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ac_math

Member

Registered: 2013-11-19

Posts: 2

Halp on hard geometric probability proof

Let n be a positive integer. A regular polygon with 2n+1 sides is inscribed in a circle. Three of the polygon's vertices are selected at random and the triangle formed by connecting these vertices is drawn. Prove that the probability that the center of the circle lies inside the triangle is (n+1)/2(2n-1).

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Halp on hard geometric probability proof

hi ac_math

Wait a mo.  I've just noticed your polygons have 2n +1 sides.  My formula was for n sides with n odd.  I'll correct what follows.

So far I've only looked at regular polygons with an odd number of sides. (To avoid the centre being on a side)

I got (n+1)/2(n-2)

That's not what you have above.  ??

I can give the working for my result if you wish.  If I've made a mistake, maybe you can spot it. 

EDITED VERSION:

now I'm getting (n+1)/(2n-1).

Much better but there's a factor of 2 missing somewhere. 

I need to supply a diagram so I'll put my answer in a new post.

SECOND EDIT: I have found where the 2 comes from.

Bob

Last edited by Bob (2013-11-20 02:05:31)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Halp on hard geometric probability proof

OK.  Here's my attempt.

My diagram shows a nonagon (2n+ 1 = 9) but my formula is for any odd number of sides.

The first point may be any.  Call it A.  Letter the remaining points B, C, D, ........ in order.

Let's suppose the second point is somewhere on the right of A (B,C,D, or E for my diagram)

(note: If it isn't, then just label the points the other way round.)

Now how many ways can I pick the third point so that the centre is included inside the triangle?

For my diagram the possibilities are:

(ABF), (ACF, ACG), (ADF, ADG, ADH), (AEF, AEG, AEH, AEI). That's 1 + 2 + 3 + 4 ways.

In general I think it is n(n+1)/2 ways.   If 2n+1 = 9, then n =4 and 4x5/2 = 10

Now to find how many triangles altogether (assuming A has already been picked)

I can choose the second point in 2n ways, and the third point in 2n-1 ways.  But this will count every triangle twice (eg. ABC and ACB) so I need to divide by 2.

Total number of triangles = 2n(2n-1)/2 = n(2n-1)

So Probability = [n(n+1)] / [2n(2n-1)]  = (n+1)/[2(2n-1)]

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

ac_math

Member

Registered: 2013-11-19

Posts: 2

Re: Halp on hard geometric probability proof

thanks