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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: I give up! (spoiler) | Comment 3 of 18 |
(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

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