# 51-Eck - 51-Eck

Regular 51-sided

The 51-Eck or Pentakontahenagon is a geometric figure , and a polygon ( polygon ). It is determined by fifty-one points and their fifty-one connections called lines , sides or edges.

## Regular 51-sided

According to Carl Friedrich Gauß and Pierre-Laurent Wantzel, the regular 51-gon is a constructible polygon , since the number of its sides is the product of a power of two with pairwise different Fermat's prime numbers (${\displaystyle 51=2^{0}\cdot 3\cdot 17}$) can be displayed.

### Sizes

Sizes of a regular 51-sided
Interior angle {\displaystyle {\begin{aligned}\alpha &={\frac {n-2}{n}}\cdot 180^{\circ }={\frac {49}{51}}\cdot 180^{\circ }\\\alpha &\approx 172{,}941176^{\circ }\end{aligned}}}

Central angle

(Center angle)

{\displaystyle {\begin{aligned}\mu &={\frac {360^{\circ }}{51}}\\\mu &\approx 7{,}058823^{\circ }\end{aligned}}}
Side length {\displaystyle {\begin{aligned}a&=R\cdot 2\cdot \sin \left({\frac {180^{\circ }}{51}}\right)\\a&\approx 0{,}1231218\cdot R\end{aligned}}}
Perimeter radius {\displaystyle {\begin{aligned}R&={\frac {a}{2\cdot \sin \left({\frac {180^{\circ }}{51}}\right)}}\\R&\approx {\frac {a}{0{,}123121}}\end{aligned}}}
Inscribed radius {\displaystyle {\begin{aligned}r&=R\cdot \cos \left({\frac {180^{\circ }}{51}}\right)\\r&\approx 0{,}998103\cdot R\end{aligned}}}
height {\displaystyle {\begin{aligned}h&=R+r=R\cdot \left(1+\cos \left({\frac {180^{\circ }}{51}}\right)\right)\\h&\approx 1{,}998103\cdot R\end{aligned}}}
Area {\displaystyle {\begin{aligned}A&=51\cdot R^{2}\cdot \sin \left({\frac {180^{\circ }}{51}}\right)\cdot \cos \left({\frac {180^{\circ }}{51}}\right)\\A&\approx 3{,}133651\cdot R^{2}\end{aligned}}}

## Mathematical relationships

### Interior angle

The interior angle ${\displaystyle \alpha }$is enclosed by two adjacent side edges. The variable is in the general formula for regular polygons ${\displaystyle n}$for the number of corner points of the polygon. In this case the variable is number${\displaystyle 51}$ to use.

${\displaystyle \alpha ={\frac {n-2}{n}}\cdot 180^{\circ }={\frac {51-2}{51}}\cdot 180^{\circ }={\frac {49}{51}}\cdot 180^{\circ }\approx 172{,}941176^{\circ }}$

### Central angle

The central angle or central angle${\displaystyle \mu }$ is made up of two neighboring perimeter radii ${\displaystyle R}$locked in. In the general formula is for the variable${\displaystyle n}$ the number ${\displaystyle 51}$ to use.

${\displaystyle \mu ={\frac {360^{\circ }}{n}}={\frac {360^{\circ }}{51}}\approx 7{,}058823^{\circ }}$

### Side length and perimeter radius

The 51-gon is divisible into fifty-one isosceles triangles, so-called partial triangles . From half of such a partial triangle, i.e. from a right-angled triangle with the cathetus (half the side length)${\displaystyle {\frac {a}{2}}}$, the hypotenuse (perimeter radius) ${\displaystyle R}$ and half the central angle ${\displaystyle {\frac {\mu }{2}}}$one obtains the side length with the help of the trigonometry in the right triangle${\displaystyle a}$ as follows

