Heptagon after Archimedes

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Construction by Archimedes, animation, 10 s pause at the end

The heptagon according to Archimedes is a continuation of the so-called construction of Archimedes , a generally known approach to the construction of the regular heptagon in the professional world. The heptagon according to Archimedes , like any regular heptagon, cannot be represented exactly with the classic tools of compasses and unmarked rulers alone (see Constructible Polygon ), but with an aid for dividing the angle into three (see Regular Polygon ), for example a marked ruler ( Neusis- Construction ). Archimedes (287–212 BC) published the construction in his work "Siebeneck im Kreis". According to the tradition of Arab scholars, the book, which was written in Greek, was lost. It was only around 1100 years later, i.e. in the 9th century , that Thabit ibn Qurra (826-901) translated Archimedes 'work into Arabic and thus received evidence of Archimedes' construction for posterity (see section Proof ). In the end, another 1100 years passed before Carl Schoy (1877–1925) translated Archimedes' book, which is about dividing the circle into 7 equal parts , into German.

Construction of Archimedes

In a square with any side length, a straight line is drawn from the point until it intersects the extension of the side of the square at the point , so that: ${\displaystyle AWRC}$ ${\displaystyle W}$ ${\displaystyle {\overline {AC}}}$ ${\displaystyle M}$

${\displaystyle \left(1\right)}$  Area of  area of ${\displaystyle \triangle {WRU}=}$ ${\displaystyle \triangle {CTM}.}$

Archimedes' geometrical construction is mainly based on the division of a route, for which, as we are told, he used the construction method of insertion (Neusis). The way in which he carried out this insertion in order to obtain the exact division points has not been passed down to us. An insertion with the help of a ruler, the edge of which is rotated around the point in such a way that a line provides the condition described above, and with areas of the same size, is obviously not effective. ${\displaystyle W}$ ${\displaystyle {\overline {WM}}}$ ${\displaystyle \triangle {WRU}}$ ${\displaystyle \triangle {CTM}}$

The following explanations, presented in modern language , are based heavily on the description of Thabit ibn Qurra, which essentially consists of two steps.

Image 1: Step 1, outline sketch

As a first step (Fig. 1), we make a basic sketch of the route with its dividing points and . Therein is the side length of the square, and at the same time it should apply: ${\displaystyle {\overline {AM}}}$ ${\displaystyle D}$ ${\displaystyle C}$ ${\displaystyle {\overline {AC}}}$ ${\displaystyle {\overline {AD}}<{\overline {AC}}}$ ${\displaystyle {\overline {AM}}>{\overline {AC}},}$

${\displaystyle \left(2\right)}$   ${\displaystyle {\overline {CM}}^{2}={\overline {AC}}\cdot {\overline {AD}}}$ and

${\displaystyle \left(3\right)}$   ${\displaystyle {\overline {AD}}^{2}={\overline {CD}}\cdot {\overline {DM}}}$

Image 2: Step 2, outline sketch with extension

As a second step (Fig. 2), we expand the basic sketch that we just created. Over the distance First, using the isosceles triangle built. If one connects the point with results which also isosceles triangle After Thābit ibn Qurra have - at exactly determined division point and end point - the angle at the crests and each of the angular distance and at the vertices ( supplementary angle, side angle ), and each angular width Thus, the Angle of the central angle of the heptagon. ${\displaystyle {\overline {AC}}}$ ${\displaystyle {\overline {CM}}}$ ${\displaystyle ACB}$ ${\displaystyle B}$ ${\displaystyle M,}$ ${\displaystyle CMB.}$ ${\displaystyle D}$ ${\displaystyle M}$ ${\displaystyle M}$ ${\displaystyle B}$ ${\displaystyle \mu ={\tfrac {\pi }{7}}}$ ${\displaystyle C}$ ${\displaystyle A}$ ${\displaystyle 2\mu ={\tfrac {2\pi }{7}}.}$ ${\displaystyle 2\mu }$

Determine the division point D

Fig. 3: Determination of the division point using the graph of the function ${\displaystyle D}$ ${\displaystyle f(x)=x^{3}-6ax^{2}+5a^{2}x-a^{3}}$

This requires at least one additional tool, such as B. a parabola or a parabola and hyperbola or the function graph determined below .

