Balti Tee 1994 · Ülesanne 13

Solution: Let $PQRS$ be a square which has the property described in the problem. Clearly, $a > 2$. Let $P'Q'R'S'$ be the square inside $PQRS$ whose sides are at distance $1$ from the sides of $PQRS$, and, consequently, are of length $a - 2$. Since all the five disks are inside $PQRS$, their centres are inside $P'Q'R'S'$. Divide $P'Q'R'S'$ 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 $PQRS$ and the other four have centres at $P'$, $Q'$, $R'$ and $S'$.