpoint $H$ of the side $AD$! Using the Pythagorean theorem, we get that

$$
BH = CH = \frac{\sqrt{80}}{8} > \frac{\sqrt{65}}{8} = 2r > 2r'
$$

and

$$
GH = \frac{\sqrt{73}}{8} > \frac{\sqrt{65}}{8} = 2r > 2r'
$$

Thus, the point $H$ cannot be covered by any of the three circles. This is a contradiction, so the unit square indeed cannot be covered by three circles of radius smaller than $\frac{\sqrt{65}}{16}$.