Part 1
Determine the mean distance between two random points on the perimeter of a unit (convex) regular hexagon.
Part 2
Determine the mean distance between two random points on the interior of a unit (convex) regular hexagon.
Chose the first point X and label the side (A). Label the other sides clockwise (B), (C), (D), (E), (F). Let x be the distance from one corner.
The second point, Y, is equally likely to be on any of the six sides. Let y be the distance from one corner.
Case 1: Y is on (A). The solution for two points on a unit segment is known to be 1/3. I derived it with a 3-D graph.
Case 2: Y is on (B) or (F). By symmetry these are the same. The average distance can be found with the double integral int(0,1)int(0,1) sqrt((1-x+y/2)^2+(sqrt(3)y/2)^2) dydx. WolframAlpha gives this as 0.891269
Case 3: Y is on (C) or (E). By symmetry these are the same. The average distance can be found with the double integral int(0,1)int(0,1) sqrt((3/2-x-y/2)^2+3/4(1+y)^2) dydx. WolframAlpha gives this as 1.54067
Case 4: Y is on (D). The average distance can be found with the double integral int(0,1)int(0,1) sqrt(3+(x-y)^2) dydx. WolframAlpha gives this as 1.77869
[WolframAlpha didn't give exact answers, just these approximations]
The solution is then the weighted average (1/3+2*0.891269+2*1.54067+1.77869)/6 = 1.16265
Part 2. Will depend on how the random points are chosen.
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Posted by Jer
on 2022-10-14 08:15:11 |
Part 1
