Thirty-corner

Regular thirty-four corner

The Dreißigeck or Triakontagon is a geometric figure , and a polygon ( polygon ). It is determined by thirty corner points and their thirty connections called lines , sides or edges.

Variations

The thirty-four corner can be represented as:

• concave triangle in which at least one interior angle is greater than 180 °.
• convex triangle in which all interior angles are less than 180 °. A convex triangle can be regular or irregular.
• Chord triangle in which all corners lie on a common perimeter , but the lengths of the sides are possibly unequal.
• Regular thirty-corner: it is determined by thirty points that lie on a virtual or real circle. The neighboring points are always at the same distance from each other and are connected by strings that are lined up, also called sides or edges .
• Regularly rolled over thirty-four corner: This results when at least one is skipped over each time when connecting the thirty corner points and the chords thus created are of the same length. Such regular stars are noted with Schläfli symbols , indicating the number of corner points and connecting every -th point. ${\displaystyle \left\{n/k\right\}}$ ${\displaystyle n}$ ${\displaystyle k}$

There are only three regular thirty ray stars.

The "star" with the symbols {30.2} and {30/28} are regular fifteen-corner , {30.3} and {30/27} regular tens corner , {30.5} and {30/25} regular hexagons , {30/6} and {30/24} regular pentagons , {30/10} and {30/20} equilateral triangles . The stars with the symbols {30/4} and {30/26}, {30/8} and {30/22} as well as {30/14} and {30/16} are regular fifteen-ray stars, {30/9} and {30/21} regular ten-ray stars and finally {30/12} and {30/18} regular pentagrams .

• Regular overturned triangles
• 

  ${\displaystyle \left\{30/7\right\}{,}\ \left\{30/23\right\}}$

• 

  ${\displaystyle \left\{30/11\right\}{,}\ \left\{30/19\right\}}$

• 

  ${\displaystyle \left\{30/13\right\}{,}\ \left\{30/17\right\}}$

Regular thirty-four corner

According to Carl Friedrich Gauß and Pierre-Laurent Wantzel, the regular triangle is a constructible polygon , since the number of its sides can be represented as the product of a power of two with pairwise different Fermat's prime numbers ( ). ${\displaystyle 30=2^{1}\cdot 3\cdot 5}$

Sizes

The quantities and their general formulas are described in detail in Fifteen, Mathematical Connections .

Sizes of a regular triangle
Interior angle	${\displaystyle {\begin{aligned}\alpha &={\frac {n-2}{n}}\cdot 180^{\circ }={\frac {28}{30}}\cdot 180^{\circ }\\&=168^{\circ }\end{aligned}}}$
Central angle (Center angle)	${\displaystyle {\begin{aligned}\mu &={\frac {360^{\circ }}{30}}\\\mu &=12^{\circ }\end{aligned}}}$
Side length	${\displaystyle {\begin{aligned}a&=R\cdot {\frac {1}{4}}\cdot \left({\sqrt {6\cdot \left(5-{\sqrt {5}}\right)}}-{\sqrt {5}}-1\right)\\&={\frac {\sin(12^{\circ })}{\sin(84^{\circ })}}\cdot R\\&=R\cdot 2\cdot \sin \left({\frac {180^{\circ }}{30}}\right)\approx 0{,}2090569\cdot R\end{aligned}}}$
Perimeter radius	${\displaystyle {\begin{aligned}R&=a\cdot {\frac {1}{2}}\cdot \left(2+{\sqrt {5}}+{\sqrt {15+6\cdot {\sqrt {5}}}}\right)\\&={\frac {\sin(84^{\circ })}{\sin(12^{\circ })}}\cdot a\\&={\frac {a}{2\cdot \sin \left({\frac {180^{\circ }}{30}}\right)}}\approx 4{,}783386\cdot a\end{aligned}}}$
Inscribed radius	${\displaystyle {\begin{aligned}r&=a\cdot {\frac {1}{4}}\cdot \left({\sqrt {15}}+3\cdot {\sqrt {3}}+{\sqrt {2}}\cdot {\sqrt {25+11\cdot {\sqrt {5}}}}\right)\\&={\frac {\sin ^{2}(84^{\circ })}{\sin(12^{\circ })}}\cdot a\approx 4{,}757182\cdot a\end{aligned}}}$
height	${\displaystyle {\begin{aligned}h&=2\cdot r\approx 9{,}514364\cdot a\end{aligned}}}$
Area	${\displaystyle {\begin{aligned}A&=a^{2}\cdot {\frac {15}{4}}\cdot \left({\sqrt {15}}+3\cdot {\sqrt {3}}+{\sqrt {2}}\cdot {\sqrt {25+11\cdot {\sqrt {5}}}}\right)\\&=15\cdot {\frac {\sin ^{2}(84^{\circ })}{\sin(12^{\circ })}}\cdot a^{2}\approx 71{,}357733\cdot a^{2}\end{aligned}}}$

