# Tetracontagon

Regular tetracontagon
A regular tetracontagon
Type Regular polygon
Edges and vertices 40
Schläfli symbol {40}, t{20}, tt{10}, ttt{5}
Coxeter diagram
Symmetry group Dihedral (D40), order 2×40
Internal angle (degrees) 171°
Dual polygon Self
Properties Convex, cyclic, equilateral, isogonal, isotoxal

In geometry, a tetracontagon or tessaracontagon is a forty-sided polygon or 40-gon.[1][2] The sum of any tetracontagon's interior angles is 6840 degrees.

## Regular tetracontagon

A regular tetracontagon is represented by Schläfli symbol {40} and can also be constructed as a truncated icosagon, t{20}, which alternates two types of edges. Furthermore, it can also be constructed as a twice-truncated decagon, tt{10}, or a thrice-truncated pentagon, ttt{5}.

One interior angle in a regular tetracontagon is 171°, meaning that one exterior angle would be 9°.

The area of a regular tetracontagon is (with t = edge length)

${\displaystyle A=10t^{2}\cot {\frac {\pi }{40}}}$

${\displaystyle r={\frac {1}{2}}t\cot {\frac {\pi }{40}}}$

The factor ${\displaystyle \cot {\frac {\pi }{40}}}$ is a root of the octic equation ${\displaystyle x^{8}-8x^{7}-60x^{6}-8x^{5}+134x^{4}+8x^{3}-60x^{2}+8x+1}$.

The circumradius of a regular tetracontagon is

${\displaystyle R={\frac {1}{2}}t\csc {\frac {\pi }{40}}}$

As 40 = 23 × 5, a regular tetracontagon is constructible using a compass and straightedge.[3] As a truncated icosagon, it can be constructed by an edge-bisection of a regular icosagon. This means that the values of ${\displaystyle \sin {\frac {\pi }{40}}}$ and ${\displaystyle \cos {\frac {\pi }{40}}}$ may be expressed in radicals as follows:

${\displaystyle \sin {\frac {\pi }{40}}={\frac {1}{4}}({\sqrt {2}}-1){\sqrt {{\frac {1}{2}}(2+{\sqrt {2}})(5+{\sqrt {5}})}}-{\frac {1}{8}}{\sqrt {2-{\sqrt {2}}}}(1+{\sqrt {2}})({\sqrt {5}}-1)}$
${\displaystyle \cos {\frac {\pi }{40}}={\frac {1}{8}}({\sqrt {2}}-1){\sqrt {2+{\sqrt {2}}}}({\sqrt {5}}-1)+{\frac {1}{4}}(1+{\sqrt {2}}){\sqrt {{\frac {1}{2}}(2-{\sqrt {2}})(5+{\sqrt {5}})}}}$

## Construction of a regular tetracontagon

Regular tetracontagon with given circumcircle, animation

### Circumcircle is given

1. Construct first the side length HC1 of a pentagon.
2. Transfer this on the circumcircle, there arises the intersection C39.
3. Connect the point C39 with the central point M, there arises the angle C39MC1 with 72°.
4. Halve the angle C39MC1, there arise the intersection C40 and the angle C40MC1 with 9°.
5. Connect the point C1 with the point C40, there arises the first side length a of the tetracontagon.
6. Finally you transfer the segment C1C40 (side length a) repeatedly counterclockwise on the circumcircle until arises a regular tetracontagon.

