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 Three Points in a Square (Posted on 2008-03-17)
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?

 No Solution Yet Submitted by Brian Smith Rating: 3.2000 (5 votes)

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 re(4): Not faster, simpler, or better | Comment 13 of 18 |
(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 A-Z

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 2008-03-18 16:49:16

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