Construction with compass and ruler for a given circumference

Fig. 1: Thirty-square, without construction of the fifteenth-corner

At first glance, it seems obvious to first draw one side of the fifteen corner with its circumference and then halve the central angle in order to get the side length of the thirty-one. However, if you look at the drawing of the fifteenth corner , it is easy to see that the central axis is already a bisector between the corner points and consequently the lines as well as the side lengths and the angle would be the central angle of a thirty-four corner . Instead, you only need to find the side length of the pentagon to find the central angle and the first side length of the triangle. The following construction (see Figure 1) uses this possibility. ${\displaystyle \mu }$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle E_{3}}$ ${\displaystyle E_{4}.}$ ${\displaystyle {\overline {AE_{3}}}}$ ${\displaystyle {\overline {AE_{4}}}}$ ${\displaystyle E_{3}MA}$

It begins with the drawing of the circle around the center of the drawing the central axis and orthogonal , it follows the halving of the track in this case arise, the first vertices and the resulting Dreißigecks. The arc around the point with the radius follows; the point of intersection is Now you draw a short circular arc around the corner point with the radius until it intersects the perimeter . The distance is the desired side length of the triangle with its central angle. Now take the side length in the compass, mark the remaining corner points counter- clockwise on the circumference and finally connect the neighboring corner points with each other. The regular thirty-one corner is now complete. ${\displaystyle k_{1}}$ ${\displaystyle M,}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle {\overline {MC}}.}$ ${\displaystyle {\overline {AM}}}$ ${\displaystyle D,}$ ${\displaystyle E_{5}}$ ${\displaystyle E_{25}}$ ${\displaystyle D}$ ${\displaystyle {\overline {CD}}}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle F.}$ ${\displaystyle E_{25}}$ ${\displaystyle {\overline {CF}}}$ ${\displaystyle E_{1}}$ ${\displaystyle {\overline {AE_{1}}}}$ ${\displaystyle a}$ ${\displaystyle \mu =12^{\circ }.}$ ${\displaystyle a}$

Construction with compass and ruler with a given side length

Fig. 2: Regular triangle with a given side length

The construction in Figure 2 is almost the same as that of the fifteen-sided with a given side length .

First, the ends of the side are designated with the first corner points (left) or , then the distance is extended beyond by approximately the same amount. There follows a circular arc having the radius around the point the orthogonal and the circular arc to also with the radius while the intersections occur and Now, a half-line from by drawn; it halves the length of the side in The next arc with the radius is drawn around , this results in the intersection on the extension. The route is thus divided according to the golden ratio with external division. ${\displaystyle a}$ ${\displaystyle E_{1}}$ ${\displaystyle E_{2}}$ ${\displaystyle {\overline {E_{1}E_{2}}}}$ ${\displaystyle E_{1}}$ ${\displaystyle {\overline {E_{1}E_{2}}}}$ ${\displaystyle E_{1},}$ ${\displaystyle {\overline {E_{1}A}}}$ ${\displaystyle E_{2},}$ ${\displaystyle {\overline {E_{1}E_{2}}};}$ ${\displaystyle B}$ ${\displaystyle C.}$ ${\displaystyle C}$ ${\displaystyle B}$ ${\displaystyle a}$ ${\displaystyle D.}$ ${\displaystyle |AD|}$ ${\displaystyle D}$ ${\displaystyle F}$ ${\displaystyle {\overline {E_{2}F}}}$

