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.
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
Solving UVW (partial spoiler)