For an exact construction (Fig. 3) you first draw the square with any side length - preferably with the help of dynamic geometry software (DGS) - and then initially only extend it beyond. In order to get the triangles and areas of the same size, it is sufficient to determine the dividing point. The missing point is then easy to find. ${\displaystyle AWRC}$ ${\displaystyle a={\overline {AC}}}$ ${\displaystyle {\overline {AC}}}$ ${\displaystyle C}$ ${\displaystyle \triangle {WRU}}$ ${\displaystyle \triangle {CTM}}$ ${\displaystyle D}$ ${\displaystyle M}$

Let it be and such that it also applies ${\displaystyle {\overline {AD}}=x,}$ ${\displaystyle {\overline {AC}}=a}$ ${\displaystyle {\overline {CM}}=y,}$

${\displaystyle \left(4\right)}$   ${\displaystyle y^{2}=a\cdot x}$ and

${\displaystyle \left(5\right)}$   ${\displaystyle x^{2}=\left(a-x\right)\cdot \left(a+y-x\right)}$

from it y

${\displaystyle \;{\begin{aligned}\left(6\right)\;\;x^{2}&=\left(a-x\right)\cdot \left(a+y-x\right)\\0&=a^{2}+ay-2ax-xy\\2ax-a^{2}&=y\cdot \left(a-x\right)\\y&=\left({\frac {a\left(2x-a\right)}{a-x}}\right)=a\cdot \left({\frac {2x-a}{a-x}}\right)\\\end{aligned}}}$

${\displaystyle y}$ used in ${\displaystyle \left(4\right)}$

${\displaystyle \;\;\;\;{\begin{aligned}\left(7\right)\;ax&=a^{2}\cdot \left({\frac {2x-a}{a-x}}\right)^{2}=a^{2}\cdot {\frac {\left(2x-a\right)^{2}}{\left(a-x\right)^{2}}}\\\end{aligned}}}$

The equation multiplied by and then divided by gives: ${\displaystyle \left(7\right)}$ ${\displaystyle \left(a-x\right)^{2}}$ ${\displaystyle a}$

${\displaystyle {\begin{aligned}\left(8\right)\;\;\;x\cdot \left(a-x\right)^{2}&=a\cdot \left(2x-a\right)^{2},\\x\cdot \left(a^{2}-2ax+x^{2}\right)&=a\cdot \left(4x^{2}-4xa+a^{2}\right),\\xa^{2}-2ax^{2}+x^{3}&=4ax^{2}-4xa^{2}+a^{3},\\\end{aligned}}}$

from this the cubic equation follows

${\displaystyle \;\;\;\;{\begin{aligned}\left(9\right)\;x^{3}&-6ax^{2}+5a^{2}x-a^{3}=0\end{aligned}}.}$

The function has two zeros within the distance . One closer to the point and one at the point. The third zero is outside the line ${\displaystyle f(x)=x^{3}-6ax^{2}+5a^{2}x-a^{3}}$ ${\displaystyle {\overline {AM}}}$ ${\displaystyle A}$ ${\displaystyle D.}$ ${\displaystyle {\overline {AM}}.}$

If then are the coordinates of the relevant point of the function graph ${\displaystyle A=\left(0\mid 0\right),}$ ${\displaystyle D}$

${\displaystyle D=\left(a\cdot 0{,}643104132\ldots \mid 0\right).}$

Because of this also results ${\displaystyle {\overline {CM}}=y={\sqrt {a\cdot x}}}$

${\displaystyle M=\left(a\cdot 1{,}801937735\ldots \mid 0\right).}$

proof

Figure 4: Proof by division of a circle,
side length ${\displaystyle s={\overline {CM}}={\overline {AB}}=a\cdot 0.801937735\ldots }$

The following division of the circle into seven equally long arcs should serve as a possible proof of the correctness of Archimedes' construction.

