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Mean Diagonals (Posted on 2018-11-23) Difficulty: 2 of 5
Let Mn be the arithmetic mean of the lengths of the diagonals of regular n-gon with a circumradius of π(pi).

Compute limit of Mn as n tends to infinity.

No Solution Yet Submitted by Danish Ahmed Khan    
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another way, which gets messy.... | Comment 2 of 4 |

Consider odd n. (I'll let someone else do even n).

There are n-3 diagonals emanating from each point. Call these n-3 diagonals a "set".

The lengths in each set come in pairs. E.g., a set of diagonals of a pentagon is one pair of equal length, while a set for a heptagon is two pairs of equal length. 

We only need to compute the mean of the unique lengths in each set due to symmetry (every unique length appears n times in the figure. There are (n-3)/2 unique lengths. 

Each diagonal is the base of an isosceles triangle where the third point is the figure's center and the sides are circumradii of length n pi.  

The angle from the center is, for n=5, (2/5)  [2 pi]. Going to n=7, 9, ...

the center angles are (2/7, 3/7) [2 pi], (2/9, 3/9, 4/9) [2 pi], ...

What length diagonals do these triangles yield? Drop an angle bisector from the center. Half the length becomes the base of a right triangle.

For the pentagon, where we divide the angle theta = (2/5) [2 pi] by two, the length l is:

l = 2 hypotenuse sin(theta/2)

l = 2 n pi sin (2 pi /5)

So, in general, 

Mn = [1/(num of lengths)]  (sum  {i= 1--> num of lengths}  length_i)

= 2/(n-3) (2 n pi) (sum {i= 1 --> (n-3)/2} sin( [1+i] pi/n ))

and, then, ah, ....

Edited on November 23, 2018, 6:28 pm
  Posted by Steven Lord on 2018-11-23 14:41:33

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