Rectangular Ground with Square tiles

Question

A rectangular ground is to be filled with square tiles of unit area. The border of the ground should be of red coloured tiles and inside white. If the number of red and white tiles are the same, find the minimum dimensions of the ground.

Clearly, the ground is $0 \times 0$. There are no red tiles, and no white tiles, and since $0 = 0$, this satisfies the conditions. — Peter Shor, Jul 04 '13 at 13:19
Peter Shor's quick answer notwithstanding it is a cute diophantine problem to determine the possible dimensions $a$, $b$, and $r$ of this pattern, where $a$ and $b$ denote the length and the width of the white inside rectangle and $r$ the width of the red frame. — Christian Blatter, Jul 04 '13 at 20:07

score 2 · Answer 1 · answered Jul 08 '13 at 12:14

2

The diophantine equation is $ab = 2(a-2)(b-2)$. To solve it, first suppose both of the numbers are at least 7. Since $\left(\frac{7}{5}\right)^2 < 2$, in this case there are no solutions. Thus, at least one of the numbers must be 6 or less. This leaves you with a very small number of linear equations (plug in $a$ $=$ 3, 4, 5, 6), which you can solve to find all the solutions. Then select the correct one.

answered Jul 08 '13 at 12:14

Peter Shor

3,314

This assumes that the border is one tile wide. Otherwise you have $ab = 2(a-2r)(b-2r)$ where $r$ is the width of the red tile border. That said, I think the minimal solution required by the OP is achieved when $r=1$. – Peter Phipps Jul 08 '13 at 14:22
If $r\geq 2$, it is easy to prove that the minimum area has to be $81$ (since both edges must be at least $4r+1$). This is enough to show that the $r=1$ solution is minimal. – Peter Shor Jul 09 '13 at 16:53

score 1 · Answer 2 · edited Sep 18 '13 at 22:46

Note that the square titles is of unit area !!! So the number of white tiles will be $(l-2)(b-2)$ and red tiles will be $lb-(l-2)(b-2)$ and both should be equal so by equating we will get $lb=2(l-2)(b-2)$ area must be positive.clearly l and b must be greater than 2 ! On deducing that equation we will arrive at $l+b=\frac{lb-8}{4}$, which implies {lb>8} and $lb-8=4k$ (k-constant) we get $lb=4k+8$ by substituting different values of k say 1,2,3,4 we will get lb and cross check with $l+b=\frac{lb-8}{4}$.

score 1 · Answer 3 · edited Sep 18 '13 at 22:43

1

The answer is $48$. Satisfying the below condition

$$lb = 2(l-2)(b-2)$$

edited Sep 18 '13 at 22:43

Stefan4024

35,843

answered Sep 18 '13 at 22:18

Abdul Rahman

11

Rectangular Ground with Square tiles

3 Answers3