Find the smallest number $a$ such that a square of side $a$ can contain five disks of radius 1 so that no two of the disks have a common interior point.
Let $P Q R S$ be a square which has the property described in the problem. Clearly, $a>2$. Let $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ be the square inside $P Q R S$ whose sides are at distance 1 from the sides of $P Q R S$, and, consequently, are of length $a-2$. Since all the five disks are inside $P Q R S$, their centres are inside $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$. Divide $P^{\prime} Q^{\prime} R^{\prime} S^{\prime}$ into four congruent squares of side length $\frac{a}{2}-1$. By the pigeonhole principle, at least two of the five centres are in the same small square. Their distance, then, is at most $\sqrt{2}\left(\frac{a}{2}-1\right)$. Since the distance has to be at least 2 , we have $a \geq 2+2 \sqrt{2}$. On the other hand, if $a=2+2 \sqrt{2}$, we can place the five disks in such a way that one is centred at the centre of $P Q R S$ and the other four have centres at $P^{\prime}, Q^{\prime}$, $R^{\prime}$ and $S^{\prime}$.