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 Summing the fiftieth power (Posted on 2012-11-29)
Find the sum of the fiftieth powers of all sides and diagonals of a regular 100-gon inscribed in a circle of radius R.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 halfway through the general solution | Comment 2 of 3 |
N = the number of sides of the regular polygon
R = the radius of the circle
p = the power
S = the requested sum
Sv = sum of segments from one vertex (^p)

Each vertex is connected to each other vertex by either a side or a diagonal. Summing up all these segments (to power p) of all vertices will give twice the requested sum. And because it is a regular polygon:
2*S = N * Sv => S = N * Sv / 2

Now let's look at a general term in Sv. The triangle defined by the two vertices and the circle center will have sides of length R, R and Lk (length of the segment) and an angle k*2*pi/N between sides of length R. Thus, from the cosine theorem in a triangle we get:
Lk = sqrt( R^2 + R^2 - 2* R^2 * cos(k*2*pi/N) )
= sqrt( 2 * R^2 * ( 1 - cos( 2 * k * pi / N ) )

But ( 1 - cos( 2 * k * pi / N ) ) = 2 * [sin( k * pi / N )]^2
Therefore:
Lk = sqrt( 2 * R^2 * 2 * [sin( k * pi / N )]^2 )
= 2 * R * sin( k * pi / N)

The sum of segments from one vertex [to the power p] is:
Sv = SUM ( [2 * R * sin( k * pi / N)]^p ) for k = 0...N-1
The case k = 0 is the case of the segment between the vertex and itself, which is a length 0 segment and does not influence the sum.

Thus I repeat the two important formulas:
Sv = (2*R)^p * SUM( [sin( k * pi / N)]^p ) for k = 0...N-1
S = N * Sv / 2

All is left is transforming the sum of the power of sines into something more manageable, and then compute for the particular values N = 100, p = 50, R = ??.

 Posted by Michael Kornrade on 2013-01-10 12:24:37

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