Problem 900

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 $\frac{\sqrt{65}}{16}$.