File:01 Neuneck-60°.svg

Summary[edit]

Description01 Neuneck-60°.svg	Deutsch: Neuneck (Nonagon) bei gegebenem Umkreis mit Winkeldreiteilung, Näherungskonstruktion English: Nonagon for a given circumference with an angle trisection, an approximation construction
Date	20 September 2022
Source	Own work
Author	Petrus3743
SVG development InfoField	The SVG code is valid.   This trigonometry was created with GeoGebra by Petrus3743.   This SVG trigonometry uses the path text method.

Licensing[edit]

I, the copyright holder of this work, hereby publish it under the following license:

This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.

You are free:

• to share – to copy, distribute and transmit the work
• to remix – to adapt the work

Under the following conditions:

• attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
• share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original.

https://creativecommons.org/licenses/by-sa/4.0CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 truetrue

Konstruktion[edit]

(Siehe hierzu auch die Animation)

1. Kreis ${\displaystyle k_{1}}$ mit beliebigem Radius ${\displaystyle r={\overline {OA}}}$ um Mittelpunkt ${\displaystyle O.}$
2. Winkelschenkel ${\displaystyle {\overline {OA}}}$ und Winkelschenkel ${\displaystyle {\overline {OB}}}$ schließen den Winkel ${\displaystyle \alpha =60^{\circ }}$ im Scheitel ${\displaystyle O}$ ein.
3. Kreis ${\displaystyle k_{2}}$ um ${\displaystyle O}$ mit Radius ${\displaystyle {\overline {OC}}={\tfrac {1}{3}}{\overline {OA}}}$; die Verlängerung des Winkelschenkels ${\displaystyle {\overline {BO}}}$ schneidet Kreis ${\displaystyle k_{2}}$ in ${\displaystyle D.}$
4. Durchmesser ${\displaystyle {\overline {EF}}}$ des Kreises ${\displaystyle k_{2}}$ mit ${\displaystyle \angle {EOA=45^{\circ }}.}$
5. Bestimme Punkt ${\displaystyle G}$ auf Kreis ${\displaystyle k_{2}}$ so, dass ${\displaystyle |EG|={\overline {EO}}.}$
6. Strecke ${\displaystyle {\overline {OF}}}$ in ${\displaystyle H}$ halbieren, die anschließende, nicht eingezeichnete, Mittelsenkrechte von ${\displaystyle |GH|}$ schneidet ${\displaystyle {\overline {OE}}}$ in ${\displaystyle I,}$ ergibt ${\displaystyle |IG|={\overline {IH}}.}$
7. Linie durch ${\displaystyle E}$ und ${\displaystyle D}$; deren Enden liegen nahe am Kreis ${\displaystyle k_{1}.}$
8. Parallele zu ${\displaystyle {\overline {ED}}}$ ab ${\displaystyle I}$ erreicht Kreis ${\displaystyle k_{2}}$ in ${\displaystyle J.}$
9. Bestimme Punkt ${\displaystyle K}$ mithilfe eines nicht eingezeichneten Kreisbogens (Radius ${\displaystyle |JE|}$ um ${\displaystyle J}$, auf Linie durch ${\displaystyle E}$ und ${\displaystyle D}$ so, dass ${\displaystyle {\overline {JK}}=|JE|.}$
10. Linie ab ${\displaystyle J}$ durch ${\displaystyle K}$ erreicht Kreis ${\displaystyle k_{1}}$ in ${\displaystyle L,}$ anschließend Linie ab ${\displaystyle L}$ bis ${\displaystyle O.}$

Hinweis: Die Punkte ${\displaystyle K}$ und ${\displaystyle J}$ liegen nicht auf der Strecke ${\displaystyle {\overline {LO}}.}$

11. Parallele zu ${\displaystyle {\overline {LO}}}$ ab ${\displaystyle D}$ erreicht Kreis ${\displaystyle k_{2}}$ in ${\displaystyle M.}$
12. Bestimme Punkt ${\displaystyle N}$ mithilfe eines nicht eingezeichneten Kreisbogens (Radius ${\displaystyle {\overline {MD}}}$) um ${\displaystyle M}$, auf Linie durch ${\displaystyle E}$ und ${\displaystyle D}$ so, dass die werdende Strecke ${\displaystyle {\overline {MN}}={\overline {MD}}.}$
13. Linie ab ${\displaystyle M}$ durch ${\displaystyle N}$ erreicht Kreis ${\displaystyle k_{1}}$ in ${\displaystyle P.}$
14. Bestimme Punkt ${\displaystyle Q}$ so, dass Winkel ${\displaystyle POQ=30^{\circ }.}$
15. Bestimme Punkt ${\displaystyle R}$ durch Verdoppelung des Kreisbogens ${\displaystyle OAQ}$ und verbinde ${\displaystyle O}$ mit ${\displaystyle R.}$

• Der konstruierte Winkel ${\displaystyle AOR=\beta }$ ist nahezu gleich zwei Drittel des Winkels ${\displaystyle \alpha =60^{\circ }}$, d. h. nahezu ${\displaystyle 40^{\circ }.}$

Ergebnis, bezogen auf den Einheitskreis r = 1 [LE][edit]

In GeoGebra werden max. 15 Nachkommastellen angezeigt.