{\displaystyle {\begin{aligned}a&=R\cdot 2\cdot \sin \left({\frac {\mu }{2}}\right)\\&=R\cdot 2\cdot \sin \left({\frac {180^{\circ }}{51}}\right)\\a&\approx 0{,}1231218\cdot R,\end{aligned}}}

the circumferential radius is obtained by forming ${\displaystyle R}$

{\displaystyle {\begin{aligned}R&={\frac {a}{2\cdot \sin \left({\frac {180^{\circ }}{51}}\right)}}\\R&\approx {\frac {a}{0{,}1231218}}\end{aligned}}}

The inscribed radius ${\displaystyle r}$ is the height of a partial triangle, perpendicular to the length of the side ${\displaystyle a}$of the 51-sided. If the same right-angled triangle is used for the calculation as for the side length, the inscribed radius applies${\displaystyle r}$

{\displaystyle {\begin{aligned}r&=R\cdot \cos \left({\frac {\mu }{2}}\right)=R\cdot \cos \left({\frac {180^{\circ }}{51}}\right)\\r&\approx 0{,}998103\cdot R\end{aligned}}}

### height

The height ${\displaystyle h}$ of a regular 51-sided result from the sum of the incircle radius ${\displaystyle r}$ and radius ${\displaystyle R}$.

${\displaystyle h=R+r=R+R\cdot \cos \left({\frac {180^{\circ }}{51}}\right)=R\cdot \left(1+\cos \left({\frac {180^{\circ }}{51}}\right)\right)}$
${\displaystyle h\approx 1{,}998103\cdot R}$

### Area

The area of ​​a triangle is generally calculated ${\displaystyle A_{\Delta }={\frac {1}{2}}a\cdot h_{a}}$. The results of the side length are used to calculate the 51 gon${\displaystyle a}$ and the incircle radius ${\displaystyle r}$ used what ${\displaystyle r}$ for the height ${\displaystyle h_{a}}$ is used.

${\displaystyle a=R\cdot 2\cdot \sin \left({\frac {180^{\circ }}{51}}\right)}$
${\displaystyle h_{a}=r=R\cdot \cos \left({\frac {180^{\circ }}{51}}\right)\;}$ from this it follows for the area of ​​a partial triangle
{\displaystyle {\begin{aligned}A_{\Delta }&={\frac {1}{2}}\cdot R\cdot 2\cdot \sin \left({\frac {180^{\circ }}{51}}\right)\cdot R\cdot \cos \left({\frac {180^{\circ }}{51}}\right)\end{aligned}}\;} summarized it results
${\displaystyle A_{\Delta }=R^{2}\cdot \sin \left({\frac {180^{\circ }}{51}}\right)\cdot \cos \left({\frac {180^{\circ }}{51}}\right)}$
${\displaystyle A_{\Delta }\approx 0{,}0614441\cdot R^{2}}$

and for the area of ​​the entire 51 gon

${\displaystyle A=51\cdot A_{\Delta }=51\cdot R^{2}\cdot \sin \left({\frac {180^{\circ }}{51}}\right)\cdot \cos \left({\frac {180^{\circ }}{51}}\right)}$
${\displaystyle A\approx 3{,}133651\cdot R^{2}}$

## construction

As described above in the regular 51-square , the 51-square can be represented as a construction with compass and ruler . Since the number of its vertices results from the multiplication of the two Fermat prime numbers${\displaystyle 3}$ and ${\displaystyle 17}$results, the regular 51-sided can be found by an extension of an already known construction of the 17- sided. The two polygons triangle and seventeen-sided (their number of sides corresponds to the Fermat prime numbers${\displaystyle 3}$ or. ${\displaystyle 17}$) are placed on top of each other symmetrically with regard to their central angles in the common circumference, as is the case with z. B. Johannes Kepler in his work WORLD HARMONIK in the construction of Fünfzehnecks shows [1] .

In principle, one of the three methods described in Siebzehneck can be selected as the basis for the construction . For reasons of the very little effort required, the method of Duane W. DeTemple , [2] from 1991, is used.

