Originally by: ReaperOfSly

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Originally by: xOm3gAx

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Edited by: xOm3gAx on 03/11/2007 02:43:30
Edited by: xOm3gAx on 03/11/2007 02:42:32

Originally by: ReaperOfSly

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Edited by: ReaperOfSly on 03/11/2007 00:21:32

Originally by: Sqalevon

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Edited by: Sqalevon on 03/11/2007 00:16:31

Originally by: ReaperOfSly

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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).

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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

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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.

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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 =)

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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.

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Originally by: CCP Abraxas

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Her boyfriend's way hot, too; tall and tanned. And I say this as a very hetero male who doesn't ever dream of the man on cold, dark nights.

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Originally by: vanBuskirk

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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?

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Originally by: ReaperOfSly

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Originally by: vanBuskirk

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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?

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Depends what the radius of the hole is.

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Originally by: vanBuskirk

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Edited by: vanBuskirk on 03/11/2007 20:11:43

Originally by: ReaperOfSly

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Originally by: vanBuskirk

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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?

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Depends what the radius of the hole is.

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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.

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Originally by: ReaperOfSly

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Originally by: vanBuskirk

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Edited by: vanBuskirk on 03/11/2007 20:11:43

Originally by: ReaperOfSly

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Originally by: vanBuskirk

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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?

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Depends what the radius of the hole is.

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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.

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It STILL depends on the radius of the hole

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Originally by: vanBuskirk

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The problem as given does in fact have a unique solution. Odd, but true.

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Originally by: xOm3gAx

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Edited by: xOm3gAx on 02/11/2007 23:34:24

Originally by: Scorpyn

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Originally by: xOm3gAx

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(a(a-b)/(a-b)=((a-b)/(a-b)(a+b))

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This line is where the error is in case anyone was wondering.

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* Let a=b.
* Square both sides: a¦=b¦
* a¦-b¦=0 ... [1]
* (a+b)(a-b)=0 ... [2]
* a¦-b+b=0
* a¦-a+b=0 ... a=b, so substitute a instead of b.
* a(a-b)=0 ... [3]
* [2] and [3] are the same expressions: a(a-b)=(a+b)(a-b)
* divide both sides by (a-b): a=a+b
* a equals b: a=a+a
* divide both sides by a, 1=1+1
* 1=2
* Therefore, 1+1=1+2=3

However, this does not work because if you divide by (a-b), since a = b, you are dividing by 0; and division by 0 is undefined.

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