, 2011

a. Set the square on the coordinate plane with vertices $(\pm\frac{1}{2}, \pm\frac{1}{2})$. Then the three circles with diameter $\frac{\sqrt{65}}{8}$ and centres $(0, \frac{7}{16})$, $(\frac{1}{4}, -\frac{1}{16})$ and $(-\frac{1}{4}, -\frac{1}{16})$ covers the unit square.

b. Label the square $ABCD$, and suppose that three sets $S_1$, $S_2$, $S_3$ of diameter less than $\frac{\sqrt{65}}{8}$ covers it. Then two of the vertices, without loss of generality $A$ and $B$, belong to the same set, say $S_1$. Then clearly $C$ and $D$ do not belong to $S_1$.

Suppose that $C$ and $D$ are both elements of $S_2$. Consider edge $AD$. Since $a + \frac{1}{8^2} = \frac{65}{64}$, $S_1$ can only contain points on $AD$ less than distance $\frac{1}{8}$ from $A$, and similarly $S_2$ can only contain points on $AD$ less than distance $\frac{1}{8}$ from $D$, and so neither the midpoints of $AD$ nor $BC$ belongs to $S_1 \cup S_2$. Hence both midpoints must belong to the set $S_3$, but a similar argument as above shows that $S_3$ can cover at most $2 \times \frac{1}{8} = \frac{1}{4}$ units of segment $AD$. Since $2 \times \frac{1}{8} + \frac{1}{4} = \frac{3}{8} < 1$, this configuration cannot cover $ABCD$. Hence we may suppose that $C \in S_2$ and $D \in S_3$.

As before, $S_1$ can only contain points at a distance less than $\frac{1}{8}$ from $A$, $S_3$ must contain the rest of segment $AD$. In particular, if $X$ is the point on $AD$ such that $AX = \frac{1}{8}$, then $X \in S_1$. Let $E$ be the midpoint of $CD$. By Pythagoras, the distance $EX = \frac{65}{8}$, and hence $E$ cannot belong to $S_3$. A similar argument shows that $E$ cannot belong to $S_2$ either, and clearly $E$ does not belong to $S_1$. Hence it is impossible to cover all of $ABCD$ with sets of diameter less than $\frac{\sqrt{65}}{8}$.