Three points are chosen at random inside a square. Each point is chosen by choosing a random xcoordinate and a random ycoordinate.
A triangle is drawn with the three random points as the vertices. What is the probability that the center of the square is inside the triangle?
(In reply to
I give up! (spoiler) by Steve Herman)
yup! When you substitute x=y=1/2 in the formula, the terms with natural logs drop out and you're left with
4(12/2^6  30/2^5 + 3/2^4  24/2^5 + 60/2^4  6/2^3 + 25/2^4  43/2^3 + 4/2^2  13/2^3 + 13/2^2  1/2 + 1/4  1/2)(1/2)^2
= (12/2^6  54/2^5 + 88/2^4  62/2^3 + 17/2^2  3/4)
= 1/4

Posted by Charlie
on 20080317 16:28:55 