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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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Some Thoughts possible approach | Comment 3 of 4 |

After a few false starts...

The formula for the number of diagonals is n(n-3)/2, as is confirmed by OEIS A000096

Starting with, say, a regular undecagon in the unit circle, the different length values including the sides are 0.563465..., 1.081282..., 1.51150098..., 1.81926620..., and 1.9796452... However, the side value is not needed for current purposes, leaving 4 diagonals for each of the 11 vertices for a total of 44 in all. 11*8/2=44, as expected.

Those values correspond to 2*sin(pi/11) - for the unused side - 2*sin(2pi/11), 2*sin(3pi/11), 2*sin(4pi/11), and 2*sin(5pi/11). Checking: 0.563465114..., 1.081281635..., 1.511499149..., 1.819263991...,1.979642884....

So for the total length of the diagonals at a single vertex of any 'oddagon', we have:

sum for k = 2 to (n-1)/2 {2sin(pi*k/n)} where n is the number of vertices. Call the sum, S.

Then we need to multiply S by the number of vertices, n, and divide by n(n-3)/2 to obtain the mean length.

The mean is therefore 2S/(n-3). Examples:

for n=2 to 44 sum {2sin(pi*n/89)} Mean = 1.315876594
    
for n=2 to 90 sum {2sin(pi*n/181)} Mean = 1.294276112
    
for n=2 to 498 sum {2sin(pi*n/997)} Mean = 1.277068583
    
for n=2 to 8000 sum {2sin(pi*n/16001)} Mean = 1.273478254

for n=2 to 50000 sum {2sin(pi*n/100001)} Mean = 1.273277741

and the sums appear to converge nicely on 4/pi. Reassuringly, the same result occurs if the two points are picked at random on the circumference: https://oeis.org/A088538, and see

http://mathworld.wolfram.com/CircleLinePicking.html 


Edited on November 24, 2018, 7:16 am
  Posted by broll on 2018-11-24 06:44:42

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