2 qs (1 Viewer)

[Templar]

and you're certain that we don't need additional theory??? (eg. what exactly is a convex polyhedron?)

and other things like the Euler Formula: V + F - E = 2 ?

Convex polyhedron is a polyhedron where any 2 points in or on it can be joined by a line that is totally or the surface or within polyhedron. Basically any polyhedron constructed by convex polygons as faces.

I thought Euler's Formula was taught in the syllabus somewhere, although I could be mistaken. But no, it's not essential, and I believe that path is actually quite messy should you pursue it.

 

[who_loves_maths]

Originally Posted by Templar
Convex polyhedron is a polyhedron where any 2 points in or on it can be joined by a line that is totally or the surface or within polyhedron. Basically any polyhedron constructed by convex polygons as faces.

the text in bold is not necessarily true is it?

Originally Posted by Templar
I thought Euler's Formula was taught in the syllabus somewhere, although I could be mistaken. But no, it's not essential, and I believe that path is actually quite messy should you pursue it.

no, Euler's Formula is not taught at the high school level - there is no graph theory in the syllabus... you should know this, you only left high school at the end of last year?

well, seeing another approach to doing this question would require some time, since ppl are still getting to know the basics about convex polyhedrons, etc...

this would be something like a special Maths Comp. question at the senior high school level.


P.S. do you need graph theory for this question AT ALL? things like simple planar graphs, etc...?

Edit: and Euler inequalities ?

 

[SeDaTeD]

Well, basic gist is, suppose there are F faces. Each face is adjacent to k other faces, where 3 <= k <= F-1, ie. it is a k-gon. thus there are F-3 possible types of polygons (as in each face could be either a 3-gon, 4-gon etc). Since there are F faces, by the pigeon hole principle at least 2 must have the same number of edges.

 

[who_loves_maths]

Originally Posted by SeDaTeD
Well, basic gist is, suppose there are F faces. Each face is adjacent to k other faces, where 3 <= k <= F-1, ie. it is a k-gon. thus there are F-3 possible types of polygons (as in each face could be either a 3-gon, 4-gon etc). Since there are F faces, by the pigeon hole principle at least 2 must have the same number of edges.

can you please explain this in detail? it doesn't make sense to me at this stage.

For example, the text in bold, when you say "Each face is adjacent to k other faces" and then go on to say "it is a k-gon" - what do you actually mean? do you mean that EACH flat face is a 2D k-gon? if so, then in this case you have already generalised that each face has the same 'k' number of sides...
or, do you mean that the polyhedron has a total of 'k' faces, hence "k-gon"? which also doesn't make sense since 'k' doesn't equal to 'F'...

also, shouldn't the inequality 3 <= k <= F-1 actually be 3 <= k <= F-2 ? since if k = F -1, then the polyhedron is not convex ---> it's not even closed ?!

so far you argument doesn't seem that strong to me... so plz explain if i have misunderstood anything that you were trying to say.

 

[who_loves_maths]

meanwhile, can you plz answer my question Templar?

do you need graph theory at all for this question? Euler inequalities?

 

[SeDaTeD]

Oh i didn't mean that the k is the same in all cases, it can be different for each face. Each face is a polygon with a number of edges, some number between 3 and F-1, since a polygon has at least 3 sides and there are at most F-1 other sides it can be adjacent to. Since there are F-3 different types of polygon faces (as in 2 polygons with the same number of faces are considered the same type), and F faces in total, at least 2 must have the same number of faces.
Yeh it was a bit sloppy that explanation, but it wasn't intended to be very formal.

 

[Templar]

SeDaTeD's proof is strong enough. Wording could have been better, but nobody cares.

As shown, you don't need Euler's inequality or graph theory to do this, although it does make it easier to understand than the proof given perhaps, by treating each face as a node.It's realy just logical reasoning (if proof by contradiction) or pigeon hole principle.

Take each face to be a node. If there are no 2 nodes with the same number of paths, one node will have >=3 paths, >=4 etc up to >=n+3 for n nodes. But there are only n-1 nodes to connect to and since it is convex, there can only be maximum one path between 2 nodes.

That is basically what he said, just reworded.

As for the syllabus...I never knew what was in the syllabus, only a general idea. I've come across too much other stuff to distinguish which is from where.

 

[who_loves_maths]

okay, well i've just given more thought about the question, and just came online and was about to post when i saw you guys are already a step ahead of me here...

graph theory is not need AT ALL here, so there is no need to allude to 'paths' or 'nodes', etc... Templar.

also, i still think the wording needs to be a little more clearer in order for ppl who do not usually touch this sort of mathematics at school to be able to understand the whole situation. in mathematics, good writing and communication skills are very important in being able to convey ideas and concepts, if no one understands you then you might as well not have spoken at all.

anyhow, since i'm making a post here, i am just going to post my solution to that problem anyways, and yes like you said, it's entirely logical:


the smallest polygon (2D) has three sides - a triangle. the simplest polyhedron (3D) is then made up entirely of triangles - a triangular pyramid or tetrahadron - that has four faces.
so now make the observation that a polyhedron with uniform k-gon faces (that is, ALL faces are k-gons) must have AT LEAST (k +1) faces.

the proof for this is simple:
it's best to visualise the polyhedron sitting on a flat surface - that is, it has a base, and since all faces are k-gons then the base will have k-sides or edges. from this base, we can build the polyhedron - every edge connects two faces, so from the k-sides of the base there must extend k-faces into space to complete a 3D figure. hence, along with the base itself as a face, the polyhedron must have at least (k+1) faces.

now, back to the original question, the hypothetical polyhedron in question must not have ANY two faces which have the same number of edges... to prove this is impossible, let's make the face of this hypothetical polyhedron that has the MOST number of edges, 'n', the base of the polyhedron.
since that face is part of the polyhedron and is now the base, then the polyhedron must have AT LEAST (n +1) faces.

now, as mentioned before, the simplest of polygons, ie. faces, is one with three sides. therefore, the total possible differently-sided faces of the hypothetical polyhedron are ones with sides: {3, 4, 5, 6, ...., n} and since no same-sided face can repeat, then the largest number of different faces is (n -2) [ie. the number of elements in that set].

but, there are at least (n+1) faces, and (n-2) < (n+1) ; and so, there are (n+1) -(n-2) = 3 faces left over that must have the same number of sides/edges as one, two, or three other faces of the polyhedron.

hence, the hypothetical polyhedron with all faces having different numbers of sides does not exist - meaning that ANY convex polyhedron must have at least two faces that share the same number of edges.