## Bordered mat

Posted by Chris on March 21, 2012 – 6:34 am

I’ve purloined this from the old site.

A table mat is made from red and white beads. There is an inner rectangle made entirely using white beads. It has a border (on all four sides) made from a single layer of red beads. Which mats can have as many white beads as red beads?

All the beads are the same size.

March 21st, 2012 at 7:48 am

There are only two possible mats.

March 21st, 2012 at 8:03 am

I agree, only 2 mats possible.

This is algebra…surprisingly from Chris.

let x and y be the dimensions of the inner (white) rectangle. The number of white beads is x*y. The number of red beads around the rectangle would be x+y+x+y+4 corner beads = 2x+2y+4. We need to find where these two are equivalent. x*y=2x+2y+4

I’ve already done the simplifying and found the solutions. They are so simple that you probably could have drawn out several dots on some paper and counted.

March 21st, 2012 at 11:36 pm

For a “square” mat the equation in post 2 gives x = y < 5.

So if y is the shorter side it can only be 2, 3 or 4.

y = 2 doesn't work so we can only have dimensions of 8 x 3 or 6 x 4.

March 22nd, 2012 at 6:58 am

Hi Wiz, you got the right sizes last time. You only got one right this time.

I think you made a simple slip up.

I’m hoping to see a proof (i.e. not involving trial and error) that there are only two possible mats. I suspect that DP has one.

March 22nd, 2012 at 11:19 am

In post 2, DP showed that x*y=2x+2y+4 where x and y are the dimensions of the inner rectangle. This may be rearranged as (x-2)y = 2x + 4, and implies that x has to be strictly larger than 2 (because the right hand side is strictly positive). The same symmetrical argument holds for y.

Rearranging again the equation leads to

y = (2x+4) / (x-2) ≥ 3

which implies 2x + 4 ≥ 3x -6

and simplifies to x ≤ 10

Therefore, the only possible solutions are such that 3 ≤ x ≤ 10. The only integer possibilities are 3 x 10 and 4 x 6.

March 22nd, 2012 at 11:57 am

Hi Zorglub. I’ve replaced your \geq and \leq with ≥ and ≤.

You can use “& g e ;” and “& l e ;” (without the quotes or spaces) to get those symbols.

You’ve got the right answer, and despite some hidden trial and error, I’d say you demostrated that no other results are possible.

I’ll add that it’s not the way I know to do it.

March 22nd, 2012 at 1:12 pm

Thanks for the &ge and &le.

I am looking forward to see your demonstration.

March 22nd, 2012 at 7:47 pm

For consistency with the previous posts, let x and y be the dimensions of the inner (white) rectangle. Assume y ≥ x. Then we require

xy = 2x + 2y + 4 => xy – 2x – 2y + 4 = 8 => (x – 2)(y – 2) = 8 = 2*2*2

By the fundamental theorem of arithmetic we must have one of the following:

x – 2 = 1, y – 2 = 8 => x = 3, y = 10

or

x – 2 = 2, y – 2 = 4 => x = 4, y = 6

QED

March 24th, 2012 at 5:38 am

That is very nice.

March 24th, 2012 at 8:01 am

Thanks Zorglub. The basic idea wasn’t mine. But I have polished the original.