EVE Search - Wrap your mind around this math =)

All Channels

Out of Pod Experience

Wrap your mind around this math =)

» Click here to find additional results for this topic using Google

Monitor this thread via RSS [?]

Author	Thread Statistics \| Show CCP posts - 1 post(s)
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.02 23:40:00 - [1] You said a=b and then you divide by a-b? Dividing by zero is bad mmmkay? --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.02 23:44:00 - [2] Originally by: lofty29 NICE. That's being printed off and posted in the common room in the maths building on monday. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.02 23:45:00 - [3] Originally by: pwnedgato Originally by: ReaperOfSly You said a=b and then you divide by a-b? Dividing by zero is bad mmmkay? Pointing out something already said is bad mmmkay? It wasn't already said when I clicked reply. You people just post too damn fast --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.02 23:51:00 - [4] Here's one for the topologists. What do you get when you glue two moebius strips together along their edges (remember a moebius stip only has one edge)? Assume their material is as bendy and strechy as you like. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.02 23:57:00 - [5] Originally by: Sqalevon Originally by: ReaperOfSly Here's one for the topologists. What do you get when you glue two moebius strips together along their edges (remember a moebius stip only has one edge)? Assume their material is as bendy and strechy as you like. I'd settle for a wormhole. That would be absolutely frakking awesome. But sadly, no --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:00:00 - [6] Edited by: ReaperOfSly on 03/11/2007 00:00:46 Originally by: pwnedgato Are the 2 strips twisted the same direction? It doesn't matter. Originally by: xOm3gAx Originally by: ReaperOfSly Here's one for the topologists. What do you get when you glue two moebius strips together along their edges (remember a moebius stip only has one edge)? Assume their material is as bendy and strechy as you like. A Klein bottle. Winner! --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:02:00 - [7] Originally by: xOm3gAx Originally by: ReaperOfSly Originally by: xOm3gAx Originally by: ReaperOfSly Here's one for the topologists. What do you get when you glue two moebius strips together along their edges (remember a moebius stip only has one edge)? Assume their material is as bendy and strechy as you like. A Klein bottle. Winner! ;) But now you've stepped up to the challenge, I am now going to ***** open my third year topology books and find a fiendishly difficult one for you --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:04:00 - [8] Originally by: Sqalevon My first thought was a Sphere, and after that a Torus ( doughnut shape ), then a Trefoil knot. A sphere is the result of gluing two disks together along their edges. A Klein Bottle is like a sphere, just non-orientable. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:10:00 - [9] Let P be a regular polyhedron in which each face has p edges and for which q faces meet at each vertex. Using v-e+f=2 (where v is the total number of vertices, e the total number of edges, f the total number of faces), prove that (1/p)+(1/q)=(1/2)+(1/e). --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:13:00 - [10] Originally by: Sqalevon Frankly, I wasnt aware of its existance. I need to get more into this stuff, its fun. Notice how if you trace a line on the surface going into the opening, and follow it round, you end up on the other side of the surface, much like on a moebius strip. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:18:00 - [11] Edited by: ReaperOfSly on 03/11/2007 00:21:32 Originally by: Sqalevon Edited by: Sqalevon on 03/11/2007 00:16:31 Originally by: ReaperOfSly Let P be a regular polyhedron in which each face has p edges and for which q faces meet at each vertex. Using v-e+f=2 (where v is the total number of vertices, e the total number of edges, f the total number of faces), prove that (1/p)+(1/q)=(1/2)+(1/e). Even I can solve that one :( Afaik its a tetrahedron (1/3}+{1/3} = (1/2)+{1/6) EDIT I just found v-e+f=2 that, I guess I'll shut up now EDIT 2, after giving it 3 seconds of my time, I realised it was still the right awnser 4 vertices - 6 edges + 4 faces = 2 ROFL By the way, I should just point out that I'm not restricting the question to 3-dimensional polyhedra, so you can't use the argument that there are only 5 polyhedra in 3-space, and do "proof by exhaustion" Edit: well the point is it hold for the tetrahedron and all other polyhedra also. The question is why this works. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:28:00 - [12] I'll give you a hint. Try "pulling the object apart" so that you have a set of regular polygons with p edges and p vertices each which make up the faces of the polyhedron. Then notice how when you fit these shapes back together again, you halve the total number of edges (each edge has precisely two faces meeting at it), and the number is vertices is divided by q (each vertex has precisely q faces meeting at it). --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 00:38:00 - [13] Originally by: Sqalevon It's 1:30 I'll give it some thought tomorrow, and might even come with something intresting myself :) In the meanwhile Imagine a rope around the equator, and add 1 meter, how much do you think the rope will floot above the earths surface when the extra space is evenly spread. Just under 18cm, unless I'm having a brain-fart? That's 1/(2*pi) meters. