Left-Sided Rhombuses in a Larger Rhombus

Date: 05/22/2000 at 14:34:55
From: Stuart Wright
Subject: Formulas for individual rhombuses inside a larger rhombus

This question is similar to one you have previously answered, though 
different. My math coursework is to work out a formula for the number 
of individual rhombuses inside a larger rhombus, say 3x3 or 4x4. These 
formulas must be for number of right-sided rhombuses, left-sided 
rhombuses, and vertical rhombuses (diamonds.) I already have the table 
and formulas for right-sided; let me show it to you.

     Size of rhombus     1x1    2x2     3x3     4x4     5x5     6x6
           1x1            1      0       0       0       0       0
           2x2            4      1       0       0       0       0
           3x3            9      4       1       0       0       0
           4x4           16      9       4       1       0       0
           5x5           25     16       9       4       1       0
           6x6           36     25      16       9       4       1
         Formula        n^2   (n-1)^2 (n-2)^2 (n-3)^2 (n-4)^2 (n-5)^2

If you imagine a rhombus 3x3, it has a potential number of 26 
rhombuses inside: 21 individual 1x1 rhombuses (including right + left 
+ vertical sided), 4 2x2 rhombuses and 1 3x3 rhombus. However, I 
require a table and formula for left-sided and vertical rhombuses like 
the one above, and I have no idea what it is. If you could provide me 
with that information I would be very grateful. Also, we have to 
explain how the formula works. If you can, could you give me an idea 
of that?

Thanks in advance.

Date: 05/22/2000 at 17:13:01
From: Doctor Peterson
Subject: Re: Formulas for individual rhombuses inside a larger rhombus

Hi, Stuart.

I'd rather tell you how the formula works, and let you find it. So 
let's start by looking at the formula you already have.

I presume your figure is something like this:

             +---+---+---+---+
            / \ / \ / \ / \ /
           +---+---+---+---+
          / \ / \ / \ / \ /
         +---+---+---+---+     N = 4
        / \ / \ / \ / \ /
       +---+---+---+---+
      / \ / \ / \ / \ /
     +---+---+---+---+

What you counted are shapes like

       +---+          +---+---+
      /   /          /       /
     +---+    and   +       +
     K = 1         /       /
                  +---+---+
                    K = 2

How did you count them? I would do it by seeing how many places there 
are to put, say, the upper left corner. For K = 1, there are N 
positions left-to-right, and N positions top-to-bottom, so there are 
N*N = N^2 in all. If K (the size of the small rhombus) is increased by 
1, that takes away one possible position in each direction; so for any 
K there are (N+1-K)^2 of them. That gives the numbers you gave. For 
example, with K = 2, the (4+1-2)^2 = 3^2 = 9 points marked with "o" 
are possible locations of the upper left corner:

             o---o---o---+---+
            / \ / \ / \ / \ /
           o---o---o---+---+         +---+---+
          / \ / \ / \ / \ /         /       /
         o---o---o---+---+         +       +
        / \ / \ / \ / \ /         /       /
       +---+---+---+---+         +---+---+
      / \ / \ / \ / \ /             K=2
     +---+---+---+---+

Do the same for each of the other orientations of the rhombus. For 
example, here are the places where the top corner of the "vertical" 
rhombus with K = 1 might be:

             o---o---o---o---+
            / \ / \ / \ / \ /
           o---o---o---o---+          +
          / \ / \ / \ / \ /          / \
         o---o---o---o---+          +   +
        / \ / \ / \ / \ /            \ /
       +---+---+---+---+              +
      / \ / \ / \ / \ /              K=1
     +---+---+---+---+

The formula will be similar to the other, but will be a rectangular 
number rather than a square. Also, you lose not one but two possible 
rows when you add one to K, so there will be a multiplication by 2 in 
the formula.

Let me know if you need more help; I'm hoping this will give you what 
you need to be able to find the other formulas yourself.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/