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Finding Rhombi in a Rhombus

Date: 9/25/95 at 10:51:26
From: Anonymous

How can I work out a formula for finding how many rhombuses 
there are in a rhombus? (say 2cm*2cm or 3cm*3cm and so on, etc.)

Date: 9/25/95 at 11:36:55
From: Doctor Ken
Subject: Re: (no subject)


I'll assume you are asking how many rhombi of 1cm side length there are
in a rhombus of side length 2, 3, 4, and so on.  Well, try this: squish your
big rhombus 

   /     /     /
 /     /     /

into a rectangle:

 |     |     |
 |     |     |

Now is it easier to see?

-Doctor Ken,  The Geometry Forum

Date: 9/26/95 at 22:25:42
From: Arthur Smith
Subject: Re: Rhombuses

 Hi Doctors,

I do not think I made myself clear. I should have said all rhombuses that 
could be found. So, a 2cm*2cm would have 9 rhombuses, a 3cm * 3cm 26
rhombuses, a 4cm*4cm would have 60 rhombuses and a 5cm*5cm would have 111
rhombuses and so on.  This includes all rhombuses that can be found.. 
Sizes of rhombus in larger rhombus in cm  

     Size of rhombus in cm  

                 1*1   2*2    3*3   4*4   5*5      Total
 1 by 1cm          1     0      0     0     0         1
 2 by 2            8     1      0     0     0         9
 3 by 3           21     4      1     0     0        26
 4 by 4           40     5      4     1     0        60
 5 by 5           65    32      9     4     1       111 (we are certain this 
                           		                 one is right)

And so on... 

We hope to find a formula for working out all the rhombuses whether they are
10cm*10cm or 1000cm*1000cm;whatever the size.

Thanks in advance,

Date: 9/27/95 at 12:46:34
From: Doctor Andrew
Subject: Re: Rhombuses

First, this problem could be done with squares, which are simpler cases
of rhombi.  This should make it easier to visualize.  I think we've figured
out that you are only counting rhombi with sides of a length that is a 
multiple of 1cm.  We're having trouble recreating your table, though.  We 
count only 4 1x1 rhombi in a 2x2 rhombus.  Perhaps you are also counting 1x2
parallelograms as well.  In a rhombus all sides must have equal length.

Here's the solution to the rhombus problem (as we undertstand it) as
well as a solution for the number of parallelograms in a rhombus.  Feel free
to just read part of it to get you started in the right direction:

First, let's find how many segments of length x are on a side of length n:

This diagram shows a side divided into n 1cm segments:


How many segments of length 1 are there?

n. Right, they're already marked for you.

How many segments of length 2?

If you start at the left and count them, there will be a segment beginning
at each mark, except for the last one where it will go off the end.  So
there n - 1.

How many of length 3?

You can count n -2 before you go off the edge.

So how many of length x?

You'll can see that the pattern is that there are n - (x+1) segments
of length x in the segment of length n.

If you are going to create a rhombus of side length x, you need to pick
a segment of length x from one side of the larger rhombus and another
segment of length x from the adjacent side.  How many such rhombi
are there?  Well, how many choices do you have?

You've go (n - (x + 1)) choices on each side.  So you have

(n - (x + 1)) * (n - (x + 1)) = (n - (x + 1))^2 [note: x^2 means x squared]

total choices.   The total number of rhombi in the larger rhombus is then
the sum of the number of each size of rhombus:

total = (n - 0 + 1)^2 + (n - 1 + 1)^2 + (n - 2 + 1)^2 + ... (n - n + 1)^2

 = (n+1)^2 + n^2 + (n-1)^2 + .. 1^2

 = (n+1)((n+1)+1)(2(n+1)+1) / 6   [1]

using a known formula that states: 1^2 + 2^2 + ... + x^2 = x(x+1)(2x+1)/6.

You can reduce [1] to:

total rhombi = (n+1)(n+2)(2n+3)/6.

So, if you are looking for rhombi (parallelograms with equal side lengths)
this is the solution.  If you are looking for parallelograms in general
the solution can also be found:

The total number of choices of segments on one side is the sum of the 
number of choices for segment length:

choices = (n - 0 + 1) + (n - 1 + 1) + (n - 2 + 1) + ... (n - n + 1) 
= n+1 + n + n-1 + ... 1 = n(n+1)/2  (the sum of the averages)

There are this many choices on the adjacent side as well so the total
number of choices is the number of choices squared:

[n(n+1)/2] * [n(n+1)/2] = n^2(n+1)^2/4

So this is the number of parallelograms with sides that are multiples of 
1cm in a rhombus.

