How many panels on a soccer ball? - Numberphile
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Date: 2022-04-08
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Comments and reviews: 9
Jay
It's soccer and that word originated in Europe. Football already has plenty of other sports: Aussie Rules (footy, American football, even rugby football. I don't really care what people say except for two things. One is if you're talking to an American remember that 'football' is the NFL and that's what we'll think of rather than soccer--you have to be cognizant of that just like Brits call the elevator a lift. Two is acting like we just made up the term soccer and that it's wrong when we didn't come up with it and it isn't wrong. Just like how aluminum and aluminium are both accepted.
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It's soccer and that word originated in Europe. Football already has plenty of other sports: Aussie Rules (footy, American football, even rugby football. I don't really care what people say except for two things. One is if you're talking to an American remember that 'football' is the NFL and that's what we'll think of rather than soccer--you have to be cognizant of that just like Brits call the elevator a lift. Two is acting like we just made up the term soccer and that it's wrong when we didn't come up with it and it isn't wrong. Just like how aluminum and aluminium are both accepted.
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HaileISela
what this tiling obscures is the fact that structurally all hexagons and pentagons need to be triangulated. So while the first sketch was of a spherical -octahedron-, the football and all of its variants is a spherical -icosahedron- and those have always, without exception, a dozen vertexes with five triangles around the vertex. the increasing number of hexagons is due to the higher frequency of triangulation. This is all basic synergetics and therefore basic principles of spacetime structure.
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what this tiling obscures is the fact that structurally all hexagons and pentagons need to be triangulated. So while the first sketch was of a spherical -octahedron-, the football and all of its variants is a spherical -icosahedron- and those have always, without exception, a dozen vertexes with five triangles around the vertex. the increasing number of hexagons is due to the higher frequency of triangulation. This is all basic synergetics and therefore basic principles of spacetime structure.
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beeble2003
There's something of a cheat, here. The video doesn't prove that a football -must- have 12 pentagons and 20 hexagons. It only proves that, if you tile a sphere with pentagons and hexagons such that exactly three shapes meet at every vertex and such that every pentagon is entirely surrounded by hexagons, and every hexagon is adjacent to exactly three pentagons, then there must be 12 pentagons and 20 hexagons. The video doesn't, for example, prove that you can't have two pentagons adjacent.
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There's something of a cheat, here. The video doesn't prove that a football -must- have 12 pentagons and 20 hexagons. It only proves that, if you tile a sphere with pentagons and hexagons such that exactly three shapes meet at every vertex and such that every pentagon is entirely surrounded by hexagons, and every hexagon is adjacent to exactly three pentagons, then there must be 12 pentagons and 20 hexagons. The video doesn't, for example, prove that you can't have two pentagons adjacent.
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Ffxfan197
I don't get how the donut's Euler Characteristic is 0. If I put one vertex on the outside of the donut opposite one another, I would get 2 vertices, 2 edges (Around the outside of the donut where it is at its widest, one from vertex A to vertex B, then the second one back to A from B) and then only 1 face as both sides of the edges belong to the same face, connected by the inside of the donut. That would mean 2-2+1=1.
What's wrong with my logic?
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I don't get how the donut's Euler Characteristic is 0. If I put one vertex on the outside of the donut opposite one another, I would get 2 vertices, 2 edges (Around the outside of the donut where it is at its widest, one from vertex A to vertex B, then the second one back to A from B) and then only 1 face as both sides of the edges belong to the same face, connected by the inside of the donut. That would mean 2-2+1=1.
What's wrong with my logic?
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Till
Actually, you made some assumptions:
1. Every Vertex has degree three.
Could you get another soccer ball, if the vertices have other degrees? (maybe not all the same)
2. No two pentaongs are adjacent and every hexagon is surrounded by 3 pentagons and 3 hexagons. Again, what happens, if you change this to. some hexagons are surrounded by hexagons only, or pentagons only.
I guess you get a lot more possible patterns.
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Actually, you made some assumptions:
1. Every Vertex has degree three.
Could you get another soccer ball, if the vertices have other degrees? (maybe not all the same)
2. No two pentaongs are adjacent and every hexagon is surrounded by 3 pentagons and 3 hexagons. Again, what happens, if you change this to. some hexagons are surrounded by hexagons only, or pentagons only.
I guess you get a lot more possible patterns.
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omp199
So why do Americans have to start almost every sentence and phrase with the word -so-, even when it makes no sense to do so? So I think it has started creeping into British English, too. So did you count how many times the person in this video started a sentence with -so-? So I think there may be a theorem that says that as t tends to infinity, the proportion of the language taken up with the word -so- tends to 1.
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So why do Americans have to start almost every sentence and phrase with the word -so-, even when it makes no sense to do so? So I think it has started creeping into British English, too. So did you count how many times the person in this video started a sentence with -so-? So I think there may be a theorem that says that as t tends to infinity, the proportion of the language taken up with the word -so- tends to 1.
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omp199
Yes, it would have been much quicker just to count the faces. In order to make the proof rigorous, the mathematician would have to establish that precisely three faces met at each vertex. Without making any assumptions about the shape of the football, that would mean examining every vertex, and there are 60 of them. That would take much longer than just counting the faces, because there are only 32 of those!
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Yes, it would have been much quicker just to count the faces. In order to make the proof rigorous, the mathematician would have to establish that precisely three faces met at each vertex. Without making any assumptions about the shape of the football, that would mean examining every vertex, and there are 60 of them. That would take much longer than just counting the faces, because there are only 32 of those!
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Chad
I feel like this could be accomplished with just the vertices and not the full Euler Characteristic. Every vertex touches exactly one corner of a pentagon, therefore the number of vertices is 5P. But it is also (5P+6H)/3 as seen in video. Some simple algebra concludes that for every three pentagons there are five hexagons. Although this doesn't yield the exact numerical result, it was way faster.
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I feel like this could be accomplished with just the vertices and not the full Euler Characteristic. Every vertex touches exactly one corner of a pentagon, therefore the number of vertices is 5P. But it is also (5P+6H)/3 as seen in video. Some simple algebra concludes that for every three pentagons there are five hexagons. Although this doesn't yield the exact numerical result, it was way faster.
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Oa
There's another way of arriving at the same numbers. A football is basically a truncated icosahedron (or a d20 if you're into D&D. The hexagons are the triangles of the icosahedron, so it makes sense that there are 20 of them. The number of pentagons will be the same as the number of vertices for the icosahedron; 20 triangles where each vertex is shared by 5 faces makes for 20 - 3 / 5 = 12. :)
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There's another way of arriving at the same numbers. A football is basically a truncated icosahedron (or a d20 if you're into D&D. The hexagons are the triangles of the icosahedron, so it makes sense that there are 20 of them. The number of pentagons will be the same as the number of vertices for the icosahedron; 20 triangles where each vertex is shared by 5 faces makes for 20 - 3 / 5 = 12. :)
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