Euler's Formula for Polyhedra
Date: 08/12/97 at 12:20:57 From: Jonah Knobler Subject: Euler's Formula for Polyhedra Dear Dr. Math, How would you prove Euler's formula V-E+F = 2 for all polyhedra of genus zero? I have seen proofs of this, but they all talk about graphs vs. trees and loops in graphs and other things that they don't bother to explain. So why is this formula always 2? And why does adding a hole always lower this number by 2? We don't talk about this in high school and I'm too curious to wait. - Jonah Knobler
Date: 08/18/97 at 10:26:31 From: Doctor Rob Subject: Re: Euler's Formula for Polyhedra Good question! First observation: For an n-gon in the plane, V = n, E = n, F = 1, and V-E+F = n-n+1 = 1. Second observation: If we paste together two polygons along one side, V-E+F = (V1+V2-2)-(E1+E2-1)+(F1+F2) = 1. Third observation: If we paste together any number of polygons in the same way, V-E+F = 1. This also means that we can take a polygon and cut it into smaller polygons by drawing diagonals, and still have V-E+F = 1. In fact, we can cut all polygons into triangles this way. Fourth observation: Stretching and shrinking edge lengths does not change V, E, or F, and so V-E+F, unless an edge is shrunk to length zero. Fifth observation: If we start with a genus zero polyhedron and think of it as made of rubber, and we blow it up like a balloon, we will end up with a sphere with the vertices as points on its surface, the edges as arcs of circles on its surface, and the faces represented by polygonal areas on its surface. (Think of a soccer ball.) Furthermore, V, E, and F, don't change, so V-E+F doesn't change. We then pick an interior point of one of the faces, and puncture the balloon there, and spread the surface out flat in the plane. This stretching and/or shrinking does not change V, E, or F (as observed above). The picture is the same as one in the plane with the other F-1 polygonal faces pasted together, for which V-E+(F-1) = 1 (as observed above). Thus V-E+F = 2 for genus-zero polyhedra. Given a genus-zero polyhedron, we can cut all its faces into triangles, as observed above, without changing V-E+F. If we take two triangles that are not touching at any vertex, we can cut a polyhedral hole from one to the other. We remove the two triangle faces and add six new triangle faces, six edges, and no vertices. Here is an approximate picture: A.-------------------.B |`-._ _,-'| | \ `-._ C _,-' | | \ `-' | | \ | \ | | \ | \ | | _._ \ | | _,-' Z `-._ \ | |_,-' `-._| X'-------------------`Y The original triangles are ABC and XYZ. I have drawn edges AX, AZ, BY, CY, and CZ. I couldn't draw, and you will have to imagine, edge BX. If you cut along AX and unroll it, you will get this picture: C B A.--------.--------.--------.A |\ |\ |\ | | \ | \ | \ | | \ | \ | \ | | \ | \ | \ | | \ | \ | \ | | \ | \ | \ | | \ | \ | \ | | \| \| \| X.--------.--------.--------.X Z Y The six new faces are ABX, BCY, CAZ, AZX, BXY, and CYZ. No new vertices were required. This reduces V-E+F by 2. We can repeat this processas often as we like, reducing V-E+F by 2 for each hole created. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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