Hope this helps!

-Doctor Andrew,  The Geometry Forum

Date: 01/18/2000 at 10:09:47
From: Oli Hickman
Subject: How many rhombi are in this rhombus?

I need a formula to find the total number of rhombi in this rhombus. 
It is the same format of rhombus as in the "Finding Rhombi in a 
Rhombus" (above). The only difference is that in the 1x1 rhombi there 
are diagonal lines from the top left to the bottom right of each 
1x1 rhombus. Because of this, the results in the table that was sent 
to you in 1995 are correct. Here is the table again.

Size of rhombus in cm

                     1*1   2*2    3*3   4*4   5*5      Total
     1 by 1 cm        1     0      0     0     0          1
     2 by 2           8     1      0     0     0          9
     3 by 3          21     4      1     0     0         26
     4 by 4          40     5      4     1     0         60
     5 by 5          65    32      9     4     1        111 
       /\    /\    /
      /  \  /  \  /
    /\    /\    /
   /  \  /  \  /

Here is a rough diagram of the rhombus type we are investigating. This 
is a 2x2 rhombus and as you can see it is divided into equilateral 
triangles as opposed to just smaller 1x1 rhombi. If you draw it out on 
paper more accurately with the diagonals going directly into the 
corner you will easily see 9 rhombi.

We have investigated this type of rhombus and we know there are three 
types of rhombi to be found within these rhombi. They are roughly 
drawn below.

      _               __
     /_/      /\      \_\

The first rhombus would have a diagonal running from top left to 
bottom right, the second has a horizontal line splitting it into 2 
equilateral triangles, and the third has a diagonal line running from 
top right to bottom left.

We need one formula to find the total possible number of rhombuses in 
any sized rhombus (we know rhombi are always 1x1, 2x2, etc. and 
anything else is a parallelogram).

Thanks very much.

Date: 01/18/2000 at 21:07:25
From: Doctor Peterson
Subject: Re: How many rhombi are in this rhombus?

Hi, Oli.

It intrigues me that your question produces the table that was "wrong" 
in the archived answer; the original question must have been the same 
as yours, but the questioner never corrected Dr. Ken's assumptions 
about the lines within the rhombus. As a result, his answer doesn't 
cover all the rhombuses you want. Yet it is a very good beginning. 
Just for completeness, I'll draw the correct diagram the best way I 
can, for a 4x4 rhombus, in which I've marked one 2x2 rhombus in each 

            /2\3/3\3/3\ / \ /
          /2\2/2\3/3\3/3\ /
        / \2/2\2/1\1/1\1/
      / \ / \2/1\1/1\1/

You've started out just the way we like you to: looking in our 
archives for an answer that applies to your question. As you are 
probably aware, our purpose is not to give you the answer to your 
homework -- especially not to give you a formula that you are supposed 
to find by investigation, and take away the exercise that work is 
meant to give you. What we want you to do is to use the hints we give 
(such as the example in the archived problem) to get ideas for your 
investigation. Having said that, I can give you some more hints.

When I look for a formula, I don't care much for tables of numbers. 
Sometimes you might be able to guess a pattern by looking at the 
numbers; but how could you ever be sure the pattern wouldn't change 
when you added the next row? In order to be sure of a pattern, you 
have to see how it forms: what is it about the circumstances of the 
problem that cause the pattern to arise? For that purpose, I like to 
make a table, but to concentrate not on the table itself, but on the 
process of building it. What thought process do you use to figure out 
each number? That's where the pattern shows up.

In this case, you've pointed out that there are three orientations for 
the rhombuses; so each number in your table is really the sum of three 
numbers that you probably counted separately. I would make my table 
that way: under "1x1" I would make three columns, one for each type of 
rhombus. If you look closely how you form each number, you'll be able 
to see a formula for that number. For example, one kind of rhombus 
will be just what Drs. Ken and Andrew were talking about, and their 
answers will apply to them. Read carefully - both the reason for each 
of these cells in our table being a square, and the formula for 
summing squares will be useful to you. The other two orientations will 
produce different formulas, but you can arrive at them by the same 
kind of reasoning.

Have fun - this is a good problem to investigate.

- Doctor Peterson, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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