Are the following tilings possible?

Question

Homi · Accepted Answer

Understand the Question

For a region to be tiled completely without overlaps or gaps, two conditions must be satisfied:
1. Area condition: The total area of the region must be exactly divisible by the area of a single tile.
2. Coloring / Invariant condition: When squares are colored alternately like a chessboard (black and white), each $1 \times 2$ domino covers exactly one black and one white square. Therefore, the board must have an equal count of black and white squares.

(i) Is Tiling 1 possible? ( $3 \times 3$ square with one square removed, tiled with L-shaped tiles of 3 squares)

Step 1 · Check Area Divisibility for Tiling 1

Diagram 1

Total area of a $3 \times 3$ square: $3 \times 3 = 9 \text{ squares}$

Area of the region with $1$ square removed: $9 - 1 = 8 \text{ squares}$

Area of each L-shaped tile $= 3 \text{ squares}$ .

Dividing the region's area by the tile's area: $8 \div 3 = 2 \text{ with a remainder of } 2$

Since $8$ is not divisible by $3$ , the tiling is not possible.

Answer

(i) No

(ii) Is Tiling 2 possible? ( $7 \times 8$ rectangle with two squares removed, tiled with $1 \times 2$ domino tiles)

Step 1 · Check Area Divisibility for Tiling 2

Total area of a $7 \times 8$ rectangle: $7 \times 8 = 56 \text{ squares}$

Area of the region with $2$ squares removed: $56 - 2 = 54 \text{ squares}$

Area of each $1 \times 2$ tile $= 2 \text{ squares}$ .

Dividing the area of the region by the tile's area: $54 \div 2 = 27$

Since $54$ is divisible by $2$ , the area condition is satisfied.

Step 2 · Check Chessboard Coloring Invariant

Diagram 2

A full $7 \times 8$ chessboard has $56$ squares with an equal number of black and white squares: $56 \div 2 = 28 \text{ white squares and } 28 \text{ black squares}$

Each $1 \times 2$ domino covers exactly $1$ white square and $1$ black square.

Determining colors of the two removed squares (assuming top-left square $(1, 1)$ is white):

Top-right square at $(1, 8)$ : $1 + 8 = 9$ (odd $\implies$ black square)
Bottom-left square at $(7, 1)$ : $7 + 1 = 8$ (even $\implies$ white square)

Remaining squares: $\text{Number of white squares} = 28 - 1 = 27$ $\text{Number of black squares} = 28 - 1 = 27$

Since the number of remaining white squares equals the number of black squares, both the area and coloring conditions are satisfied.

Answer

(ii) Yes

Common Mistakes

Checking only the area condition: Area divisibility is necessary but not sufficient on its own. For domino tilings, the number of black and white squares must also balance.
Assuming opposite corners have the same color: In an odd $\times$ even board (like $7 \times 8$ ), the top-right and bottom-left corners have opposite parities and thus different colors (one black, one white).

Question 24

More questions in FIO