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Math Help - algebra problem

  1. #1
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    Jun 2008
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    algebra problem

    I'm having a problem understanding what must be an easy problem, but I can't see a way through it.

    It relates to sequences in counting the number of corners, edges and middle panes in a square or rectangular of equal sized pieces. So with a 3x3 square, there are:
    c e c............c - Corner panes
    e m e........... e - Edge panes
    c e c........... m - Middle panes

    .......Corners...Edges...Middle
    3x3.......4..........4.........1 - So it follows that:
    3x4.......4..........6.........2 - If the width 'W' is variable, the sequence can be expressed as
    3xW......4.......2w-2.....w-2

    Doing the same for a 4 sides square:

    .......Corners...Edges...Middle
    4x4.......4..........8.........4 - So it follows that:
    4x5.......4.........10........6 - If the width 'W' is variable, the sequence can be expressed as
    4xW......4.........2w.....2w-4

    My problem is how to express a formula when introducing a variable Height 'H' which would explain the sequence:

    .......Corners...Edges...Middle
    3xW.....4........2w-2.....w-2
    4xW.....4.........2w......2w-4
    5xW.....4........2w+2....3w-6
    HxW....4..........??.........??

    Any help would be appreciated. Thanks in advance.
    Last edited by Jazzy331; June 12th 2008 at 02:00 PM. Reason: Clarifying the terms W and H
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  2. #2
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    Lexington, MA (USA)
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    Hello, Jazzy331!

    A fascinating problem . . .


    My problem is how to express a formula introducing H which would explain the sequence:

    . . \begin{array}{|c|c|c|c|}\hline<br />
H \times W& \text{Corners} & \text{Edges} & \text{Middle} \\ \hline<br />
3\times W & 4 & 2(W-1) & 1(W-2) \\<br />
4\times W & 4 & 2W & 2(W-2) \\<br />
5\times W  & 4 & 2(W+1) & 3(W-2) \\<br />
\vdots & \vdots & \vdots & \vdots \\<br />
H \times W & 4 & ?? & ?? \\ \hline \end{array}

    I think I found the H-formula for the Edges and Middles.


    . . \begin{array}{|c|c|c|c|}\hline<br />
H \times W& \text{Edges} \\ \hline<br />
{\color{red}3}\times W & 2(W{\color{red}-1}) \\<br />
{\color{red}4}\times W & 2(W{\color{red}+0}) \\<br />
{\color{red}5}\times W  & 2(W{\color{red}+1}) \\<br />
\vdots & \vdots \\<br />
{\color{red}H} \times W & 2(W {\color{red}+ H-4)}\\ \hline \end{array}


    . . \begin{array}{|c|c|c|c|} \hline<br />
H \times W &  \text{Middle} \\ \hline<br />
{\color{red}3}\times W & {\color{red}1}(W-2) \\<br />
{\color{red}4}\times W & {\color{red}2}(W-2) \\<br />
{\color{red}5}\times W & {\color{red}3}(W-2) \\<br />
\vdots & \vdots \\<br />
{\color{red}H} \times W & {\color{red}(H-2)}(W-2) \\ \hline \end{array}

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  3. #3
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    Thanks Soroban - that's brilliant.

    Is there a specific method of working to get me there?

    I have followed the maths to test it and agree with the H-4 element on the Edges, but my problem with maths is always that I need to understand what I've done to get somewhere, otherwise it won't stick. Is it purely that the '1' in 5xW 2(w+1) resolves the h-4? If so, then I assume that the same would apply to the '3' in 5xW : 3(w-2) in the middles H-2 = 3.

    But why does the expression of the formula work that way? Any thoughts?

    Thanks in advance.

    Jazzy331
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