Diagonals in 3D FiguresDate: 06/21/99 at 10:48:26 From: Jamie Vassallo Subject: Diagonals in 2D and 3D shapes I am having great difficulty with math and require assistance. Could you help? I have the formula n(n-3)/2, but I don't know how to justify or prove it. Another thing is, could you please tell me the number of diagonals in various 3D shapes, such as a tetrahedron, cube and so on. Is there a formula for this or is it just coincidental? Thank you very much for your help and I hope to hear from you soon. Thank you. Jamie Date: 06/21/99 at 12:26:06 From: Doctor Peterson Subject: Re: Diagonals in 2D and 3D shapes Hi, Jamie. Let's think how we can count the diagonals in a polygon. Pick any vertex; there are N ways to do that. Now pick any vertex to go to EXCEPT the two neighbors (and the point itself, of course). How many ways are there to do that? When you finish, you'll have counted every diagonal - except that you will have counted each one twice, once starting from each end. Taking that into account will give you the formula. Now, for a general polyhedron, the same method would work except for one detail: there can be any number of "neighbors" to any vertex. But you can use a similar method to find the TOTAL number of segments that can be drawn between any two vertices; then you can just subtract from your count the number of edges in the polyhedron. Since you can make different polyhedra with the same number of vertices but different numbers of edges, you can't give an answer based only on the number of vertices. Have you looked at a tetrahedron and tried to count the diagonals? Try it - you'll find there aren't any. For a cube, there will be two on each of the six faces, and three going through the center. See if that agrees with the formula you come up with. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/22/99 at 13:32:17 From: jamie vassallo Subject: Diagonals in 3D objects Could you please help me? In a 3D object, the tetrahedron has 0 diagonals, the cube has 15 and then the pentagonal prism has either 28 or 32. Can you tell me the next 3 or 4 terms so that I can work out the sequence? Thank you very much. Jamie Vassallo Date: 06/22/99 at 17:03:30 From: Doctor Peterson Subject: Re: Diagonals in 3D objects Hi again, Jamie. There are a couple of problems in trying to figure this out by making a list. For one thing, there really is no "sequence": you can't put all the possible polyhedra in an ordered list as you can with polygons, because there are lots of different ways to connect N vertices to make a shape. Secondly, even if you can see a pattern in the jumble of shapes you consider, it will be hard to be sure it's a real pattern if you haven't given thought to the reason for the pattern. It's much easier to find patterns if you look for a pattern in the way you count, rather than just make a list of numbers and then look for a pattern there. Yet it will be worthwhile for you to make a list of shapes and try to count the edges and the diagonals of each, in order to get to know them. You'll want to get a feel for how diagonals work, so it wouldn't be helpful for me to just give you a list - you need to do some counting on your own. But I'll start your list to give you some ideas on what to look for and how to count the diagonals. Let's list several characteristics of each shape: the number of Vertices (V), Edges (E), Faces (F), and Diagonals (D). Tetrahedron V E F D --- --- --- --- 4 6 4 0 + / \\ / \ \ / \ \ / \ \ / __\__ + /____---- \ / +-------------+ Since every point is connected to every other point, there's nothing left to be a diagonal. Square pyramid V E F D --- --- --- --- 5 8 5 2 + / \\ / \ \ / \ \ / \ \ /---------\---+ / \ / +-------------+ The top point is connected to all the others, so the only diagonals are the two in the base. Triangular hexahedron (Twin tetrahedra) V E F D --- --- --- --- 5 9 6 1 + / \\ / \ \ / \ \ / \ \ / __\__ + /____---- \ / +-------------+ \ / \ / \ / \ / \ / \ / + The top and bottom points are connected to everything but each other; the "equatorial" points are connected to everything, leaving only the one diagonal. Notice that these last two polyhedra have the same number of vertices, but different numbers of edges. But the sum of the number of edges and the number of diagonals is the same for both, because that's just the total number of lines you can draw connecting any two points. Octahedron V E F D --- --- --- --- 6 12 8 3 + / \\ / \ \ / \ \ / \ \ /---------\---+ / \ \ / +-------------+ \ \ // \ \ / \\ / + Each point is connected to every point except the opposite vertex; the three pairs of opposites form three diagonals. Pentagonal pyramid V E F D --- --- --- --- 6 10 6 5 + /|\ /|||\ / ||| \ / | | | \ / | | | \ / | | | \ /___|_-+-_|__ \ +- | | -+ \ | | / \| |/ +---------+ The top point is connected to all the others, so the only diagonals are the five in the base. Triangular prism V E F D --- --- --- --- 6 9 5 6 + / |\ / | \ +------------------+ | | | | | | | | | | | | | + | | / \ | | / \| +------------------+ The diagonals are all in the three sides. I've given you three different polyhedra with six vertices. It gets a lot worse with more vertices! I'm not sure why you gave two numbers for the pentagonal prism; the correct count, according to my formula, is 30: 5 in each base, 2 in each of 5 sides, and 2 reaching from each vertex on one base to vertices on the other base through the interior. The important thing to see is that you can't just look at V and tell me what D will be. You need to know at least two things about the shape. There is a formula that relates D+E to V, which is what I suggested to you in my last response; and another that relates V+F to E. Putting these together, it turns out that D+F is determined by V. So if you know V and either E or F, you can figure out what D is. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/23/99 at 06:17:47 From: James Vassallo Subject: Thank you Dr Peterson Dear Dr Peterson, Thank you very much for your help, it is greatly appreciated, I now understand more fully what my task actually is. Thank you once again for all your time with me. Jamie Vassallo Date: 06/26/99 at 06:16:38 From: James Vassallo Subject: Pyramids Thank you for your faith in me, this is just a check, I have found your information on pyramids useful, and have come up with the formula (n-1)(n-4) n = ---------- 2 Is this correct? Just one other query; is there a link between the formula for pyramids and prisms, or is there one formula linking all 3D shapes together? Thank you. Jamie Date: 06/26/99 at 16:38:23 From: Doctor Peterson Subject: Re: Pyramids Hi, Jamie. Yes, you have the right formula for a pyramid. I'm not sure how you found it, but there's a very easy way: all the diagonals will be in the base, since the apex is already connected to all other vertices; so you just have to put n-1 into the formula for diagonals of a polygon. You can do something similar for a prism: many of the diagonals will be in one or the other of the bases, and the rest can be easily counted because they go from a vertex on the top to a vertex on the bottom. I'm not sure there would be a direct link to the pyramid formula, but they certainly aren't unrelated either. There is a very simple general formula for the total number of lines that can be drawn between any two of a set of n points in space, and this will be the sum of the number of edges of the polyhedron and the number of diagonals. You can easily derive your formula for a pyramid, or the one for a prism, from this by coming up with a formula for the number of edges. For example, for a pyramid, the number of edges is just 2(n-1), since there are n-1 edges in the base, and the same number of edges from the base to the apex. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/13/99 at 10:43:18 From: Jamie Vassallo Subject: Diagonals Dear Dr Peterson, It's me again. I have yet another problem which I hope that you can help me with. The problem has to do with the formula for prisms. I think that I may have this formula but I am unsure. The formula I have is n(n-4)/2. That is the factored version, I can't find the square on my computer, so I can't give the expanded. The way that I sort of got it was by identifying the number of diagonals in a prism, with an x + y + z method, where x = z. X and z are the diagonals in the two end polygons, and y is equal to the number of diagonals in the middle. This is what I have: n(n-3) + n(n-1) + n(n-3) ------ ------ 2 2 That is what I have. I am not sure about what the n(n-1) is, what it stands for. Is it the same as with the pyramids, the number of vertices multiplied by the number of vertices minus the apex? Because in a pyramid, no diagonals go to the apex, without going along the edges, and that is not the case in prisms. I have tried the formula above on various figures and it works. So that is the testing I've done. Then the proof would be finding it through the x + y + z method, which I am not sure of, but I think I am close to finding it. Now the area that I need your help with is checking it. Also help me, if you could, in justifying the formula. I don't understand what each part of it means. Another question that I have is that I have seen the x + y + z, not written over 2, but rewritten as 1/2 n. Am I right in assuming that this is just to get rid of the 'over 2'? Is the n(n-1) over 2, or not? That really confuses me. Thanks once again for your help and look forward to hearing your response. Jamie Vassallo Date: 07/13/99 at 12:20:03 From: Doctor Peterson Subject: Re: Diagonals