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.

Well, I am encouraged by Frank's kind remarks, and by the fact that nobody else has commented recently, and by Charlie's empirical solution which confirms my intuition. So, here goes some handwaving about triangle XYZ.

a) Let's say that we know that angle XOY is n degrees. Disregard the possibility that XOY = 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. One of those 4 sections (the "far" section) has neither X nor Y on its edge. Note that the far section forms an angle of n degrees at point O. Triangle XYZ includes O if and only if Z is in that "far" section.

b) So what is the probability that Z is in that "far" section? It is the ratio of the expected area of the "far" section to the area of the full triangle. I assert (without proof) that the expected ratio across all possible X's and Y's is just the ratio of the angle it forms at point 0 to the full 360 degrees. The probability that XYZ contains 0 is n/360.

c) And what is the expected value of n? Well, the hexagon in which X and Y randomly exist is so close to a circle that I am very confident that the expected value of XOY is 90 degrees.

d) So, the probability that XYZ contains 0 is 90/360 = 1/4. This agrees with Charlie's empirical findings.

e) So, we are done.

*Edited on ***May 5, 2008, 10:34 pm**