Euler's formula help

Asakura · 2009-01-28 04:58:10

I was reading through an old math book, and at the end of the chapter about Euler's formula (V-E+F=2) there was a Mini Challenge. If anyone could help me solve it it would be much appreciated.

1) A rhombic triacontahedron is a polyhedron with 30 congruent faces which are rhombuses, and 2 types of vertices. At vertice type 1, 3 rhombuses meet and the angles are all obtuse. At vertice type 2, 5 rhombuses meet and the angles are all acute.
i) How many edges are there and
ii) How many vertices of each type?

2) A new semi-regular polyhedron is formed by suitably truncating all the vertices of type 2 of the rhombic triacontahedron. In this case all the vertices have the same number of edges meeting at them but there are two types of faces.
i) Describe the types of faces
ii) How how many of each type there are
iii) Give the number of vertices and edges.

Help would be very much appreciated.
Thank you.

Asakura · 2009-01-28 05:05:24

the answers at the back of the book give the following

1.
i) 60
ii) 20{3}+12{5}

2.
i)/ii) 12 pentagons and 30 hexagons
iii) 80V 120E

if possible, i would like to know how can i get to these answers please.

mathsmypassion · 2009-01-28 08:00:43

1) I can give you a little idea. F = 30 so V - E = -28

But V is made up from two types of vertices. Let's say there are m of type one and n of type two.
So V = m + n
Every edge joins two vertices.
Some of them are type one, they belong to 3 edges in the same time, and the others are type two and they can be found on 5 edges.
So , actually, when you count the edges, you count 3 times every of the vertices type one, and 5 times every of the vertices type two.
I would say 2E = 3m + 5n .

I put 2E because I counted both vertices at the ends of one edge.
We have now one equation:
m + n - (3m/2 +5n/2) = -28
same as:
m + 3n = 56

We need another equation. I can't think of any other so far.

mathsmypassion · 2009-01-28 08:07:55

I got it.
Every face has 4 vertices. So 4F = 120.
So 3m + 5n = 120.
I hope you understand why.
we have now simultaneous equation.

m + 3n = 56
3m + 5n = 120

Solve it!

mathsyperson · 2009-01-28 09:28:20

Judging by the question structure, I'd guess you're meant to argue that every face has four edges and every edge is shared by two faces, concluding that E = (4x30)/2 = 60.

Then you can say from Euler's formula that V = 32 and continue from there.

Asakura · 2009-01-28 10:24:37

thank you very much, this helped a lot.

but for question 2, i understand how the shapes are pentagon and hexagon, but i'm struggling to find how many of each type there is.

mathsmypassion · 2009-01-28 16:47:58

Initially you had 30 faces. By truncating the vertices of type 2, which are 12 in total, actually we add another 12 faces (in fact pentagons ) to the 3-d shape.

Last edited by mathsmypassion (2009-01-28 16:50:13)

Math Is Fun Forum

#1 2009-01-28 04:58:10

Euler's formula help

#2 2009-01-28 05:05:24

Re: Euler's formula help

#3 2009-01-28 08:00:43

Re: Euler's formula help

#4 2009-01-28 08:07:55

Re: Euler's formula help

#5 2009-01-28 09:28:20

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#6 2009-01-28 10:24:37

Re: Euler's formula help

#7 2009-01-28 16:47:58

Re: Euler's formula help

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