wlcm: there is no circle. your proof seems to have confused me.
here's swing #2 from me, dimbulb shot down my first try.
in order for a triangle w/ points on the edges of a square to encompass the center of said square 2 things must be true.
#1: 2 points must be on opposide sides of the square.
#2: the third point must be within a certain area opposite the center. this area is defined by projecting a line between the first 2 points and the center. the distance between those 2 points is the area in which a third point will create a triangle that encompasses the center. lets call the distance between one point and a line through the center parallel to a side of the square a, and the similar distance with the other point b. so. we're back to the fact that the probility that the first and 2nd points will be on opposites faces is 1/4
P(#1) = .25
now, the P(#2) is true is a ratio between the defined length, and the complete length of the perimeter of the square. there are 2 possiblities. either the first 2 points are both on the same side of the center dividing lines, like in the first picture below. in this case, the area is defined by 2*l - a - b assuming that l is the length of a side.
if the points are on opposide sides of the center line, then the lenght is defined by 2*l - a + b. alternatively, one of the values can be considered negative. now, considering the probablities that each case will happen (1/2 for each),
P(#2) =[ 1/2 * ( 2*l -a -b) + 1/2 * (2*l -a +b)] 4*l
so, the answer will be the intersection of those 2 probablities.
P(1 n 2) = 1/4 * (above) = 1/16* (l-a)/l
That's stab #2. here's the piccies.
