You choose a random point within a regular pentagon having unit sides.
What is the average distance to the five sides?
The regular pentagn consists of 5 isosceles triangles meeting with their apices at the center. The apex angles are each 72°. The height of each triangle is .5/tan(36°), Any random point in the pentagon will be in one of these triangles, and the nearest point to the perimeter of the pentagon will be the point at the base of the given triangle.
The average distance of a point within this triangle is integ{0 to h}(x*(h-x)/h) / .5/(2*tan(36°)
>> h=.5/tand(36)
h =
0.688190960235587
A=.5/(2*tand(36)
A =
0.344095480117793
fun=@(x)x.*(h-x)/h;
>> integral(fun,0,h)
ans =
0.0789344662916632
>> ans/A
ans =
0.229396986745196
This is the average distance of a point in the pentagon from the nearest side.
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Posted by Charlie
on 2024-06-23 17:07:53 |