Summary

Description01 Dreiteilung des Winkels-180°-2.svg	Deutsch: Dreiteilung des Winkels mit Kurzbeschreibung, Näherungskonstruktion für Winkel zwischen 0° und 180°. Ein paar Konstruktionselemente stammen aus Alberts' Konstruktion. English: Trisection of an angle with brief description, proximity construction for angles between 0° and 180°. A few construction elements come from Alberts' construction.
Date	7 March 2020
Source	Own work
Author	Petrus3743
Other versions
SVG development InfoField	The SVG code is valid.   This trigonometry was created with GeoGebra by Petrus3743.   This SVG trigonometry uses the path text method.

Anmerkungen

Mit der stark vereinfachten Konstruktion wird folgendes erreicht:

• Ein großer Teil der Konstruktion liegt in der unteren Hälfte des Kreises ${\displaystyle k_{1}.}$
• Eine praktikable Dreiteilung des Winkels ab nahe ${\displaystyle 0^{\circ }}$ bis ${\displaystyle 180^{\circ }.}$ Siehe hierzu in GeoGebra

Konstruktion

1. Kreis ${\displaystyle k_{1}}$ mit beliebigem Durchmesser ${\displaystyle {\overline {AB}}}$ um Mittelpunkt ${\displaystyle O.}$
2. Winkelschenkel ${\displaystyle {\overline {OA}}}$ und Winkelschenkel ${\displaystyle {\overline {OC}}}$ schließen den Winkel ${\displaystyle \alpha }$ im Scheitel ${\displaystyle O}$ ein, ${\displaystyle {\overline {OB}}}$ und ${\displaystyle {\overline {OC}}}$ den Ergänzungswinkel ${\displaystyle \gamma .}$
3. Kreis ${\displaystyle k_{2}}$ um ${\displaystyle O}$ mit Radius ${\displaystyle {\tfrac {1}{3}}{\overline {OA}}}$; die Verlängerung des Winkelschenkels ${\displaystyle {\overline {CO}}}$ schneidet Kreis ${\displaystyle k_{2}}$ in ${\displaystyle D.}$
4. Durchmesser ${\displaystyle {\overline {EF}}}$ mit ${\displaystyle \angle {EOA=45^{\circ }}}$ und Verbindung des Punktes ${\displaystyle D}$ mit ${\displaystyle E.}$
5. 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 Mittelsenkrechte von ${\displaystyle |GH|}$ schneidet ${\displaystyle {\overline {OE}}}$ in ${\displaystyle I,}$ ergibt ${\displaystyle |IG|={\overline {IH}}.}$
7. Parallele zu ${\displaystyle {\overline {ED}}}$ ab ${\displaystyle I}$ erreicht Kreis ${\displaystyle k_{2}}$ in ${\displaystyle J.}$
8. Parallele zu ${\displaystyle {\overline {IJ}}}$ ab ${\displaystyle D,}$ Punkt ${\displaystyle K}$ darauf so, dass ${\displaystyle {\overline {JK}}={\overline {JE}}.}$
9. Linie ab ${\displaystyle J}$ durch ${\displaystyle K}$ erreicht Kreis ${\displaystyle k_{1}}$ in ${\displaystyle L,}$ anschließend Linie ab ${\displaystyle L}$ bis ${\displaystyle O.}$
10. Parallele zu ${\displaystyle {\overline {LO}}}$ ab ${\displaystyle D}$ erreicht Kreis ${\displaystyle k_{2}}$ in ${\displaystyle M.}$
11. Strecke ${\displaystyle {\overline {KE}}}$ über ${\displaystyle E}$ hinaus verlängern, Punkt ${\displaystyle N}$ darauf so, dass ${\displaystyle {\overline {MN}}={\overline {MD}}.}$
12. Linie ab ${\displaystyle M}$ durch ${\displaystyle N}$ erreicht Kreis ${\displaystyle k_{1}}$ in ${\displaystyle P.}$
13. Bestimme Punkt ${\displaystyle Q}$ so, dass Winkel ${\displaystyle POQ=30^{\circ },}$ Verbindung ${\displaystyle Q}$ mit ${\displaystyle O}$ ergibt den Winkel ${\displaystyle AOQ=\beta .}$
14. Mittelsenkrechte von ${\displaystyle |CQ|}$ schneidet ${\displaystyle k_{1}}$ in ${\displaystyle R,}$ verbinde ${\displaystyle R}$ mit ${\displaystyle O.}$
15. Bestimme Punkt ${\displaystyle S}$ so, dass Winkel ${\displaystyle QOS=120^{\circ }.}$
16. Abschließende Verbindung ${\displaystyle S}$ mit ${\displaystyle O}$ ergibt Winkel ${\displaystyle SOB=\delta .}$

• Der Winkel ${\displaystyle AOQ=\beta }$ ist nahezu gleich einem Drittel des Winkels ${\displaystyle AOC=\alpha .}$
• Der Winkel ${\displaystyle SOB=\delta }$ ist nahezu gleich einem Drittel des Winkels ${\displaystyle COB=\gamma .}$

Fehlerbetrachtung

Eine Fehleranalyse, ähnlich Alberts' Konstruktion, ist nicht vorhanden.

