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Random Point Within a Regular Pentagon (Posted on 2024-06-22) Difficulty: 3 of 5
You choose a random point within a regular pentagon having unit sides.

What is the average distance to the five sides?

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution analytic solution Comment 2 of 2 |
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.        

  Posted by Charlie on 2024-06-23 17:07:53
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