Given a rectangular area that needs to be completely paved using 1 metre square paving stones (for simplicity's sake), the number of stones that are used around the perimeter of the rectangle must equal the number used for the interior of the rectangle. In other words, the area of the perimeter stones must equal the area of the interior.

What are the dimensions of such a rectangular area?

What I know:

    - There are two possible answers. They are 12-by-5 and 8-by-6.
    - Because you are using 1 metre square paving stones, the dimensions of the rectangular area therefore must be a whole number.

I managed to work out the first dimension as I was given that the perimeter was equal to 30, and solving for simultaneous equations resulted in 12-by-5. The 8-by-6 was given as another alternative answer and claimed to be the only other such possible dimension.

So my question is, is there a way to find these two possible answers algebraically without going into something convoluted like going through every possible whole number or something?

~Sorry if this is in the wrong area or if this question has already been asked before. Cheers~ ^_^


Let's say the dimension of the area is $a \times b$ with $a \ge b$.

The area of the interior is $(a-2)(b-2)$. If this is half of the whole area, one has

$$2(a-2)(b-2) = ab \iff ab-4a-4b+8 = 0 \iff (a-4)(b-4) = 8$$ This leads to $$(a-4, b-4) = (8,1) \text{ or } (4,2) \quad\implies\quad (a,b) = (12,5) \text{ or } (8,6)$$

  • $\begingroup$ Omgsh, I went in a complete circle trying to figure out something convoluted like $(x-12)(y-5)(x-8)(y-6)=0$ and forgot about simply finding the possible answers from a non-zero equality, you're a life saver man... <3 $\endgroup$ – sk9c00 Jul 27 '17 at 5:07

Even without seeing the algebraic approach by achille hui you can quickly put limits on the number of possibilities.

Taking the longer side as $a$ and the shorter as $b$, it's clear that $b>4 $ otherwise there is no way for the interior to reach half the total area - at $b=4$, all the interior stones match to the adjacent bordering pavers on the long side and there are none left to match to the $b$ edges' pavers.

Then you can check $b=5$ and $b=6$ for the solutions you already found - the $a$ sides must be $2$ more than space occupied by $2b$ pavers on the appropriate number of interior rows not adjacent to the perimeter, $a-2 = 2b / (b-4)$ . However $b=7$ gives the non-integer solution $a=2+14/3 = 20/3<b$ indicating that we have already exhausted the solutions.


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