Die dargestellte Konstruktion wurde mit der Dynamische-Geometrie-Software (DGS) GeoGebra angefertigt; darin werden in diesem Fall die Winkelgrade meist mit signifikanten dreizehn Nachkommastellen angezeigt. Die sehr kleinen Fehler des Winkels ${\displaystyle \beta }$ bzw. ${\displaystyle \delta }$, sprich, die Differenzwerte aus ${\displaystyle {\frac {\alpha }{3}}-\beta }$ bzw. ${\displaystyle {\frac {\gamma }{3}}-\delta ,}$ werden von GeoGebra stets mit ${\displaystyle 0^{\circ }}$ angezeigt.

Betrachtet man die Grafik in GeoGebra, in sehr kleinen Schritten, die zu- oder abnehmenden Winkelweiten des Winkels ${\displaystyle \beta }$ bzw. ${\displaystyle \delta }$ mithilfe des Schiebereglers oder der Animation, ist vereinzelt eine max. Abweichung ${\displaystyle 1\cdot 10^{-13}\mathrm {^{\circ }} }$ vom SOLL-Wert ${\displaystyle {\frac {\alpha }{3}}}$ bzw. ${\displaystyle {\frac {\gamma }{3}}}$ ablesbar.

Verdeutlichung des absoluten Fehlers

Der in GeoGebra ablesbare Differenzwert von max. ${\displaystyle 1\cdot 10^{-13}\mathrm {^{\circ }} }$ entspricht einem absoluten Fehler ${\displaystyle F}$ der – nicht eingezeichneten – Sehne ${\displaystyle |AQ|}$ bzw. ${\displaystyle |BS|,}$ der sich wie folgt ergibt:

${\displaystyle F=2\cdot \sin \left({\frac {1\cdot 10^{-13}\mathrm {^{\circ }} }{2}}\right)=0{,}000\;000\;000\;000\;001\;745\ldots =1{,}745\ldots \cdot 10^{-15}\;[\mathrm {LE} ]}$

Hätten die Winkelschenkel die Länge gleich 1 Milliarde km (das Licht bräuchte für diese Strecke ≈ 56 Minuten, das ist etwas weniger als 7-mal die Entfernung Erde – Sonne), wäre der absolute Fehler der beiden – nicht eingezeichneten – Sehnen ${\displaystyle |AQ|}$ bzw. ${\displaystyle |BS|}$ ≈ 1,7 mm.

Fehlerüberprüfung eines frei gewählten Winkels > 0° ... 180°

• Siehe GeoGebra

Remarks

With the greatly simplified construction, the following is achieved:

• Much of the construction is in the lower half of the circle ${\displaystyle k_{1}.}$
• A practical trisection of the angle from close ${\displaystyle 0^{\circ }}$ to ${\displaystyle 180^{\circ }.}$ See also in GeoGebra

