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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.


This post is under “Mathemagic” and has 10 respond so far.
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10 Responds so far- Add one»

  1. 1. Zorglub Said:

    There are only two possible mats.

  2. 2. DP Said:

    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.

  3. 3. Wizard of Oz Said:

    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.

  4. 4. Chris Said:

    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.

  5. 5. Zorglub Said:

    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.

  6. 6. Chris Said:

    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.

  7. 7. Zorglub Said:

    Thanks for the &ge and &le.
    I am looking forward to see your demonstration.

  8. 8. Chris Said:

    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

  9. 9. Zorglub Said:

    That is very nice.

  10. 10. Chris Said:

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

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