Suppose a regular 17-gon has side length a a a
a = 2 sin ( π 17 ) . a=2\sin\left(\dfrac{\pi}{17}\right). a = 2 sin ( 17 π ) . Using complex numbers, one can obtain an algebraic expression for a a a
Let
x = e 2 π i 17 . x=\mathrm{e}^{\frac{2\pi\mathrm{i}}{17}}. x = e 17 2 π i . Since
x 0 , x 1 , x 2 , … , x 16 x^0,x^1,x^2,\ldots,x^{16} x 0 , x 1 , x 2 , … , x 16 are all roots of the equation
x 17 − 1 = 0 , x^{17}-1=0, x 17 − 1 = 0 , we have
x 0 + x 1 + x 2 + ⋯ + x 16 = 0 , x^0+x^1+x^2+\cdots+x^{16}=0, x 0 + x 1 + x 2 + ⋯ + x 16 = 0 , or
x 1 + x 2 + x 3 + ⋯ + x 16 = − 1. x^1+x^2+x^3+\cdots+x^{16}=-1. x 1 + x 2 + x 3 + ⋯ + x 16 = − 1. Now let
s = x 1 + x 9 + x 9 2 + x 9 3 + ⋯ + x 9 7 = x 1 + x 9 + x 13 + x 15 + x 16 + x 8 + x 4 + x 2 , s ′ = x 3 + x 3 3 + x 3 5 + ⋯ + x 3 15 = x 3 + x 10 + x 5 + x 11 + x 14 + x 7 + x 12 + x 6 . \begin{aligned}
s
&=
x^1+x^9+x^{9^2}+x^{9^3}+\cdots+x^{9^7}\\
&=
x^1+x^9+x^{13}+x^{15}+x^{16}+x^8+x^4+x^2,\\
s^{\prime}
&=
x^3+x^{3^3}+x^{3^5}+\cdots+x^{3^{15}}\\
&=
x^3+x^{10}+x^5+x^{11}+x^{14}+x^7+x^{12}+x^6.
\end{aligned} s s ′ = x 1 + x 9 + x 9 2 + x 9 3 + ⋯ + x 9 7 = x 1 + x 9 + x 13 + x 15 + x 16 + x 8 + x 4 + x 2 , = x 3 + x 3 3 + x 3 5 + ⋯ + x 3 15 = x 3 + x 10 + x 5 + x 11 + x 14 + x 7 + x 12 + x 6 . In obtaining these expressions, we have used
together with
9 2 = 4 × 17 + 13 , 9 3 = 42 × 17 + 15 , ⋮ 3 3 = 1 × 17 + 10. \begin{aligned}
9^2&=4\times17+13,\\
9^3&=42\times17+15,\\
&\vdots\\
3^3&=1\times17+10.
\end{aligned} 9 2 9 3 3 3 = 4 × 17 + 13 , = 42 × 17 + 15 , ⋮ = 1 × 17 + 10. Multiplying s s s s ′ s^{\prime} s ′
s s ′ = − 4. ss^{\prime}=-4. s s ′ = − 4. Now mark the points
x 0 , x 1 , … , x 16 x^0,x^1,\ldots,x^{16} x 0 , x 1 , … , x 16 in the complex plane.
We see that these points are uniformly distributed on the unit circle. Moreover,
x 1 and x 16 , x 9 and x 8 , x 13 and x 4 , x 15 and x 2 x^1\text{ and }x^{16},
\qquad
x^9\text{ and }x^8,
\qquad
x^{13}\text{ and }x^4,
\qquad
x^{15}\text{ and }x^2 x 1 and x 16 , x 9 and x 8 , x 13 and x 4 , x 15 and x 2 are pairs of complex conjugates.
This shows that s s s
x 1 , x 2 , x 4 , x 5 , x^1,x^2,x^4,x^5, x 1 , x 2 , x 4 , x 5 , we may conclude that s s s
Therefore,
s = 1 2 ( 17 − 1 ) , s ′ = − 1 2 ( 17 + 1 ) . s=\dfrac{1}{2}(\sqrt{17}-1),
\qquad
s^{\prime}=-\dfrac{1}{2}(\sqrt{17}+1). s = 2 1 ( 17 − 1 ) , s ′ = − 2 1 ( 17 + 1 ) . We now further split s s s s ′ s^{\prime} s ′
p = x 1 + x 13 + x 16 + x 4 , p ′ = x 9 + x 15 + x 8 + x 2 , q = x 3 + x 5 + x 14 + x 12 , q ′ = x 10 + x 11 + x 7 + x 6 . \begin{aligned}
p&=x^1+x^{13}+x^{16}+x^4,
&
\qquad
p^{\prime}&=x^9+x^{15}+x^8+x^2,\\
q&=x^3+x^5+x^{14}+x^{12},
&
\qquad
q^{\prime}&=x^{10}+x^{11}+x^7+x^6.
\end{aligned} p q = x 1 + x 13 + x 16 + x 4 , = x 3 + x 5 + x 14 + x 12 , p ′ q ′ = x 9 + x 15 + x 8 + x 2 , = x 10 + x 11 + x 7 + x 6 . It is easy to verify that
p + p ′ = s , p p ′ = − 1 , q + q ′ = s ′ , q q ′ = − 1. \begin{aligned}
p+p^{\prime}&=s,
&
\qquad
pp^{\prime}&=-1,\\
q+q^{\prime}&=s^{\prime},
&
\qquad
qq^{\prime}&=-1.
\end{aligned} p + p ′ q + q ′ = s , = s ′ , p p ′ q q ′ = − 1 , = − 1. Hence
p = 1 2 ( s + s 2 + 4 ) , q = 1 2 ( s ′ + s ′ 2 + 4 ) . p
=
\dfrac{1}{2}
\left(
s+\sqrt{s^2+4}
\right),
\qquad
q
=
\dfrac{1}{2}
\left(
s^{\prime}+\sqrt{s^{\prime2}+4}
\right). p = 2 1 ( s + s 2 + 4 ) , q = 2 1 ( s ′ + s ′2 + 4 ) . Next, let
r = x 1 + x 16 , r ′ = x 13 + x 4 . r=x^1+x^{16},
\qquad
r^{\prime}=x^{13}+x^4. r = x 1 + x 16 , r ′ = x 13 + x 4 . Clearly,
r + r ′ = p , r r ′ = q . r+r^{\prime}=p,
\qquad
rr^{\prime}=q. r + r ′ = p , r r ′ = q . Therefore,
r = x 1 + x 16 = 2 cos 2 π 17 = 1 2 ( p + p 2 − 4 q ) . r
=
x^1+x^{16}
=
2\cos\dfrac{2\pi}{17}
=
\dfrac{1}{2}
\left(
p+\sqrt{p^2-4q}
\right). r = x 1 + x 16 = 2 cos 17 2 π = 2 1 ( p + p 2 − 4 q ) . Finally,
a = 2 − r . a=\sqrt{2-r}. a = 2 − r . Using straightedge-and-compass constructions, we can construct
s , s ′ , p , q , r , s,\quad s^{\prime},\quad p,\quad q,\quad r, s , s ′ , p , q , r , and hence obtain the side length a a a
