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The ancient Greek mathematicians knew how to construct a regular polygon with 3, 4, or 5 sides, [1]: p. xi and they knew how to construct a regular polygon with double the number of sides of a given regular polygon. [1]: pp. 49–50 This led to the question being posed: is it possible to construct all regular polygons with compass and straightedge?
As 17 is a Fermat prime, the regular heptadecagon is a constructible polygon (that is, one that can be constructed using a compass and unmarked straightedge): this was shown by Carl Friedrich Gauss in 1796 at the age of 19. [1] This proof represented the first progress in regular polygon construction in over 2000 years. [1]
Articles related to constructible regular polygons, i.e. those amenable to compass and straightedge construction. Carl Friedrich Gauss proved that a regular polygon is constructible if its number of sides has no odd prime factors that are not Fermat primes, and no odd prime factors that are raised to a power of 2 or higher.
[2]: p. 1 They could also construct half of a given angle, a square whose area is twice that of another square, a square having the same area as a given polygon, and regular polygons of 3, 4, or 5 sides [2]: p. xi (or one with twice the number of sides of a given polygon [2]: pp. 49–50 ).
However, it is constructible using neusis, or an angle trisector. The following is an animation from a neusis construction of a regular tridecagon with radius of circumcircle O A ¯ = 12 , {\displaystyle {\overline {OA}}=12,} according to Andrew M. Gleason , [ 1 ] based on the angle trisection by means of the Tomahawk (light blue).
Dih 15 has 3 dihedral subgroups: Dih 5, Dih 3, and Dih 1. And four more cyclic symmetries: Z 15, Z 5, Z 3, and Z 1, with Z n representing π/n radian rotational symmetry. On the pentadecagon, there are 8 distinct symmetries. John Conway labels these symmetries with a letter and order of the symmetry follows the letter. [3]
There are three regular star polygons, {16/3}, {16/5}, {16/7}, using the same vertices, but connecting every third, fifth or seventh points. There are also three compounds: {16/2} is reduced to 2{8} as two octagons , {16/4} is reduced to 4{4} as four squares and {16/6} reduces to 2{8/3} as two octagrams , and finally {16/8} is reduced to 8{2 ...
It is also constructible with compass, straightedge and angle trisector. The impossibility of straightedge and compass construction follows from the observation that 2 cos 2 π 7 ≈ 1.247 {\displaystyle \scriptstyle {2\cos {\tfrac {2\pi }{7}}\approx 1.247}} is a zero of the irreducible cubic x 3 + x 2 − 2 x − 1 .