Tiling a Framed Rectangle - Can you solve it?

17. Tiling a Framed Rectangle

You have an equal number of black and white square tiles.

You lay the black tiles in a solid rectangle. Around this rectangle you place the white tiles so that they form a border exactly one tile thick on all four sides, using every white tile and leaving no gaps or leftovers.

If every tile you own is used in this construction, what are all the possible total numbers of tiles you could have?

Added 6 October 2012

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Hint:

Write equations for the area of the inner (black) rectangle and the surrounding (white) border, then look for integer solutions.

Solution:

Let the inner (black) rectangle have dimensions m × n (in tiles). Its area is mn tiles.
The outer rectangle, after the one-tile-thick white border is added, has dimensions (m + 2) × (n + 2). Hence the number of white tiles is

(m + 2)(n + 2) − mn = 2m + 2n + 4.

Because the numbers of black and white tiles are equal, we require

mn = 2m + 2n + 4.

Rewriting,

(m − 2)(n − 2) = 8.

The positive factor pairs of 8 give all solutions:

m − 2 = 1, n − 2 = 8 ⇒ (m, n) = (3, 10)
m − 2 = 2, n − 2 = 4 ⇒ (m, n) = (4, 6)
m − 2 = 4, n − 2 = 2 ⇒ (m, n) = (6, 4) (same total as previous)
m − 2 = 8, n − 2 = 1 ⇒ (m, n) = (10, 3) (same total as first)

The corresponding numbers of black (and hence white) tiles are:

3 × 10 = 30 ⇒ total tiles = 2 × 30 = 60
4 × 6 = 24 ⇒ total tiles = 2 × 24 = 48

Therefore the only possible totals are 48 or 60 tiles.

Comments (2)

Allstar 4 March 2013

spoiler alert
48, 24 black and 24 white. The black rectangle is 4x6.

Dzallen 14 July 2016

There are only 2 solutions: 48 tiles with 4*6 black rectangle, or 60 tiles with 3*10 black rectangle

The difference in tiles between the white tiles and the outer most layer of the black rectangle is 8 tiles (2 tiles for each side). Therefore the inner black rectangle must have 8 tiles, which can be arranged as 2*4 or 1*8, giving the whole black rectangle as 4*6 or 3*10.

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