Problem 6

The key geometric fact is that no tile can receive mappings from two distinct black cells. Indeed, a tile is an axis-parallel rectangle. If two mapped cells used the same direction, the rectangle would contain the whole axis-parallel strip between their two side-adjacencies; the corresponding broken path through consecutive cells of $C$ or $A$ crosses that strip, so the rectangle would either cover a black cell or cross a path boundary, both impossible. If the two directions are different, the two relevant path arcs meet only at $c$; rectangle convexity again forces the tile across that crossing, and the only possible common source is the single cell $c$ itself. Thus all mappings from distinct black cells point to distinct tiles. There are $n-4$ successful first mappings: exactly one can fail at each of the first row, last row, first column, and last column. Every cell of $(A\cup C)\setminus\{c\}$ supplies one additional successful direction, giving $a+b-2$ more mappings, while $c$ supplies three additional directions. Consequently the covering has at least $(n-4)+(a+b-2)+3=n+a+b-3$ tiles. From $ab\ge n$ and AM-GM, $a+b\ge2\sqrt{ab}\ge2\sqrt n=2k$, so the number of tiles is at least $n+2k-3$. For $n=2025=45^2$, this lower bound is $2025+90-3=2112$, matching the explicit construction.