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 Capture the Flag (Posted on 2008-05-04)

Let O designate the centre of an equilateral triangle. Points U-Z are chosen at random within the triangle. We have learnt that points U,V,W are each nearer to a (possibly different) vertex than to O; while X,Y are each closer to O than to any of the vertices.

Show that triangle XYZ is more likely than triangle UVW to contain the point O within its interior.

 See The Solution Submitted by FrankM Rating: 3.5000 (2 votes)

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 Solving UVW (partial spoiler) | Comment 1 of 11
Points U, V and W are in one of the three equilateral triangles which include the triangle's three vertices and have sides with a length equal to 1/3 of the length of the side of the full triangle.

Points X and Y are in the hexagon which is formed after "removing" the three equilateral triangles.

Triangle UVW includes O if and only if they are nearer to three different vertices (i.e. each one is in a different smaller equilateral triangle).  V is in a different triangle than U with probability 2/3, and W is in the third with probability 1/3, so the probability that UVW includes O = (2/3)*(1/3) = 2/9
 Posted by Steve Herman on 2008-05-04 15:26:12

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