Puzzle of the Week for 26 April 1999: Solution

The pigeonhole principle may be helpful in grappling with this problem. In brief, the pigeonhole principle states that if there are more pigeons than pigeonholes, and each pigeon is in a pigeonhole, there must be at least one pigeonhole containing more than one pigeon.

Here is one way to apply the pigeonhole principle to the table-arranging problem. If the room were divided into four squares, each two meters on a side, and there were more than four tables, at least one square would have to contain two tables. Yet the maximum distance between any two points in a two-meter square is between the ends of a diagonal, which (using the Pythagorean theorem) is less than three meters (about 2.83 meters, in fact).

This line of reasoning suggests that four tables is the maximum. This is true if all tables are on the same plane, as would be the case if the floor of the room were flat. If, however, four tables are placed in the corners of the room, a fifth table can be placed on a raised platform in the center. The platform would need to be at least one meter in height (in practice, higher, since the centers of the corner tables would be some distance away from the corners of the room). Since the room is four meters in each direction, it's even possible to hang four more tables from the ceiling above the first four, and still keep them separated by three meters center-to-center, for a total of nine tables. Doubtless the Mathopolis City Council will be rewriting their fire safety rules once they hear about this!