Hi, Jamie. Here's your equation: n(n-3) + n(n-1) + n(n-3) ------ ------ 2 2 The first and third parts, as you explained, are the number of diagonals in the top and bottom. You're using n here, of course, not for the total number of vertices in the prism, but for the number of vertices in the base (which is the natural number to use here, but has to be clearly stated to avoid confusion). The middle part is the number of "body diagonals" crossing from the top to the bottom. It can be easily explained: you can choose one of n vertices for the top end of such a diagonal, and then have n-1 choices for the bottom end, since any vertex except the one attached to the top end by an edge of the prism will give you a diagonal. This gives n(n-1). You don't have to divide by two here, as you do in working out the total number of lines you can draw between two of n points (or the number of handshakes between n people), because each diagonal is counted only once: the diagonal from point A on the top and point B' on the bottom is different from the one between points B and A'. When you count handshakes, A shaking with B is the same as B shaking with A, so you divide by 2. To get from this to your n(n-4)/2, you need two steps. n(n-3) + n(n-1) + n(n-3) ------ ------ 2 2 n^2 - 3n + 2n^2 - 2n + n^2 - 3n 4n^2-8n = ------------------------------- = ------- 2 2 = 2n^2 - 4n = 2n(n-2) Putting n = v/2, this becomes 4(v/2)^2 - 8(v/2) v^2 - 4v v(v-4) ----------------- = -------- = ------ 2 2 2 as you said. I suspect that this is where those 1/2 n's came from that you saw: n must have been what I'm calling v, and n/2 was therefore used where I put n. The fact that you didn't mention that the n in this formula is different from the n in the longer version tells me that you probably don't understand the details about the formulas, but just got them from two different sources and are trying to make them fit together. I hope you'll think through this carefully so you really know how it all works, and can reconstruct it for yourself. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/19/99 at 06:12:12 From: Jamie Vassallo Subject: Re: Diagonals Dear Dr Peterson, You sent me that email showing that: n(n-3) + n(n-1) + n(n-3) ------ ------ 2 2 = n^2 - 3n + 2n^2 - 2n + n^2 - 3n ------------------------------- 2 What I am unsure of is, how does the n(n-1) become the 2n^2 - 2n? I've looked at multiplying by one half, but that just gives me n(n-1)/2. I know that the first n (outside of the bracket) must be multiplied by 2 somewhere, but I don't follow how and why. The last query that I have is: at the bottom of your last e-mail, you said "putting n = v/2." Why do you do this? That is what I can't understand. Thanks for your help and I look forward to hearing from you. Jamie Date: 07/19/99 at 09:02:48 From: Doctor Peterson Subject: Re: Diagonals Hi, Jamie. I've never asked you or your classmates how much you know of algebra, though I've wondered whether some of you might have trouble with some of the tools you need to work comfortably with these formulas. I wrote assuming you'd ask if you didn't follow some steps. I'll be happy to explain a little more deeply. Here's what I wrote: n(n-3) + n(n-1) + n(n-3) ------ ------ 2 2 n^2 - 3n + 2n^2 - 2n + n^2 - 3n 4n^2-8n = ------------------------------- = ------- 2 2 = 2n^2 - 4n = 2n(n-2) Here are the steps I left out between the first and second expressions: n(n-3) + 2n(n-1) + n(n-3) n(n-3) + 2n(n-1) + n(n-3) ... = ------ ------- ------ = ------------------------- = ... 2 2 2 2 Do you see what I did? Before adding the three expressions, I had to write them all with the same denominator. Since the middle part didn't have a denominator, I just multiplied it by 2/2. Then I combined them into one fraction, after which I could multiply out each part using the distributive property, and then add like terms. (I deliberately left out these steps, because if you're good at algebra I'd be talking down to you, as well as doing work for nothing, and if you aren't, it would be a good exercise to figure out what I was doing rather than have every step shown.) As for v and n: changing variables can be pretty hard to follow; it requires that you be really comfortable with the idea of variables themselves. What's happening is that you want to get an expression for the number of diagonals in terms of the total number of vertices in the prism, whereas the formulas we'd been working with up to that point were based on n being the number of vertices in each of the two polygons that form the prism. For example, for a triangular prism, n is 3, but the total number of vertices in the prism is 2*3 = 6. Our n(n-3)/2 is the number of diagonals in one base of the prism, which is a polygon with n vertices. But the number of vertices in the whole prism is v = 2n, so in order to get a formula using v, we replace n everywhere with its equivalent in terms of v, n = v/2. I'd actually prefer leaving the formula in terms of n, which seems to me a natural way to describe a prism; but your goal is to find formulas for different polyhedra in a form that can be compared from one to the next, so you wanted to get them all in terms of v. I hope that helps; if not, keep asking. Good questions from someone who cares about understanding are always welcome! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/19/99 at 13:10:19 From: Jamie Vassallo Subject: Re: Diagonals This is my last query; then I've figured the whole thing out. Can you tell me how many diagonals there are in a dodecahedron and an icosahedron? Or can you tell me where on the Internet I can find that answer? Jamie Date: 07/19/99 at 16:59:34 From: Doctor Peterson Subject: Re: Diagonals Hi, Jamie. It's time to introduce you to a different approach to the prism, which you can apply to these regular polyhedra. Regularity means that some or all features of a shape are true everywhere. A prism is not a regular polyhedron, because not all faces are the same shape and not all edges are the same length, and so on; but it is regular in one way: at every vertex you find the same number of edges, namely two going to adjacent vertices in the same base, and one going to the corresponding vertex in the other base. Luckily, that's the kind of regularity that helps most with this problem. You can use this to do the same thing you did originally in counting the diagonals of a polygon. If there are v vertices in all, you can pick any of them to start a diagonal, and then you go from there to any vertex except itself and the three other vertices to which it is connected, giving you (v-4) choices for each of the (v) starting vertices. Since you will have counted each vertex twice this way, the total number of diagonals is v(v-4) D = ------ 2 Sound familiar? You can use the same method for any regular polyhedron. See what you get! (You'll find that regular polyhedra are pretty boring.) Then see if you can actually visualize where these diagonals are and see that you counted them correctly. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/20/99 at 08:21:53 From: Jamie Vassallo Subject: Re: Diagonals Dear Dr Peterson, Thank you for the e-mails. I find that I now understand it all, except for two small areas, which I hope you can help me with. The first is: using the prism formula and the notes you gave me, I worked out the number of diagonals. Can you check my data, see if they are correct, as I have two conflicting sources, the Internet and also an encyclopedia: Vertices Edges Faces Diagonals Dodecahedron 12 30 12 48 Icosahedron 20 30 20 160 If those are correct, which I believe them to be, I have come up with this logic, please tell me if I'm right. v(v-j) ------ 2 1. v This represents the total number of vertices in the polyhedron 2. v-j The total number of vertices minus j, where j is equal to the number of diagonals unable to be drawn to from any given starting vertex. For instance in the cube, from any given vertex, you are unable to draw diagonals to 3 vertices as they are connected with edges. Then you can't draw a diagonal to the vertex from where you started. So it would be the total number of vertices minus 4. This value would be (for all except two polyhedra) the shape of which the polyhedron is made from plus 1. The exceptions are the cube, where the 1 need not be added; and the octahedron, where it is needs to be added to 2. 3. 2 It is placed over two because by using this method you count each diagonal twice. I would be grateful to hear from you soon. Jamie Vassallo Date: 07/20/99 at 08:50:01 From: Doctor Peterson Subject: Re: Diagonals Hi, Jamie. This sounds fine except for the details of how you find j. I would express the formula not in terms of j, which is not an obvious number, but in terms of something like the number of edges that meet at each vertex. It has nothing to do with the shape of the faces. You might like to make a chart for all the regular polyhedra and a few prisms, showing not only V, E, and F, but also the number of edges per face and the number of edges per vertex. Then show what your "j" is, and see how it's related. Or just think carefully about what it means to be unable to draw a diagonal to a vertex. While you're studying polyhedra, you might want to look at several sites on that subject; there are a lot of fascinating facts about them, some of which might be relevant: George Hart's Virtual Polyhedra http://www.georgehart.com/virtual-polyhedra/vp.html Alexander Bogomolny's Regular Polyhedra http://www.cut-the-knot.org/do_you_know/polyhedra.shtml - 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/