Construction

1. Circle ${\displaystyle k_{1}}$ with any diameter ${\displaystyle {\overline {AB}}}$ around center ${\displaystyle O.}$
2. Angle leg ${\displaystyle {\overline {OA}}}$ and angle leg ${\displaystyle {\overline {OC}}}$ enclose the angle ${\displaystyle \alpha }$ in apex ${\displaystyle O}$, ${\displaystyle {\overline {OB}}}$ and ${\displaystyle {\overline {OC}}}$ the supplementary angle ${\displaystyle \gamma .}$
3. Circle ${\displaystyle k_{2}}$ around ${\displaystyle O}$ with radius ${\displaystyle {\tfrac {1}{3}}{\overline {OA}}.}$ the extension of the angle leg ${\displaystyle {\overline {CO}}}$ intersects circle ${\displaystyle k_{2}}$ in ${\displaystyle D.}$
4. Diameter ${\displaystyle {\overline {EF}}}$ with ${\displaystyle \angle {EOA=45^{\circ }}}$ and connection of point ${\displaystyle D}$ with ${\displaystyle E.}$
5. Point ${\displaystyle G}$ on circle ${\displaystyle k_{2}}$ so that ${\displaystyle |EG|={\overline {EO}}.}$
6. Line segment ${\displaystyle {\overline {OF}}}$ in ${\displaystyle H}$ halved, the perpendicular bisector of ${\displaystyle |GH|}$ cuts ${\displaystyle {\overline {OE}}}$ in ${\displaystyle I,}$ results ${\displaystyle |IG|={\overline {IH}}.}$
7. Parallel to ${\displaystyle {\overline {ED}}}$ from ${\displaystyle I}$ reaches circle ${\displaystyle k_{2}}$ in ${\displaystyle J.}$
8. Parallel to ${\displaystyle {\overline {IJ}}}$ from ${\displaystyle D,}$ point ${\displaystyle K}$ on it such that ${\displaystyle {\overline {JK}}={\overline {JE}}.}$
9. Line from ${\displaystyle J}$ through ${\displaystyle K}$ reaches circle ${\displaystyle k_{1}}$ in ${\displaystyle L,}$ then line from ${\displaystyle L}$ to ${\displaystyle O.}$
10. Parallel to ${\displaystyle {\overline {LO}}}$ from ${\displaystyle D}$ reaches circle ${\displaystyle k_{2}}$ in ${\displaystyle M.}$
11. Line segment ${\displaystyle {\overline {KE}}}$ extended beyond ${\displaystyle E}$, point ${\displaystyle N}$ on it that ${\displaystyle {\overline {MN}}={\overline {MD}}.}$
12. Line from ${\displaystyle M}$ through ${\displaystyle N}$ reaches circle ${\displaystyle k_{1}}$ in ${\displaystyle P.}$
13. Determine point ${\displaystyle Q}$ so that angle ${\displaystyle POQ=30^{\circ },}$ connection ${\displaystyle Q}$ with ${\displaystyle O}$ gives the angle ${\displaystyle AOQ=\beta .}$
14. Perpendicular bisector of ${\displaystyle |CQ|}$ cuts ${\displaystyle k_{1}}$ in ${\displaystyle R,}$ connect ${\displaystyle R}$ with ${\displaystyle O.}$
15. Determine point ${\displaystyle S}$ so that angle ${\displaystyle QOS=120^{\circ }.}$
16. Final connection ${\displaystyle S}$ with ${\displaystyle O}$ gives angle ${\displaystyle SOB=\delta .}$

• The angle ${\displaystyle AOQ=\beta }$ is almost equal to one third of the angle ${\displaystyle AOC=\alpha .}$
• The angle ${\displaystyle SOB=\delta }$ is almost equal to a third of the angle ${\displaystyle COB=\gamma .}$

Error viewing

An error analysis, similar to Alberts' construction, is not available.

The construction shown was made with the dynamic geometry software (DGS) GeoGebra; in this case the degrees are usually displayed with significant thirteen decimal places. The very small errors of the angle ${\displaystyle \beta }$ beta or ${\displaystyle \delta }$, that is, the difference values from ${\displaystyle {\frac {\alpha }{3}}-\beta }$ or ${\displaystyle {\frac {\gamma }{3}}-\delta ,}$ are always displayed by GeoGebra with ${\displaystyle 0^{\circ }.}$

If you look at the graphic in GeoGebra, in very small steps, the increasing or decreasing angular widths of the angle ${\displaystyle \beta }$ or ${\displaystyle \delta }$ using the slider or animation, there is occasionally one max. deviation ${\displaystyle 1\cdot 10^{-13}\mathrm {^{\circ }} }$ of the rated value ${\displaystyle {\frac {\alpha }{3}}}$ or ${\displaystyle {\frac {\gamma }{3}}}$ readable.

Clarification of the absolute error

The difference value shown in GeoGebra of max. ${\displaystyle 1\cdot 10^{-13}\mathrm {^{\circ }} }$ corresponds to an absolute error ${\displaystyle F}$ of the  – not shown – chord ${\displaystyle |AQ|}$ or ${\displaystyle |BS|,}$ which results as follows:

${\displaystyle F=2\cdot \sin \left({\frac {1\cdot 10^{-13}\mathrm {^{\circ }} }{2}}\right)=0.000\;000\;000\;000\;001\;745\ldots =1.745\ldots \cdot 10^{-15}\;[\mathrm {unit\;of\;length} ]}$

If the angled legs had a length of 1 Billion km (the light would need ≈ 56 minutes for this distance, that's a little less than 7 times the distance earth - sun.), the absolute error of the twoden – not shown – cords would be ${\displaystyle |AQ|}$ or ${\displaystyle |BS|}$ ≈ 1.7 mm.

Trisection an angle > 0° ... 180°, for checking the error

• See GeoGebra

Licensing

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

Die Methode basiert in einigen Konstruktionselementen auf Alberts' Konstruktion.