[bananarama2]

(Original post by Maths_Lover)
I'm going to school for a while soon as I have some business to sort out. I'll do the questions when I get back.

You sounds like a mafia boss.

[Maths_Lover]

(Original post by wcp100)
You sounds like a mafia boss.

What if I am?

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Anyway, I'm off now to shoot a gangster sort some stuff out.

[Blutooth]

(Original post by Maths_Lover)
Thank you very much. I do like to see different ways of solving the same problem.

I do but I am currently going through a phase of irregular sleeping. I'll sort it out soon, hopefully.



I'm going to school for a while soon as I have some business to sort out. I'll do the questions when I get back.

Happens to the best of us. Here's something else to keep you up.

Spoiler:

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A polyhedron is a solid bounded by F plane faces, which meet in E edges and V vertices. You may assume Euler’s formula, that V − E + F = 2.

In a regular polyhedron the faces are equal regular m-sided polygons, n of which meet at each vertex. Show that:



where


By considering the possible values of h, or otherwise, prove that there are only five regular polyhedra.

[Llewellyn]

(Original post by Blutooth)
In a regular polyhedron the faces are equal regular m-sided polygons, n of which meet at each vertex.

So

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I take this to mean that for each of the m faces on the polyhedron, each m will meet the vertices n times? (So that mF counts the vertices n times?)

Edit: nvm, it does work

[Blutooth]

(Original post by Llewellyn)
So

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I take this to mean that for each of the m faces on the polyhedron, each m will meet the vertices n times? (So that mF counts the vertices n times?)

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You are correct , though I think you have deviated from the notation used in the question. M=the number of edges on each face. F= the number of faces and n= the number of edges meeting at each vertex.


Edit: Each edge belongs to two faces, so Fm=double the total number of edges. Similarly each edge belongs to two vertices so Vn= double the number of edges.

[Llewellyn]

(Original post by Blutooth)

Spoiler:

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You are correct , though I think you have deviated from the notation used in the question. M=the number of edges on each face. F= the number of faces and n= the number of edges meeting at each vertex.


Edit: Each edge belongs to two faces, so Fm=double the total number of edges. Similarly each edge belongs to two vertices so Vn= double the number of edges.

Yeah I messed up because I didn't realise there would be quite that much algebra, so when it didn't come out straight away I was a little taken aback

That's quite a fun question though, and it's interesting that there are only 5 regular polyhedra.

Anyways; this is how I did it (eventually).
Some parts have been chopped off (terrible camera ), but it still gives the general swing of the method.

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I'm guessing this was probably a BMO question, but I've been through most of the past papers and I can't recall doing it

[Ensam_Varg]

Guten tag. I had a strange dream last night...I was in a German listening exam...yet half of the words were Swedish. I also don't do A-Level German. ¬_¬

[Blutooth]

(Original post by Llewellyn)
Yeah I messed up because I didn't realise there would be quite that much algebra, so when it didn't come out straight away I was a little taken aback

That's quite a fun question though, and it's interesting that there are only 5 regular polyhedra.

Anyways; this is how I did it (eventually).
Some parts have been chopped off (terrible camera ), but it still gives the general swing of the method.

Spoiler:

Show

I'm guessing this was probably a BMO question, but I've been through most of the past papers and I can't recall doing it

Nice solution. Surprisingly the question came from a STEP iii paper. Most of the STEP questions I've tried have been quite a bit more algebra intensive than this, but FWIW when I first did this question last year I found it hard. Anyway you seemed to make quick work of it. I was gonna offer you guys a BMO2 question but not sure where is the right level of difficulty where questions are fun and not just demoralising.

[bananarama2]

(Original post by Blutooth)
Nice solution. Surprisingly the question came from a STEP iii paper. Most of the STEP questions I've tried have been quite a bit more algebra intensive than this. Anyway you seemed to make quick work of it. I was gonna offer you guys a BMO2 question but not sure where is the right level of difficulty where questions are fun and not just demoralising.

[Llewellyn]

(Original post by Blutooth)
Nice solution. Surprisingly the question came from a STEP iii paper. Most of the STEP questions I've tried have been quite a bit more algebra intensive than this, but FWIW when I first did this question last year I found it hard. Anyway you seemed to make quick work of it. I was gonna offer you guys a BMO2 question but not sure where is the right level of difficulty where questions are fun and not just demoralising.

Very, very few people (even on this forum) would make light work of a BMO2 question. I know I'm not one of them but that doesn't mean you can't post it as I suspect some people in this thread could give it a good go.

[Blutooth]

(Original post by wcp100)

I can't tell whether that smiley is rolling around in laughter or dying agony having seen his marks from the latest BMO2 paper.

Edit: Why do I come out with the lamest jokes imaginable?

[bananarama2]

Where are these BMO2 questions then?

[Maths_Lover]

(Original post by wcp100)
Where are these BMO2 questions then?

Someone's keen.

[bananarama2]

(Original post by Maths_Lover)
Someone's keen.

I've done up to and including question 4.

[Maths_Lover]

(Original post by Blutooth)
Happens to the best of us. Here's something else to keep you up.

Spoiler:

Show

A polyhedron is a solid bounded by F plane faces, which meet in E edges and V vertices. You may assume Euler’s formula, that V − E + F = 2.

In a regular polyhedron the faces are equal regular m-sided polygons, n of which meet at each vertex. Show that:



where


By considering the possible values of h, or otherwise, prove that there are only five regular polyhedra.

Everytime I try eliminating one of E or V in my expression for F, I end up with the other one. I've been going in circles for a while, so I'm missing something important obviously.

I'll get there eventually... hopefully.

[Maths_Lover]

(Original post by wcp100)
I've done up to and including question 4.

Excellent.

There was an item of business that I forgot to sort out at school.

I thought I there was something I hadn't done...

In addition, my bicycle is soaked.

[bananarama2]

(Original post by Maths_Lover)
Excellent.

There was an item of business that I forgot to sort out at school.

I thought I there was something I hadn't done...

In addition, my bicycle is soaked.

Tutut. It's the horses head in someones bed then?

That happens when it rains!

[Llewellyn]

I've found an interesting question

[bananarama2]

(Original post by Llewellyn)
I've found an interesting question

Oh dear...go on....

[Llewellyn]

Prove that a positive integer n exists, such that the decimal representation of begins and ends with the digits "2001".

i.e.


Things you may want to consider:
-Modulo
-The pigeonhole principle
-Rewriting the question

But not necessarily in that order of course