Now an arc with the radius that intersects the half-line is drawn around . The line generated by this corresponds to the radius of a fifteen corner. The calculation of the circumferential radius is described in detail in the article Fifteen . The distance taken in the circle and drawn around a short arc through the half-straight line results in the center of the circumference of a fifteen-triangle not drawn in with its central angle ${\displaystyle E_{2}}$ ${\displaystyle {\overline {E_{2}F}}}$ ${\displaystyle G}$ ${\displaystyle {\overline {CG}}}$ ${\displaystyle {\overline {CG}}}$ ${\displaystyle {\overline {CG}}}$ ${\displaystyle E_{2}}$ ${\displaystyle H}$ ${\displaystyle \mu _{1}=24^{\circ }.}$

It continues with the arc around the point with the radius that intersects the half-line . Because of the central angle theorem, the angle at the apex is half as large as the central angle of a fifteen-triangle. Because of this, the center of the triangle you are looking for is with its central angle. Now just draw the circumference around the center , take the side length 29 times counterclockwise onto the circumference and connect the neighboring corner points with each other, then the regular triangle is constructed. ${\displaystyle H}$ ${\displaystyle {\overline {E_{1}H}},}$ ${\displaystyle M}$ ${\displaystyle {\overline {E_{1}H}}={\overline {HM}}}$ ${\displaystyle \mu }$ ${\displaystyle M}$ ${\displaystyle \mu _{1}}$ ${\displaystyle M}$ ${\displaystyle \mu =12^{\circ }.}$ ${\displaystyle M}$ ${\displaystyle a}$

Diagonals

Image 3: triangle,
diagonals up to (diameter) ${\displaystyle d_{2}}$ ${\displaystyle d_{15}}$

Each thirty-four corner has 405 diagonals. There are 27 possible end points for each of the 30 corners at which a diagonal can begin. This number has to be divided by 2 so that no diagonal is counted twice. This is how the mentioned diagonals result . But they are just of different lengths. In general, that diagonal is referred to that runs over the sides of the polygon. B. the diagonal over fourteen sides. ${\displaystyle {\tfrac {30\cdot 27}{2}}=405}$ ${\displaystyle \left\lfloor {\tfrac {30-2}{2}}\right\rfloor =14}$ ${\displaystyle d_{n}}$ ${\displaystyle n}$ ${\displaystyle d_{14}}$

Golden cut

Since 5 of the corners of the regular triangle form a regular pentagon , the diagonal is over 12 sides and the diagonal is in the ratio of the golden ratio to each other. ${\displaystyle d_{12}}$ ${\displaystyle d_{6}}$

This ratio can also be found at with , these diagonals are part of a regular decagon , and at with . is one side of the regular hexagon and thus has the same length as the circumferential radius of the triangle. So also forms with the golden ratio as a ratio. ${\displaystyle d_{5}}$ ${\displaystyle d_{3}}$ ${\displaystyle d_{9}}$ ${\displaystyle d_{5}}$ ${\displaystyle d_{5}}$ ${\displaystyle R}$ ${\displaystyle R}$ ${\displaystyle d_{9}}$

Other conditions

The diagonal is in proportion to . ${\displaystyle d_{10}}$ ${\displaystyle d_{5}}$ ${\displaystyle {\sqrt {3}}}$

Occurrence

Wiener Riesenrad in the Minimundus
with 30 corners and 15 wagons

• The Vienna Giant Ferris Wheel has the shape of a regular triangle. Although only 15 wagons have been attached since 1945, it was originally built for 30 wagons.
• The layout of the Sarrasani Circus is a regular thirty-four corner.

See also

• Tendon polygon
• Regular polygon

literature

• H. Maser: Die Teilung des Kreises ..., Article 365. , in Carl Friedrich Gauss' studies on higher arithmetic , published by Julius Springer, Berlin 1889; Göttingen Digitization Center, University of Göttingen; accessed on March 15, 2018.

Web links

Wiktionary: Dreßigeck  - explanations of meanings, word origins, synonyms, translations

• Regular thirty-four corner, math alpha

Individual evidence

1. ↑ Eric W. Weisstein : Triacontagon. In: Wolfram MathWorld . Retrieved March 13, 2018 . 
2. ↑ Ing. Ludw. Fischer: Modern circus. Messages from various fields. In: kobv.de - Cooperative Library Association Berlin-Brandenburg. Journal of the Austrian Association of Engineers and Architects No. 23, 1913, pp. 410–411, PDF file: Pages 58–59 , accessed on March 11, 2018 .