Finding Rhombi in a RhombusDate: 9/25/95 at 10:51:26 From: Anonymous Subject: 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) Hello! 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, Arthur 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 orientation: +---+---+---+---+ /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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/