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        riddles >> easy >> Triangles, Pentagons, and Hexagons
        
(Message started by: ThudanBlunder on May 8th, 2007, 8:36am)
Title: Triangles, Pentagons, and Hexagons
Post by ThudanBlunder on May 8th, 2007, 8:36am
1) If you choose a number at random from Pascal's triangle (and not just a finite segment of it) what is the probability that it will be odd?

2) In order to make a regular polyhedron from pentagons and hexagons how many pentagons will you need?
Title: Re: Triangles, Pentagons, and Hexagons
Post by towr on May 8th, 2007, 8:43am
1) The same probability that a random integer is odd, depending on your method of random selection. ::)

2) [hide]always 12, if I recall correctly[/hide]
Title: Re: Triangles, Pentagons, and Hexagons
Post by Grimbal on May 8th, 2007, 8:46am
1) Something like [hide] (3/4)infinity[/hide]

2) There is a solution in [hide]12[/hide], well known by even the less math-inclined people.
Title: Re: Triangles, Pentagons, and Hexagons
Post by ThudanBlunder on May 8th, 2007, 9:21am

on 05/08/07 at 08:43:48, towr wrote:
1) The same probability that a random integer is odd, depending on your method of random selection. ::)

That sounds like 1/2 to me.   :-/
Title: Re: Triangles, Pentagons, and Hexagons
Post by towr on May 8th, 2007, 9:53am

on 05/08/07 at 09:21:26, ThudanBlunder wrote:
That sounds like 1/2 to me.   :-/
It should sound like "whatever you want"
Consider the set of (non-negative) integers {4i+1, 4i+3, 2i | i http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbn.gif}. If you pick uniformly random from the set with i's up to n, then odd numbers are twice as likely as even numbers.
Picking (uniformly) randomly from an infinite set often has these problems.

If you pick the typical way of constructing Pascals's triangle as basis for growing the multiset, the limit of picking uniformly random from it should give the same probability as Grimbal.
Title: Re: Triangles, Pentagons, and Hexagons
Post by Grimbal on May 8th, 2007, 9:57am
I think my solution is the most natural one, even though there really isn't a natural way to choose a number randomly in Pascal's triangle.  So Towr's answer is more correct.
Title: Re: Triangles, Pentagons, and Hexagons
Post by ThudanBlunder on May 8th, 2007, 10:07am

on 05/08/07 at 09:57:45, Grimbal wrote:
I think my solution is the most natural one, even though there really isn't a natural way to choose a number randomly in Pascal's triangle.  So Towr's answer is more correct.

Yes, I should have phrased the question differently.
Title: Re: Triangles, Pentagons, and Hexagons
Post by SMQ on May 8th, 2007, 10:12am

on 05/08/07 at 08:46:58, Grimbal wrote:
12) There is a solution in [hide]12[/hide], well known by even the less math-inclined people.

And, assuming the number of pentagons is indeed constant, the number needed with zero hexagons should be a dead giveaway to the math-inclined. ;)

--SMQ
Title: Re: Triangles, Pentagons, and Hexagons
Post by Eigenray on May 8th, 2007, 12:39pm

on 05/08/07 at 08:46:58, Grimbal wrote:
1) Something like [hide] (3/4)infinity[/hide]

Let p(n) be the probability that a number, picked uniformly from the first n rows of Pascal's triangle, is odd.  If one looks at p(2x)*[hide]2(4/3)x[/hide], the graph quickly converges to the following (with period 1).

[How come Firefox and IE treat bgcolor="#252525" as #272A2F?]
Title: Re: Triangles, Pentagons, and Hexagons
Post by Grimbal on May 8th, 2007, 1:32pm

on 05/08/07 at 08:46:58, Grimbal wrote:
2) There is a solution in [hide]12[/hide], well known by even the less math-inclined people.

Now that I think of it, few of them would actually know how many pentagons there are.  Or even that there are any pentagons.  ;D
Title: Re: Triangles, Pentagons, and Hexagons
Post by JohanC on May 8th, 2007, 1:58pm

on 05/08/07 at 08:36:06, ThudanBlunder wrote:
2) In order to make a regular polyhedron from pentagons and hexagons how many pentagons will you need?

For some kind of proof, I think we need to start with a well-known formula from
[hide]Descartes (but named after Euler)[/hide].

[hide]That formula states:
V+F-E=2
Or "vertices + faces - edges = 2" for closed genus 0 polyhedra.
Next to that we probably also need to state that for every regular polyhedron consisting of pentagons and hexagons exactly 3 edges meet in every vertex. More than 3 would make the angle sum more than 360º. And less than 3 would be degenerate.
Putting things together:
- suppose we have h hexagons and p pentagons
- then we get E=(6*h+5*p)/2 and F=h+p
- the 3 edges meeting in every vertex can be written as V=E*2/3
- rewriting things we get
       F - E/3 = 2
  or  h+p - (6*h+5*p)/2/3 = 2
  or  6h+6p - (6*h+5*p) = 12
  thus p=12[/hide]

By the way, is there a special reason you put these two questions together?
Other than that Blaise Pascal and [hide]René Descartes[/hide] are both famous philosophers living in 17th century France?
Title: Re: Triangles, Pentagons, and Hexagons
Post by ThudanBlunder on May 10th, 2007, 5:17am

on 05/08/07 at 13:58:02, JohanC wrote:
For some kind of proof, I think we need to start with a well-known formula from
[hide]Descartes (but named after Euler)[/hide].