### Preliminary considerations

Fig. 1: Seventeen-corner after Duane W. DeTemple

In the drawing of the seventeenth-corner by Duane W. DeTemple (Fig. 1) it is easy to see the perpendicular from the center${\displaystyle Q'}$ not only cuts the arc of a circle ${\displaystyle c_{2},}$but also the periphery. If this intersection is called${\displaystyle P_{34}}$ marked, it is right next to the corner point ${\displaystyle P_{11}.}$ This results in the central angle ${\displaystyle P_{34}OP_{0}}$with the angular width ${\displaystyle 120^{\circ }}$of an equilateral triangle , which is added geometrically clockwise to the central angle of the seventeenth corner.

Hence for

Central angle ${\displaystyle \theta _{1}}$of the district sector ${\displaystyle OP_{11}P_{0}}$
${\displaystyle \theta _{1}={\frac {6}{17}}\cdot 360^{\circ }=127{,}0588235294117647\ldots ^{\circ },}$
Central angle ${\displaystyle \theta _{2}}$ of the district sector ${\displaystyle OP_{11}P_{34}}$
${\displaystyle \theta _{2}={\frac {6}{17}}\cdot 360^{\circ }-120^{\circ }=7{,}0588235294117647\ldots ^{\circ },}$ because of
Central angle ${\displaystyle \mu }$ des 51-Ecks
${\displaystyle \mu ={\frac {360^{\circ }}{51}}=7{,}0588235294117647\ldots ^{\circ }}$ also applies
${\displaystyle \theta _{2}=\mu .}$

Thus the route is ${\displaystyle {\overline {P_{11}P_{34}}}}$ one side length ${\displaystyle a}$ and ${\displaystyle P_{34}}$ a corner point of the 51 corner you are looking for.

The position of the corner point ${\displaystyle P_{34}}$ of the 51-sided result from the number of side lengths ${\displaystyle a_{A}}$ those in the central angle ${\displaystyle 120^{\circ }}$ are included

${\displaystyle a_{A}={\frac {120^{\circ }}{7{,}0588235294117647}}=17,}$ it follows
starting from the corner point not counted ${\displaystyle P_{0},}$ clockwise the 17th corner corresponds to the corner ${\displaystyle P_{34}}$ which is counted counterclockwise.

The 17th corner point of the 51 corner is therefore based on the central axis ${\displaystyle {\overline {QP_{0}}}}$, exactly opposite the 34th corner point.

### Construction description

The identifiers in Figure 2, which have been changed compared to the original, correspond to those used today.