On a straight line, first the given distances and are plotted, then the isosceles triangle is drawn and the points are connected with and with . After determining the center of the circumference with the help of the two perpendiculars through and , the circumference is drawn. This is followed by the extensions of the lines and until they intersect in or around the area. Now it is connected with , which results in the intersection point that is immediately connected with. ${\displaystyle {\overline {AM}},}$ ${\displaystyle {\overline {AD}}}$ ${\displaystyle {\overline {AC}}}$ ${\displaystyle ACB}$ ${\displaystyle B}$ ${\displaystyle M}$ ${\displaystyle B}$ ${\displaystyle D}$ ${\displaystyle O_{1}}$ ${\displaystyle C}$ ${\displaystyle D,}$ ${\displaystyle {\overline {BD}}}$ ${\displaystyle {\overline {BC}}}$ ${\displaystyle E}$ ${\displaystyle Z}$ ${\displaystyle A}$ ${\displaystyle Z}$ ${\displaystyle H,}$ ${\displaystyle C}$

The illustration (Fig. 4) shows ( circular arc): ${\displaystyle arc=}$

${\displaystyle (10)\;MC=CB}$in fact follows: ${\displaystyle \triangle {MBC}}$

${\displaystyle (11)\;\angle {BMC}=\angle {MBC},}$

${\displaystyle (12)\;arc\;O_{1}ZM=arc\;O_{1}BA,}$ consequently:

${\displaystyle (13)\;{\overline {CD}}\cdot {\overline {DM}}={\overline {DB}}^{2}={\overline {AD}}^{2}}$ (→ fulfills (3)) and

${\displaystyle (14)\;\triangle {MBD}\sim \triangle {CBD},}$ because

${\displaystyle (15)\;\angle {DMB}=\angle {CBD},}$d. H.

${\displaystyle (16)\;arc\;O_{1}EZ=arc\;O_{1}BA.}$ So are

${\displaystyle (17)\;arc\;O_{1}BA,arc\;O_{1}ZM}$and three arches of equal length. ${\displaystyle arc\;O_{1}ZE}$

In addition:

${\displaystyle (18)\;{\overline {AZ}}\parallel {\overline {BM}}}$ and

${\displaystyle (19)\;\angle {BMC}=\angle {DBC}=\angle {HAC},}$

${\displaystyle (20)\;{\overline {BD}}={\overline {AD}};\;{\overline {CD}}={\overline {HD}};\;{\overline {BC}}={\overline {AH}},}$

This means that the four points and lie on the same circle with the center ${\displaystyle A,B,C}$ ${\displaystyle H}$ ${\displaystyle O_{2}.}$

Because of the similarity of the two triangles

${\displaystyle (21)\;\triangle {ACB}}$and follows: ${\displaystyle \triangle {HBA}}$

${\displaystyle (22)\;{\overline {AC}}\cdot {\overline {AD}}={\overline {CB}}^{2}={\overline {CM}}^{2}}$ (→ fulfills (2)),

as well as the similarity of the two triangles

${\displaystyle (23)\;\triangle {BHC}}$and follows: ${\displaystyle \triangle {CBD}}$

${\displaystyle (24)\;{\overline {HB}}\cdot {\overline {BD}}={\overline {BC}}^{2}.}$

Furthermore:

${\displaystyle (25)\;{\overline {CA}}={\overline {HB}}}$ and ${\displaystyle \angle {BCD}=\angle {CHB}=2\cdot \angle {BMC},}$

${\displaystyle (26)\;\angle {CHD}=\angle {DAB}=2\cdot \angle {BMC},}$ consequently:

${\displaystyle (27)\;arc\;O_{1}MB=2\cdot arc\;O_{1}BA,}$ because of

${\displaystyle (28)\;\angle {ABD}=\angle {DAB},}$ is also

${\displaystyle (29)\;arc\;O_{1}AE=2\cdot arc\;O_{1}BA,}$

so everyone is one of the arches

${\displaystyle (30)\;arc\;O_{1}MB}$ and ${\displaystyle \;arc\;O_{1}AE=2\cdot O_{1}BA.}$

Thus the circle is divided into seven equally long parts, ${\displaystyle MBAEZ}$ ${\displaystyle q.\;e.\;d..}$

Further construction for a given circumference or a given side length

Given radius

Starting from the construction of Archimedes (Figure 3), is drawn as shown in Figure 5, first from the angle vertex , the half-line with the angular width is followed by the removal of the given radius radius on the half-lines from this result, the center of the circumscribed circle. Now draw around the area around the heptagon you are looking for with the radius . If the area intersects the line in Figure 5, the side length is found. If the circumference does not intersect the line , the half- line is then used to extend it until it intersects the circumference and thus provides the side length . Finally, the side length is plotted five times counter- clockwise and the remaining sides of the heptagon are drawn in. ${\displaystyle A}$ ${\displaystyle g_{1}}$ ${\displaystyle {\tfrac {\mu }{2}}.}$ ${\displaystyle r}$ ${\displaystyle g_{1}}$ ${\displaystyle A;}$ ${\displaystyle O}$ ${\displaystyle O}$ ${\displaystyle r={\overline {AO}}.}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle E,}$ ${\displaystyle s}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle g_{2}}$ ${\displaystyle E}$ ${\displaystyle s}$ ${\displaystyle s}$

