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l33tminion) wrote2006-10-21 12:53 am
l33tminion) wrote2006-10-21 12:53 am
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Soccer Ball Theorem Proof
Here's the answer to that homework problem I mentioned earlier. To restate the problem:
A ball is sewn together from pentagonal and hexagonal pieces of material. Each edge of each polygon lines up exactly with an edge of another polygon. Three polygons meet at each corner of each polygon. Prove that the ball must contain exactly 12 pentagons.
The answer to this problem is simple if we use Euler's formula, which applies to planar graphs. A graph is a collection of points (vertices), which may be connected by lines (edges). In a planar graph, all the vertices must be in a plane (on a flat surface), and none of the edges can cross. Euler's formula is as follows:
r = e - v + 2
In this equation, r is the number of regions (the number of areas the graph splits the plane into), e is the number of edges, and v is the number of vertices.
We can make a planar graph out of the ball by removing one of the polygons and stretching the remaining material flat, placing vertices at the corners of each polygon and edges at the sides of each polygon. None of the edges cross, as the polygons are sewn together evenly at their sides.
Each region inside the graph represents a polygon on the stretched-flat sheet of material, and the region outside of the graph represents the polygon that was removed. Thus, the number of regions defined by the graph is equal to the total number of polygons. If P is the number of pentagons and H is the number of hexagons:
r = P + H
The number of edges of the graph is the total number of polygon sides divided by 2, since 2 polygon sides meet at each edge. Thus:
e = (5P+6H)/2
The number of vertices of the graph is the total number of polygon corners divided by 3, since 3 polygon corners meet at each edge. Thus:
v = (5P+6H)/3
We can plug those values into Euler's formula:
Therefore, the ball must contain exactly 12 pentagons, QED.
A ball is sewn together from pentagonal and hexagonal pieces of material. Each edge of each polygon lines up exactly with an edge of another polygon. Three polygons meet at each corner of each polygon. Prove that the ball must contain exactly 12 pentagons.
The answer to this problem is simple if we use Euler's formula, which applies to planar graphs. A graph is a collection of points (vertices), which may be connected by lines (edges). In a planar graph, all the vertices must be in a plane (on a flat surface), and none of the edges can cross. Euler's formula is as follows:
r = e - v + 2
In this equation, r is the number of regions (the number of areas the graph splits the plane into), e is the number of edges, and v is the number of vertices.
We can make a planar graph out of the ball by removing one of the polygons and stretching the remaining material flat, placing vertices at the corners of each polygon and edges at the sides of each polygon. None of the edges cross, as the polygons are sewn together evenly at their sides.
Each region inside the graph represents a polygon on the stretched-flat sheet of material, and the region outside of the graph represents the polygon that was removed. Thus, the number of regions defined by the graph is equal to the total number of polygons. If P is the number of pentagons and H is the number of hexagons:
r = P + H
The number of edges of the graph is the total number of polygon sides divided by 2, since 2 polygon sides meet at each edge. Thus:
e = (5P+6H)/2
The number of vertices of the graph is the total number of polygon corners divided by 3, since 3 polygon corners meet at each edge. Thus:
v = (5P+6H)/3
We can plug those values into Euler's formula:
| r = e - v + 2 | (Euler's formula) |
| P + H = (5P+6H)/2 - (5P+6H)/3 + 2 | (Plug in values for r, e, and v) |
| 6P + 6H = 15P + 18H - 10P - 12H + 12 | (Multiply both sides by 6) |
| 6P + 6H = 5P + 6H + 12 | (Group terms) |
| P = 12 | (Subtract 5P+6H from both sides) |
Therefore, the ball must contain exactly 12 pentagons, QED.
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