Three points have been chosen randomly from the vertices of a n-sided regular polygon.

Determine the probability (in terms of n) that they form (a) an acute triangle; (b) a right triangle?

Without loss of generality consider A to be fixed, with B and C chosen at random to form the triangle. B can be chosen in n - 1 ways and C in n - 2 ways so that the number of different triangles is (n - 1)(n - 2)/2.

 For ABC to be acute angled, all three sides must subtend angles less than 180 degrees at the centre of the circumcircle; in other words the centre must be inside the triangle. If CD is a diameter, this means that D must lie within the arc AB which must be less than a semicircle, with the following consequences.
(a)
When n is even: If AB spans x sides of the polygon (x can be from 1 to n/2 - 1) then C can be chosen in x - 1 ways, giving the following number of possible triangles

            Sum(x=1 to n/2-1) of (x - 1)  =  0 + 1 + 2 + ... + (n/2 - 2)

                                                            = (n/2 - 2)(n/2 - 1)/2

Therefore:         P(acute angled)  =  (n/2 - 2)(n/2 - 1)/[(n - 1)(n - 2)]

                                                =  (n - 4)/[4(n - 1)]                    n even

When n is odd: If AB spans x sides of the polygon (x can be from 1 to (n - 1)/2 then C can be chosen in x ways, giving the following number of possible triangles:

            Sum(x=1 to (n-1)/2) of x      =  1 + 2 + 3 + ... + (n - 1)/2

                                                            = (n - 1)(n + 1)/8

Therefore:         P(acute angled) = (n - 1)(n + 1)/[4(n - 1)(n - 2)]

                                                =  (n + 1)/[4(n - 2)]                   n odd

(b)
ABC will be right angled if and only if one of its sides is a diameter. This is only possible when n is even; then the probability that AB is a diameter is 1/(n - 1) and this must be the same for BC and CA. These three possibilities are mutually exclusive so we can add the three probabilities to find the probability of AB or BC or CA being a diameter (given that ABC is a triangle).

            When n is even:             P(right angled)   =  3/(n - 1)

            When n is odd:               P(right angled)   = 0

Results
n          P(acute angled)              P(right angled)
3          1                                  0
4          0                                  1
5          1/2                               0
6          1/10                              3/5
7          2/5                               0
8          1/7                               3/7

Edited on August 15, 2011, 1:03 am

Posted by Harry on 2011-08-15 01:00:49