A Chord Frenzy

For General Case:

Let $u_{0} = 2r \sin(\dfrac{\pi}{n})$ and for $(1 \leq j \leq k - 1)$ let $u_{j} = \sqrt{u_{0}^2 + u_{j - 1}^2 - 2 a_{0} a_{j - 1}\cos(\dfrac{(n - 2)\pi}{n} - (j - 1)\dfrac{\pi}{n})}$

and $P = (\prod_{j = 0}^{k - 1} u_{j})^2.$

Program(written in Free Pascal):

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program                 forbrillant;

uses mathunit;

var r:real;
    n,k:longint;


{General case:}

procedure        getinput;


begin

writeln('Enter the radius of the circle');

readln(r);

writeln('Enter number of sides of n-gon');

read(n);

writeln('Enter k');

readln(k);

end;

function               prod(k:longint):real;

type arraytype = array[0 .. 1000] of real;

var j:longint;

    u:arraytype;

    s:real;

begin

u[0]:= 2* r * sin(pi/n);

for j:= 1 to k - 1 do

begin

u[j]:= sqrt(power(u[0],2) + power(u[j - 1],2) - 2 * u[0] * u[j - 1] *

cos((n - 2) * pi/n - (j - 1) * pi/n));

end;

s:= 1;

for j:= 0 to k - 1 do

begin

s:= s * u[j];

end;

prod:= power(s,2);

end;



begin

getinput;

writeln(prod(k):0:4);

readln;

end. 

Using the program and $r = 2$ , $n = 23$ , and $k = 11$ we obtain $P = (\prod_{j = 0}^{10} u_{j})^2 = \boxed{96468992}$ .