We want to cover an m×m square with n non-overlapping axis-parallel polyominos such that both the interior and the boundary of the square are divided into n equal areas and n equal lengths, respectively. Obviously, m and n cannot be arbitrary for there to be a solution. For example, when n=4, there’s a solution for any m that’s a multiple of 2, and it’s not hard to construct the solution: just divide the square into four equal squares of side m/2.
Here’s a slightly nontrivial example with n = 3 and m = 6:
Indeed, the polyominos divide both the interior and the boundary of the square into three equal parts of area 12 and length 8 respectively. It is easy to check that m = 3 doesn’t have a solution, so m = 6 is the minimum possible value for which there’s a solution.
- For n = 5, what’s the smallest value of m for which there’s a solution?
- Bonus: What about general n?
Clarification: While a polyomino is connected in the interior, the boundary it shares with the big square need not be connected. In that case, the length of the shared boundary is the sum of lengths of all components.