Fig. 2: Extension of the construction of the 17-sided according to Duane W. DeTemple to the construction of the 51-sided by adding the corners of the equilateral triangle (${\displaystyle P_{0}}$ - ${\displaystyle P_{17}}$ - ${\displaystyle P_{34}}$) and removing the missing points
• Draw a straight line ${\displaystyle x}$ (analytically an X-axis) and determine a point ${\displaystyle M_{1}}$ on the later center point of the polygon (analytically a coordinate origin).
• Drawing a circle as a perimeter (analytically a unit circle) ${\displaystyle C_{1}}$ one ${\displaystyle M_{1}}$. There are two points of intersection , the corner point${\displaystyle P_{0}}$ of the polygon and the counterpoint ${\displaystyle B}$.
• Establishing the vertical ${\displaystyle y}$ (analytically a Y-axis) on the straight line ${\displaystyle x}$ in ${\displaystyle M_{1}}$. The point of intersection results${\displaystyle Y_{0}}$.
• Halving the route ${\displaystyle {\overline {BM_{1}}}}$ in ${\displaystyle M_{2}}$.
• Establishing the vertical on the straight line in ${\displaystyle M_{2}}$. The two intersections with${\displaystyle C_{1}}$ are the cornerstones ${\displaystyle P_{17}}$ and ${\displaystyle P_{34}}$ des 51-Ecks.
• Draw the circular arc ${\displaystyle C_{2}}$ one ${\displaystyle M_{2}}$ with the radius ${\displaystyle {\overline {M_{2}P_{0}}}}$. The intersection with the vertical is${\displaystyle M_{Cc1}}$.
• Now is around ${\displaystyle M_{Cc1}}$the first Carlyle circle ${\displaystyle C_{C1}}$ through the point ${\displaystyle Y_{0}}$drawn; the intersections are${\displaystyle X_{Cc1,1}}$ and ${\displaystyle X_{Cc1,2}}$.
• The distance ${\displaystyle {\overline {M_{1}X_{Cc1,1}}}}$is halved. You get${\displaystyle M_{Cc2}}$.
• Draw a second Carlyle circle ${\displaystyle C_{C2}}$ one ${\displaystyle M_{Cc2}}$ by ${\displaystyle Y_{0}}$. The points of intersection with x are the points${\displaystyle X_{Cc2,1}}$ and ${\displaystyle X_{Cc2,2}}$ (the latter is not shown because it is no longer needed).
• The distance ${\displaystyle {\overline {M_{1}X_{Cc1,2}}}}$is halved. You get${\displaystyle M_{Cc3}}$.
• Draw a third Carlyle circle ${\displaystyle C_{C3}}$ one ${\displaystyle M_{Cc3}}$ by ${\displaystyle Y_{0}}$. The points of intersection with x are the points${\displaystyle X_{Cc3,1}}$ and ${\displaystyle X_{Cc3,2}}$ (The latter is also not shown, as it is no longer needed).
• Removal of the route ${\displaystyle {\overline {M_{1}X_{Cc2,1}}}}$ on ${\displaystyle y}$ of ${\displaystyle Y_{0}}$off off. You get point${\displaystyle Y_{1}}$
• Connect the dots ${\displaystyle Y_{1}}$ and ${\displaystyle X_{Cc3,1}}$ with a route.
• Halve the route ${\displaystyle {\overline {Y_{1}X_{Cc3,1}}}}$. You get point${\displaystyle M_{Cc4}}$.
• Draw a fourth Carlyle circle ${\displaystyle C_{C4}}$ one ${\displaystyle M_{Cc4}}$ by ${\displaystyle Y_{0}}$. The points of intersection with x are the points${\displaystyle X_{Cc4,1}}$ and ${\displaystyle X_{Cc4,2}}$ (the latter not labeled as it is no longer needed).
• Redraw an arc of a circle ${\displaystyle X_{Cc4,1}}$ with the perimeter radius ${\displaystyle {\overline {M_{1}B}}}$. The intersections with the perimeter${\displaystyle C_{1}}$ are the two too ${\displaystyle P_{0}}$ adjacent points of the 17-sided and thus the points ${\displaystyle P_{3}}$ and ${\displaystyle P_{48}}$ des 51-Ecks.
• By repeatedly removing the route ${\displaystyle {\overline {P_{0}P_{3}}}}$ on the perimeter ${\displaystyle C_{1}}$, starting with ${\displaystyle P_{0}}$, you get the missing points of a regular 17-sided. Up to this point the construction corresponds to that of the 17-sided.
• By repeatedly removing the route ${\displaystyle {\overline {P_{0}P_{3}}}}$ on the perimeter ${\displaystyle C_{1}}$, starting from the points ${\displaystyle P_{17}}$ (blue) and ${\displaystyle P_{34}}$ (red), you get all the missing corner points of the 51 corner, which can be connected to each other to form the 51 corner.

## Occurrence

architecture

RWE tower in Essen

The cross-section of the RWE tower in Essen is a regular 51-sided.

## Individual evidence

1. Johannes Kepler: WORLD HARMONICS. XLIV. Sentence., Page of the Fifteenth, Page 44. In: Google Books. R. OLDENBURG VERLAG 2006, translated and introduced by MAX CASPAR 1939, p. 401 , accessed on February 21, 2018 .
2. Duane W. DeTemple: Carlyle Circles and the Lemoine Simplicity of Polygon Constructions. (Memento vom 11. August 2011 im Internet Archive). The American Mathematical Monthly, Vol. 98, No. 2 (Feb., 1991), S. 101–104 (JSTOR 2323939) aufgerufen am 16. Februar 2018.