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 Perpendiculars Picking Probability (Posted on 2005-11-24)
You choose a random point, uniformly, within an equilateral triangle.

What's the average distance to the three sides?

 See The Solution Submitted by Old Original Oskar! Rating: 3.8000 (5 votes)

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 Solution | Comment 1 of 6

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
 Posted by Ken Haley on 2005-11-24 09:44:07

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