Connecting FA, FB, FC, EA and EB, it is obvious that EA = E 'a = P and EB = E 'b = B. Let the diameter ef = 1 and the angle EFC = α.

Then the angle EFB = π/n-α, the angle EFA = 2π/n-α, and the triangles EFA, EFB and EFC are right triangles, so

EA=E'A=p=EFsinEFA=sin(2π/n-α)，EB=E'B=b=EFsinEFB=sin(π/n-α)，EC=c=sinα

That is, p=sin(2π/n)cosα-cos(2π/n)sinα.

b=sin(π/n)cosα-cos(π/n)sinα

c=sinα

From the last two items, we can find that

sinα= c cosα=(b+ c cos(π/n))/sin(π/n)

Substitute p = 2sin (π/n) cos (π/n) cos α-cos (2π/n) sin α.

= 2 sin(π/n)cos(π/n)(b+ c cos(π/n))/sin(π/n)-cos(2π/n)c

= 2cos(π/n)(b+ c cos(π/n))-cos(2π/n)c

=2cos(π/n)b+2c(cos(π/n))^2-cos(2π/n)c

=2cos(π/n)b+(2(cos(π/n))^2-cos(2π/n))c

=2cos(π/n)b+c

So n = 3, 2cos (π/n) = 2cos (π/3) = 2 *1/2 =1,p = b+C.

N=4, 2cos(π/n)=2cos(π/4)=2* radical number 2/2= radical number 2, p= radical number 2b+c.

N = 12, 2cos (π/n) = 2cos (π/12) = 2 * (square root of 6+radical number 2)/4 = (square root of 6+radical number 2)/2, p = [(square root of 6+radical number 2)/.