Let XYZ be an equilateral triangle, centered at O. A point P is chosen at random, inside the triangle. Find the probability that point P is closer to O than to any of the vertices. (In other words, find the probability that OP is less than XP, YP, and ZP.)
Let the side length of the equilateral triangle be 3. Then the points that are closer to the center of the triangle than to X, Y, Z is the circle centered at O with radius 1, so the probability is (pi)/(sqrt(3)/4*9) = (4*pi*sqrt(3))/27.
I am answering this one for you, if asker guest is still interested that would be a bonus.
Here is my pic.
You can see the equilateral triangle.
The desired region is in green.
There are actually 18 congruent triangles here.
12 of them are in the green zone.
So the prob of being in the green (desired) zone is 12/18 = 2/3
I have not included much explanation but if you cannot work it out and you want more explanation then just ask.