A beautiful proof and plot: Gaussian positive 17-edged

One night, 19-year-old is reading Bo's Gauss tutor because of negligence will be more than 2000 years unsolved problem--ruler to do 17 side shape left to Gauss, Gauss more leisurely to bite pen head write homework, and then face serious, mom this problem a little BT AH! Think, think, all night long, accompanied by the dawn of the morning light, a Gaussian pencil throw, chest a sigh of relief. The heart said, alas, recently IQ has dropped, think I 9 years old count 1+2+3......+100 also useless so long time ah, such essay incredibly spent a night time! The next day to guide, Bo Guide surprised, said to him, this is Akimide did not do out of the problem Ah! You're a genius!

The drawing steps and proofs are attached below.

First, based on such a simple theorem, the line segment A, B, the line segment C satisfies c^2 + ac + b = 0 (c is real roots, the segment length is definitely a real number), we are able to make C. One of the basic ideas used in this theorem is to use algebraic methods to establish the connection between line segments, which is also the core idea of the cos (2Π/17).

Order: A = 2 (cos (2Π/17) + cos (4Π/17) + cos (8Π/17) + cos (16Π/17)) ①

A1 = 2 (cos (6Π/17) + cos (10Π/17) + cos (12Π/17) + cos (14Π/17)) ②

Through the and differential product, induction formula, we will get a + a1 =-1, a*a1 =-4, can be restored by the establishment of a two-time equation, using the above theorem, can be a length of a, A1 line.
Order: b = 2 (cos (2Π/17) + cos (8Π/17)) ③

B1 = 2 (cos (4Π/17) + cos (16Π/17)) ④

Through and the difference product, induces the formula, we will obtain B + B1 = A, b*b1 =-1, can do the length is B, B1 the line segment.

Order: c = 2 (cos (6Π/17) + cos (10Π/17)) ⑤

C1 = 2 (cos (12Π/17) + cos (14Π/17)) ⑥

Through the and differential product, induction formula, we will get C + c1 = a1, c*c1 =-1, can do the length of C, C1 segment.
The ⑤, the use of and the difference of product, induction formula, to the following form. [2cos (2Π/17)] [2cos (8Π/17)] = C⑦
A line segment of the length cos (2Π/17) can be made when the ③⑦ is in the United State. (Note that you need to compare the size of two roots) to make a positive 17-edged shape.

1. Give a circle O, make two vertical diameter AB, CD. 2. Make e-point on OA to Oe=1/4ao, Link ce.3. As the ∠ceb of the line ef. 4. As the ∠feb of the cross-line eg, the Co in P. 5. Make ∠geh=45°, and turn on CD at Q. 6. With CQ as the diameter of the circle, intersection ob in K. 7. With P as the center, PK radius for the circle, the CD in L, M. 8. The vertical line of the CD is over M, l, and the circle O is N, R. 9. The midpoint S of the Arc nr, and the Circle O is divided into 17 equal parts by the SN radius.

A beautiful proof and plot: Gaussian positive 17-edged