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
re(2): Not faster, simpler, or better by Charlie)
If all three points lie on the same straight line (in Euclidean space), there would be no triangle. But if two points were on one side of the square, there are infinitely many points on the opposing side, and all three would create triangles. If one of the points was at the center, there would be infinitely many lines going through that point (and even infinitely many with it as a vertex), and serving as one side of an infinite variety of triangles. You seem to say there are more points inside a triangle than there are on its perimeter. How would you prove that? The Alephs of absurdity engulf us. If by a "point" we really mean a "small area" (e.g. if someone "drew" the three "dots") we'd need to ask if the "center" was also to be construed similarly as having dimensions. Mathematical "lines" like "points" have no area. ---- This all seems persnickety, but my point is that the problem does not seem clearly formulated. I believe I read a comment that the center was more likely to be in one of those random triangles than was a point elsewhere in the square, yet several other comments seem to have suggested the probability is 1:4 (which would be half of the range of areas, from near zero to 1/2 the area of the square, i.e. my guess). ---- May I commend the landscape paintings of Ma Yuan, with the alternate suggestion that anyone asked to place three dots in a square "at random" (and without further specification of the use to be made of the dots) would be very likely (psychologically) to space them so as to generate a triangle which would include the center: not to do so could even be a diagnostic test for neural irregularities.
re(3): Not faster, simpler, or better
