Let A1,A2,...A2n be distinct points on the circle with radius r. They divide the circle into 2n arcs with equal length. Find the value of A1A2^2,A1A3^2+A1A4^2+...+A1A2n^2

Answer 1

if A1A2 is an arc,then there is no trigonometry.

and sum of lengths squared = π²R²*((1/n)²+(2/n)²+...+(2n/n)²)=(π²R²/n²)*(1+2²+...+(2n)²)

=(π²R²/n²)*[(2n*(2n+1)(4n+2)/6)]

If the segments are cords much trigonometry must be used. Note since there are 2n points regularly spaced in circle, an even numbered side polygon can be incscribed, and the A1A(n+1) cord will always be diameter of circle. My approach is to add the squares of the cords A1A2 through A1A(n+1), then multiply that expresiion times two (the cords on either side of the diameter are symmetric). Then I subtact the diameter (A1A(n+1)=2R) squared since it was counted twice and that expression is the sum of the cord lengths squared. R is circle radius.

It is a fun enough problem, and the answer is fun too, answer is 4R²*n.

I checked for n=2 (2n=4), a square inside ciircle, and n=3 (2n=6), a regular hexagon inside

circle and A1A2²+A1A3²+A1A4²=8R² for square,

and A1A2²+A1A3²+A1A4²+A1A5²+A1A6²=12R² for hexagon.

to start. A1A2=2*R*sin(π/2n) & A1A3=2*R*sin(2π/2n), ... & A1A(n+1)=2*R*sin((nπ/2n).

you can pull 4R² out of sum as common factor.

have 4R²*[(sin²(π/2n)+sin²(2π/2n)+...+sin²((n-1)π/2n)+sin²((nπ/2n) as the sum of cord lengths squared individually from A1A2 to A1A(n+1) (the last term is diameter squared)

use sin²q=(1/2)*(1-cos(2q))

have 4R²*([(n/2) - (1/2)*(cos(2π/2n)+cos(4π/2n)+...+cos(2nπ/2n)] since 1/2 times n = n/2

as the sum of the cord lengths squared.individually from A1A2 to A1A(n+1)

=4R²*(n/2 - (1/2)*(cos(π/n)+cos(2π/n)+cos(3π/n)+...+cos(nπ/n)]

There are two good proofs in "Advanced Trigonometry, Durell and Robson"

which prove cos(z) +cos(z+v) +cos(z+2v)+...+cos(z+(n-1)v)

=[cos(z+((n-1)/2)v]*[sin(nv/2]/[sin(v/2)]

here put z=π/n and put v=π/n, which is allowed, then

cos(π/n)+cos(2π/n)+cos(3π/n)+...+cos(nπ/n)

=cos(π/n)+cos(π/n+π/n)+cos(π/n+2π/n)+...+cos(π/n+(n-1)π/n)

=[cos(π/n+((n-1)/2)π/n)]*[sin(n*π/2n)]/[sin(π/2n)]

=[cos(π/2n+π/2)]*[1]/[sin(π/2n)]

=-1

now have as sum of cord lengths squared individually from A1A2 to A1A(n+1) (the last term is diameter squared) is

=4R²(n/2+1/2)

to find total length of all cord squared as specified, must double and subtract diameter squared

that is

=2*4R²(n/2+1/2)-4R²

=4R²n