But didn't Descartes consider only vertices, faces, and plane angles? By omitting edges he couldn't have derived Euler's formula.

Well, I posted the puzzles together only because I thought they were too easy to deserve their own threads.
Title: Re: Triangles, Pentagons, and Hexagons
Post by JohanC on May 10th, 2007, 6:55am

on 05/10/07 at 05:17:03, ThudanBlunder wrote:
But didn't Descartes consider only vertices, faces, and plane angles? By omitting edges he couldn't have derived Euler's formula.

I got this information among other sources from
Mathworld (http://mathworld.wolfram.com/PolyhedralFormula.html) which has a high reputation but, of course, is not always correct.
However, I also encountered some historically reputed source with quite some detail at The Mathematical Association of America (http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2009%20V%20E%20and%20F%20part%202.pdf).


Quote:
I posted the puzzles together only because I thought they were too easy to deserve their own threads.


Oh, I see. Wouldn't things be more organized having only one subject per thread?
Title: Re: Triangles, Pentagons, and Hexagons
Post by ThudanBlunder on May 11th, 2007, 4:23pm

on 05/10/07 at 06:55:20, JohanC wrote:
Oh, I see. Wouldn't things be more organized having only one subject per thread?

True, but I knew that expert metagrobologists (http://en.wikipedia.org/wiki/Metagrobology) like Eigenray, towr, Grimbal, and SMQ would make short work of them (not to mention your own text-book solution to the second).

It seems to me from pages 4 and 5 of your excellent MAA (http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2009%20V%20E%20and%20F%20part%202.pdf)link that Descartes didn't quite get there.
Title: Re: Triangles, Pentagons, and Hexagons
Post by Eigenray on May 11th, 2007, 5:36pm

on 05/11/07 at 16:23:23, ThudanBlunder wrote:
True, but I knew that expert metagrobologists (http://en.wikipedia.org/wiki/Metagrobology) like Eigenray, towr, Grimbal, and SMQ would make short work of them.

I believe that's the first time anyone's used that particular word to describe me.

Incidentally, it seems there's no penury of pentagonal penchants among the pensive.  See the penultimate penetration [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1156455517]here[/link].

And now I'm more than a little curious about that graph I posted before (I guess there was no reason to hide it).  Here's a nicer version (plotting 50000 points).  It certainly looks self-similar, so I'm wondering exactly what iterated transformation would give it, or how to try to find it given the data.  Is there a closed form for the limit?  (At this resolution, it looks exactly the same between 200-201, say.)  But I haven't had much time to look into it.
Title: Re: Triangles, Pentagons, and Hexagons
Post by Eigenray on May 28th, 2007, 9:44pm
Let p(n) be the probability that a number, picked uniformly from the first n rows of Pascal's triangle, is odd, and let f(x) = 2p(x)*x2-r, where r = log2(3).  Then for each x, as khttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif through integers, f(2kx) converges to some value g(x).  The function g satisfies g(2x)=g(x), and for 1<x<2,

g(x) = x-r [ 1+2(x-1)r g(x-1) ].

If (x, y) lies on the graph of g, 1<x<2, then for k=1,2,3,..., so does the point (x', y'), where

x' = 1+x/2k,
y' = x'-r [ 1 + 2(x'-1)r y ].

Starting at a random point and iterating this function system, choosing a random k each time, we get the following picture (in black).  The vertical lines are at 1+1/2k, k=1,2,....  Thus the above transformation maps the whole graph into each interval.

Let g0(x)=1, and

gi+1(x) = x-r [ 1 + 2(x-1)r gi(2k(x-1)) ],

where k is the integer such that 1 < 2k(x-1) < 2.

In red, green, and blue, respectively, are shown the graphs of g1, g2, and g3, approximating the graph of g.

The minimum value of g is

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3/[3-2*(2-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2)r] = 0.80818...

What is the fractal dimension of this graph? ::)
Title: Re: Triangles, Pentagons, and Hexagons
Post by towr on Dec 22nd, 2008, 5:08am

on 05/08/07 at 12:39:37, Eigenray wrote:
[How come Firefox and IE treat bgcolor="#252525" as #272A2F?]
They don't, they just think class is more important than background.
Title: Re: Triangles, Pentagons, and Hexagons
Post by Eigenray on Dec 22nd, 2008, 6:02am

on 12/22/08 at 05:08:14, towr wrote:
They don't, they just think class is more important than background.

Ah, I'll keep that in mind next time I try to hide an image.  But then why does .windowbg have background-color #272A2F while the table has bgcolor #252525?   ::)
Title: Re: Triangles, Pentagons, and Hexagons
Post by towr on Dec 22nd, 2008, 6:37am
That's just one of those mysteries of life that might only ever be cleared up if William comes round to do so.


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