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20 13 detailed process of mathematics 25 in Chengdu senior high school entrance examination
This problem is not difficult to solve after it is drawn. Then there is' ea is misleading, it should be E'A or EA'.

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)/.