${\displaystyle {\begin{alignedat}{2}{\text{in GeoGebra konstruierte Seitenlänge}}\qquad a&=&&0{,}684040286651337\;[\mathrm {LE} ]\\-{\text{Sollwert }}\qquad -a_{SOLL}&=-&&0{,}684040286651337\dots \;[\mathrm {LE} ]\\\hline {\text{absoluter Fehler in GeoGebra nicht evaluierbar }}\qquad F&=&&0{,}000000000000000\dots \;[\mathrm {LE} ]\end{alignedat}}}$

Beispiel zur Verdeutlichung des Fehlers[edit]

• Bei einem Radius r = 1 Mrd. km (das Licht bräuchte für diese Strecke ca. 56 min) wäre der Fehler der konstruierten Seitenlänge a des Neunecks < 1 mm.

Construction[edit]

(See also the animation)

1. Circle ${\displaystyle k_{1}}$ with any radius ${\displaystyle r={\overline {OA}}}$ around centre ${\displaystyle O.}$
2. Angle leg ${\displaystyle {\overline {OA}}}$ and angle leg ${\displaystyle {\overline {OB}}}$ enclose angle ${\displaystyle \alpha =60^{\circ }}$ at vertex ${\displaystyle O}$.
3. Circle ${\displaystyle k_{2}}$ around ${\displaystyle O}$ with radius ${\displaystyle {\overline {OC}}={\tfrac {1}{3}}{\overline {OA}}}$; the extension of the angle leg ${\displaystyle {\overline {BO}}}$ intersects circle ${\displaystyle k_{2}}$ in ${\displaystyle D.}$.
4. Diameter ${\displaystyle {\overline {EF}}}$ of circle ${\displaystyle k_{2}}$ with ${\displaystyle \angle {EOA=45^{\circ }}.}$
5. Determine point ${\displaystyle G}$ on circle ${\displaystyle k_{2}}$ such that ${\displaystyle |EG|={\overline {EO}}.}$
6. Bisect the line segment ${\displaystyle {\overline {OF}}}$ in ${\displaystyle H}$, the subsequent, not drawn, median perpendicular of ${\displaystyle |GH|}$ intersects ${\displaystyle {\overline {OE}}}$ in ${\displaystyle I,}$ gives ${\displaystyle |IG|={\overline {IH}}.}$
7. Line through ${\displaystyle E}$ and ${\displaystyle D}$; their ends are close to circle ${\displaystyle k_{1}.}$.
8. Parallel to ${\displaystyle {\overline {ED}}}$ from ${\displaystyle I}$ reaches circle ${\displaystyle k_{2}}$ in ${\displaystyle J.}$
9. Determine point ${\displaystyle K}$ using an undrawn arc of a circle (radius ${\displaystyle |JE|}$ around ${\displaystyle J}$, on the line through ${\displaystyle E}$ and ${\displaystyle D}$ such that ${\displaystyle {\overline {JK}}=|JE|.}$
10. Line from ${\displaystyle J}$ through ${\displaystyle K}$ reaches circle ${\displaystyle k_{1}}$ in ${\displaystyle L,}$ then line from ${\displaystyle L}$ to ${\displaystyle O.}$

Note: The points ${\displaystyle K}$ and ${\displaystyle J}$ are not on the line ${\displaystyle {\overline {LO}}.}$

11. Parallel to ${\displaystyle {\overline {LO}}}$ from ${\displaystyle D}$ reaches circle ${\displaystyle k_{2}}$ in ${\displaystyle M.}$
12. Determine point ${\displaystyle N}$ using an undrawn arc of a circle (radius ${\displaystyle {\overline {MD}}}$) around ${\displaystyle M}$, on the line through ${\displaystyle E}$ and ${\displaystyle D}$ such that the becoming distance is ${\displaystyle {\overline {MN}}={\overline {MD}}.}$
13. Line from ${\displaystyle M}$ through ${\displaystyle N}$ reaches circle ${\displaystyle k_{1}}$ in ${\displaystyle P.}$
14. Determine point ${\displaystyle Q}$ such that angle ${\displaystyle POQ=30^{\circ }.}$
15. Determine point ${\displaystyle R}$ by doubling arc of a circle ${\displaystyle OAQ}$ and connect ${\displaystyle O}$ with ${\displaystyle R.}$

• The constructed angle ${\displaystyle AOR=\beta }$ is nearly equal to two thirds of the angle ${\displaystyle \alpha =60^{\circ }}$, i.e. nearly ${\displaystyle 40^{\circ }.}$

Result, related to the unit circle r = 1 [LE][edit]

In GeoGebra, max. 15 decimal places are displayed.

${\displaystyle {\begin{alignedat}{2}{\text{side length constructed in GeoGebra}}\qquad a&=&&0.684040286651337\;[\mathrm {LE} ]\\-{\text{setpoint}}\qquad -a_{setpoint}&=-&&0.684040286651337\dots \;[\mathrm {LE} ]\\\hline {\text{absolute error in GeoGebra cannot be evaluated }}\qquad F&=&&0.000000000000000\dots \;[\mathrm {LE} ]\end{alignedat}}}$

Example to illustrate the error[edit]

• For a radius r = 1 billion km (the light would need approx. 56 min for this distance) the error of the constructed side length a of the neuneck would be < 1 mm.