You choose a random point, uniformly, within an equilateral triangle.

What's the average distance to the three sides?

Assuming the side is length 1, the average distance is sqrt(3)/6, no matter what point you pick.

Proof: Label the original triangle ABC, and the random point in the interior R. Draw 3 lines through R parallel to AB, AC and BC. These lines intersect the original triangle at 6 points. Label them D,E,F,G,H,I so that the sequence around the triangle is: A, D, E. B, F, G, C, H, I, A.

Notice that RDE, RFG, and RHI are all equilateral, since all their sides are parallel to the sides of ABC. Also notice that the line representing the distance from R to any of the sides of ABC is the altitude of one of these triangles. Furthermore, RI + DE + RF = 1, since RI=AD and RF = EB. That means the sum of the sides of the 3 small triangles = one side of the larger triangle. By similar triangle ratios, the sum of the altitudes of these 3 small triangles must equal the altitude of the large triangle.

The altitude of the big triangle is sqrt(3)/2. So the sum of the altitudes of the three smaller triangles must also be sqrt(3)/2. But the problem calls for the average distance, which is just this sum / 3, which is sqrt(3)/6.

*Edited on ***November 24, 2005, 9:55 am**