The golden ratio

${\displaystyle {\frac {\overline {HM}}{\overline {BH}}}={\frac {\overline {BM}}{\overline {HM}}}={\frac {1+{\sqrt {5}}}{2}}=\varphi \approx 1.618}$

### Side length is given

Regular tetracontagon with given side length, animation
(The construction is very similar to that of icosagon with given side length)
1. Draw a segment AB whose length is the given side length a of the tetracontagon.
2. Extend the segment AB by more than two times.
3. Draw each a circular arc about the points A and B, there arise the intersections C and D.
4. Draw a vertical straight line from point C through point D.
5. Draw a parallel line too the segment CD from the point B to the circular arc, there arises the intersection F.
6. Draw a circle arc about the point E with the radius EF till to the extension of the side length, there arises the intersection G.
7. Draw a circle arc about the point A with the radius AG till to the vertical straight line, there arises the intersection H and the angle AHB with 36°.
8. Draw a circle arc about the point H with radius AH till to the vertical straight line, there arises the intersection I and the angle AIB with 18°.
9. Draw a circle arc about the point I with radius AI till to the vertical straight line, there arises the central point O of the circumcircle and the angle AOB with 9°.
10. Draw around the central point O with radius AO the circumcircle of the tetracontagon.
11. Finally transfer the segment AB (side length a) repeatedly counterclockwise on the circumcircle until to arises a regular tetracontagon.

The golden ratio

${\displaystyle {\frac {\overline {AB}}{\overline {BG}}}={\frac {\overline {AG}}{\overline {AB}}}={\frac {1+{\sqrt {5}}}{2}}=\varphi \approx 1.618}$

## Symmetry

The symmetries of a regular tetracontagon. Light blue lines show subgroups of index 2. The left and right subgraphs are positionally related by index 5 subgroups.

The regular tetracontagon has Dih40 dihedral symmetry, order 80, represented by 40 lines of reflection. Dih40 has 7 dihedral subgroups: (Dih20, Dih10, Dih5), and (Dih8, Dih4, Dih2, Dih1). It also has eight more cyclic symmetries as subgroups: (Z40, Z20, Z10, Z5), and (Z8, Z4, Z2, Z1), with Zn representing π/n radian rotational symmetry.

John Conway labels these lower symmetries with a letter and order of the symmetry follows the letter.[4] He gives d (diagonal) with mirror lines through vertices, p with mirror lines through edges (perpendicular), i with mirror lines through both vertices and edges, and g for rotational symmetry. a1 labels no symmetry.

These lower symmetries allows degrees of freedoms in defining irregular tetracontagons. Only the g40 subgroup has no degrees of freedom but can seen as directed edges.

## Tetracontagram

A tetracontagram is a 40-sided star polygon. There are seven regular forms given by Schläfli symbols {40/3}, {40/7}, {40/9}, {40/11}, {40/13}, {40/17}, and {40/19}, and 12 compound star figures with the same vertex configuration.

 Picture Interior angle {40/3} {40/7} {40/9} {40/11} {40/13} {40/17} {40/19} 153° 117° 99° 81° 63° 27° 9°
 Picture Interior angle Picture Interior angle {40/2}=2{20} {40/4}=4{10} {40/5}=5{8} {40/6}=2{20/3} {40/8}=8{5} {40/10}=10{4} 162° 144° 135° 126° 108° 90° {40/12}=4{10/3} {40/14}=2{20/7} {40/15}=5{8/3} {40/16}=8{5/2} {40/18}=2{20/9} {40/20}=20{2} 72° 54° 45° 36° 18° 0°

Many isogonal tetracontagrams can also be constructed as deeper truncations of the regular icosagon {20} and icosagrams {20/3}, {20/7}, and {20/9}. These also create four quasitruncations: t{20/11}={40/11}, t{20/13}={40/13}, t{20/17}={40/17}, and t{20/19}={40/19}. Some of the isogonal tetracontagrams are depicted below, as a truncation sequence with endpoints t{20}={40} and t{20/19}={40/19}.[5]

 t{20}={40} t{20/19}={40/19}

## References

1. ^ Gorini, Catherine A. (2009), The Facts on File Geometry Handbook, Infobase Publishing, p. 165, ISBN 9781438109572.
2. ^
3. ^ Constructible Polygon
4. ^ The Symmetries of Things, Chapter 20
5. ^ The Lighter Side of Mathematics: Proceedings of the Eugène Strens Memorial Conference on Recreational Mathematics and its History, (1994), Metamorphoses of polygons, Branko Grünbaum