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 09:54:00 - [14] Originally by: xOm3gAx Edited by: xOm3gAx on 03/11/2007 02:43:30 Edited by: xOm3gAx on 03/11/2007 02:42:32 Originally by: ReaperOfSly Edited by: ReaperOfSly on 03/11/2007 00:21:32 Originally by: Sqalevon Edited by: Sqalevon on 03/11/2007 00:16:31 Originally by: ReaperOfSly Let P be a regular polyhedron in which each face has p edges and for which q faces meet at each vertex. Using v-e+f=2 (where v is the total number of vertices, e the total number of edges, f the total number of faces), prove that (1/p)+(1/q)=(1/2)+(1/e). Even I can solve that one :( Afaik its a tetrahedron (1/3}+{1/3} = (1/2)+{1/6) EDIT I just found v-e+f=2 that, I guess I'll shut up now EDIT 2, after giving it 3 seconds of my time, I realised it was still the right awnser 4 vertices - 6 edges + 4 faces = 2 ROFL By the way, I should just point out that I'm not restricting the question to 3-dimensional polyhedra, so you can't use the argument that there are only 5 polyhedra in 3-space, and do "proof by exhaustion" Edit: well the point is it hold for the tetrahedron and all other polyhedra also. The question is why this works. Snuck out =P 1. Remove a triangle with only one edge adjacent to the exterior. This decreases the number of edges and faces by one each and does not change the number of vertices, so it preserves V − E + F. 2. Remove a triangle with two edges shared by the exterior of the network. Each triangle removal removes a vertex, two edges and one face, so it preserves V − E + F. Repeat these two steps, one after the other, until only one triangle remains. At this point the lone triangle has V = 3, E = 3, and F = 2 (counting the exterior), so that V − E + F = 2. This equals the original V − E + F, since each transformation step has preserved this quantity. Therefore at the start of the process it was true that V − E + F = 2. Edit: thats Euler's formula right? Edit 2: use that formula and solve for a Cubohemioctahedron =) Ehm, the question was to prove (1/p)+(1/q)=(1/2)+(1/e). Euler's theorem, v-e+f is a given in this question. Also you seem to be resticting yourself to polyhedra with triangular faces. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 14:36:00 - [15] Originally by: vanBuskirk Another one that I have heard about, given in an exam paper: A sphere has a hole drilled in it. The hole is perfectly circular in cross-section and uniform in cross-sectional area, and is centred on the sphere - in other words, the centre line of the hole contains the centre of the original sphere. The hole is 1 metre long from edge to edge, along the straight side. What is the area of the outer surface of the sphere that still remains after this operation? Depends what the radius of the hole is. --------------------------------------------------------------------
ReaperOfSly Gallente Lyrus Associates M. PIRE	Posted - 2007.11.03 22:27:00 - [16] Originally by: vanBuskirk Edited by: vanBuskirk on 03/11/2007 20:11:43 Originally by: ReaperOfSly Originally by: vanBuskirk Another one that I have heard about, given in an exam paper: A sphere has a hole drilled in it. The hole is perfectly circular in cross-section and uniform in cross-sectional area, and is centred on the sphere - in other words, the centre line of the hole contains the centre of the original sphere. The hole is 1 metre long from edge to edge, along the straight side. What is the area of the outer surface of the sphere that still remains after this operation? Depends what the radius of the hole is. Quite right; I got the problem wrong - the problem was asking for the volume of the remainder of the sphere after the cut. Hint; there are two ways of solving this problem. It STILL depends on the radius of the hole --------------------------------------------------------------------

COPYRIGHT NOTICE
EVE Online, the EVE logo, EVE and all associated logos and designs are the intellectual property of CCP hf. All artwork, screenshots, characters, vehicles, storylines, world facts or other recognizable features of the intellectual property relating to these trademarks are likewise the intellectual property of CCP hf. EVE Online and the EVE logo are the registered trademarks of CCP hf. All rights are reserved worldwide. All other trademarks are the property of their respective owners. CCP hf. has granted permission to EVE-Search.com to use EVE Online and all associated logos and designs for promotional and information purposes on its website but does not endorse, and is not in any way affiliated with, EVE-Search.com. CCP is in no way responsible for the content on or functioning of this website, nor can it be liable for any damage arising from the use of this website.