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.
(In reply to
re: Extra Credit (Triangle XUV) by Charlie)
Ah yes, thanks Charlie. I see my mistake.
2/3 of the time U and V are near different vertexes. And 1/3 of those times X is located such that XUV contains O. (1/3)*(2/3) = 2/9, which is the number I gave. (Incidentally, this is true even if X is not in the triangle, or is closer to a vertex than the center. And yes, I know I haven't given this proof yet).
What I overlooked was that XUV can contain O even if U and V are closest to the same vertex (which happens with probability 1/3). If Charlie is right (and he usually is), then in 1/30 of these cases XUV contains 0.
2/9 + (1/3)*(1/30) = 21/90 = 7/30. I have yet to calculate the 1/30, and don't see an obvious way to do it.
re(2): Extra Credit (Triangle XUV)
