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
re(3): Not faster, simpler, or better by ed bottemiller)
As an example of a point other than the center to be tested for inclusion in the triangle, take a point that's midway between the center and one of the corners of the square, at (0.75,0.75) (for a square with diagonal from (0,0) to (1,1).
In 1,000,000 trials, 92,807 of the time this point was within the triangle, for a probability of 0.092807. To allow for the vagaries of the simulation, say 0.0928 or 0.093. That's quite a bit smaller than 1/4. It's between 1/10 and 1/11.
DEFDBL AZ
FOR i = 1 TO 1000000
x1 = RND(1): y1 = RND(1)
x2 = RND(1): y2 = RND(1)
x3 = RND(1): y3 = RND(1)
m = (y2  y1) / (x2  x1)
a = y1  m * x1
test1 = y3  (m * x3 + a)
test2 = .75  (m * .75 + a)
m = (y3  y1) / (x3  x1)
a = y1  m * x1
test3 = y2  (m * x2 + a)
test4 = .75  (m * .75 + a)
m = (y3  y2) / (x3  x2)
a = y2  m * x2
test5 = y1  (m * x1 + a)
test6 = .75  (m * .75 + a)
IF test1 * test2 > 0 AND test3 * test4 > 0 AND test5 * test6 > 0 THEN
hit = hit + 1
END IF
ct = ct + 1
PRINT hit, ct, hit / ct
NEXT

Posted by Charlie
on 20080318 16:49:16 