Tile Pattern at the Community Pool
A rectangular pool deck is being tiled with border tiles around a central square of blue tiles. For a blue square that is 1 tile by 1 tile, the border uses 8 tiles. For a 2 by 2 blue square, the border uses 12 tiles. For a 3 by 3 blue square, the border uses 16 tiles. The tiler claims the rule is 'border = 4 × side + 4,' so a 10 by 10 blue square needs 44 border tiles. A second worker says the pattern grows by 4 each time, so the border for a 10 by 10 square should be found by extending a table, and gets 44 as well but is unsure why. The deck can hold at most 200 border tiles.
1. Test the claimed rule against the three given cases, then explain why the rule works using the geometry of a border around a square.
Read the worked answer
Check each case with border = 4s + 4:
- s = 1: 4(1) + 4 = 8 ✓
- s = 2: 4(2) + 4 = 12 ✓
- s = 3: 4(3) + 4 = 16 ✓
The rule fits all three cases. Geometry explains it: each side of the blue square has s border tiles along it, giving 4s tiles, but the four corner tiles are each counted once already, so actually the four sides contribute 4s tiles and the four corners add 4 more. Equivalently, the border is a ring one tile thick; going around, each side contributes s tiles and each corner contributes 1, so total = 4s + 4. The rule is not just a coincidence of three data points; it follows from the shape. So 44 border tiles for s = 10 is correct.
2. The tiler wants the largest blue square whose border still fits within 200 tiles. Solve for the side length and check your answer.
Read the worked answer
Set 4s + 4 ≤ 200.
Subtract 4: 4s ≤ 196. Divide by 4: s ≤ 49.
So the largest blue square is 49 by 49. Check: 4(49) + 4 = 196 + 4 = 200, exactly the limit. A 50 by 50 square would need 204 tiles, which exceeds the limit.
A strong answer also notes that s must be a whole number of tiles, so 49 is the maximum integer. Extending a table by adding 4 each step would reach the same place but is slower and easier to miscount; solving the inequality is more reliable.