Three points are chosen at random inside a square. Each point is chosen by choosing a random x-coordinate and a random y-coordinate.

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 2008-03-17 16:28:55