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.
Ok, Frank, you goaded me into it. Here is a solution that works, involving no handwaving.
a) Disregard the possibility that XOY or XOZ or YOZ = 0 or 180 degrees, since that probability is 0. Extend line XO in both directions, dividing the triangle into two pieces. Similarly, Extend YO in both directions, Now the triangle is divided into 4 pieces. Z lies in one of those four pieces.
b) Place X' on line OX, such that O is halfway between X' and X. Because X is within the hexagon whose center is O, X' is also in the hexagon. Similarly, Place Y' on line OY, such that O is halfway between Y' and Y.
c) No matter where Z is located, O is in one and only one of the four triangles XYZ, X'YZ, X'YZ, and X'Y'Z. (This is true even if Z is outside the triangle, although of course it isn't). For any triangle XYZ, there are three "sister" triangles, such that only one of the four "sisters" contains point O.
e) Therefore, for any point Z, exactly 1/4 of all triangles XYZ (where randomly chosen X any Y are closer to O than to a vertex) contain O.
f) Therefore, the probability that triangle XYZ contains O is exactly 1/4, which is greater than the 2/9 probability that UVW contains O.
Thanks for the hint, Frank. Nice puzzle.
The missing insight
