Let S be a set of 31 equally spaced points on a circle centered at O, and consider a uniformly random pair of distinct points A and B (A,B∈S). The probability that the perpendicular bisectors of OA and OB intersect strictly inside the circle can be expressed as m/n, where m,n are relatively prime positive integers. Find m+n?
To find the probability that the perpendicular bisectors of OA and OB intersect strictly inside the circle, we need to determine the number of favorable outcomes and the total number of possible outcomes.
First, let's find the total number of possible outcomes. Since we have a set of 31 equally spaced points on a circle, there are a total of (31 choose 2) = 31*30/2 = 465 possible pairs of distinct points (A, B).
Now, let's determine the number of favorable outcomes. We know that the perpendicular bisectors of OA and OB intersect strictly inside the circle if and only if the points A and B are not diametrically opposite on the circle.
To count the number of pairs (A, B) where A and B are not diametrically opposite, we need to consider that the circle has 31 points. We start by fixing point A and count the number of possible positions for point B such that they are not diametrically opposite.
If point A is fixed, there are 15 points on one side of the diameter passing through A, and 15 points on the other side. However, we exclude the case where B is diametrically opposite to A, which gives us a total of 15*2 - 1 = 29 possible points for B.
Since point A can be chosen in 31 different ways, the total number of favorable outcomes is 31 * 29 = 899.
Therefore, the probability is 899/465.
To express this probability as a fraction in simplest form, we can find the greatest common divisor (GCD) of 899 and 465 and divide both the numerator and denominator by the GCD.
GCD(899, 465) = 31
899/31 = 29
465/31 = 15
So the simplified fraction is 29/15.
Finally, the answer to the problem is 29 + 15 = 44.
Therefore, m + n = 44.