Figure 5: Heptagon given a given circumference, continuation of Archimedes' construction (Figure 3)

Fig. 6: Heptagon with a given side length , continuation of Archimedes' construction (Fig. 3) ${\displaystyle s}$

Side length given

Based on Archimedes' construction (Fig. 3), as shown in Fig. 6, first draw the half-line with the angular width from the apex of the angle. Now the given side length is to be plotted. If the side length is as shown in Figure 6, it can be removed directly. Otherwise, the line is lengthened beforehand using the half-straight line . After drawing a circular arc to the radius to the distance in is cut, a line is drawn from by the half-line ; it results because of the center of the heptagon sought. Finally, the perimeter is drawn around with the radius , the length of the sides is removed five times counterclockwise and the sides of the heptagon that are still missing are drawn in. ${\displaystyle A}$ ${\displaystyle g_{1}}$ ${\displaystyle {\tfrac {\mu }{2}}.}$ ${\displaystyle s}$ ${\displaystyle s={\overline {AE}}\leq {\overline {AB}},}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle {\overline {AB}}}$ ${\displaystyle g_{2}}$ ${\displaystyle A}$ ${\displaystyle {\overline {AE}}}$ ${\displaystyle {\overline {AM}}}$ ${\displaystyle F}$ ${\displaystyle E}$ ${\displaystyle F}$ ${\displaystyle g_{1}}$ ${\displaystyle \triangle {AFE}\sim \triangle {AOE},}$ ${\displaystyle O}$ ${\displaystyle O}$ ${\displaystyle r={\overline {AO}}}$ ${\displaystyle s}$

Individual evidence

1. ↑ ^{a b} Carl Schoy: The trigonometric teachings of the Persian astronomer Abu'l-Raiḥân Muḥ. Ibn Aḥmad al-Bîrûnî based on Al-qânûn al-masʻûdî , Hanover, Orient-Buchhandlung Heinz Lafaire, 1927, p. 712 (section: About the construction of the side of the regular heptagon inscribed in the circle ), digitized (PDF) on Jan P Hogendijk, University of Utrecht, Department of Mathematics; accessed on October 14, 2019.
2. ↑ H.-W. Alten, A. Djafari Naini, B. Eick, M. Folkerts, H. Schlosser, K.-H. Schlote, H. Wesemüller-Kock, H. Wußing: 4000 years of algebra, history – cultures – people, p. 85; Springer-Verlag Berlin Heidelberg 2003, 2014; Online copy (Google) , ISBN 978-3-642-38238-3 ; accessed on October 14, 2019.
3. ^ JL Berggren: Mathematics in Medieval Islam. (PDF) §4 Abu Sahl on the regular heptagon. Spektrum.de, 2011, p. 85 , accessed on October 14, 2019 . 
4. ↑ H.-W. Alten, A. Djafari Naini, B. Eick, M. Folkerts, H. Schlosser, K.-H. Schlote, H. Wesemüller-Kock, H. Wußing: 4000 years of algebra, history – cultures – people, p. 86; Springer-Verlag Berlin Heidelberg 2003, 2014; Online copy (Google) , ISBN 978-3-642-38238-3 ; accessed on October 14, 2019.
5. ↑ ^{a b c} H.-W. Alten, A. Djafari Naini, B. Eick, M. Folkerts, H. Schlosser, K.-H. Schlote, H. Wesemüller-Kock, H. Wußing: 4000 years of algebra, history-cultures-people, p. 87 Springer-Verlag Berlin Heidelberg 2003, 2014; Online copy (Google) , ISBN 978-3-642-38238-3 ; accessed on October 14, 2019.
6. ^ Carl Schoy: The trigonometric teachings of the Persian astronomer Abu'l-Raiḥân Muḥ. Ibn Aḥmad al-Bîrûnî based on Al-qânûn al-masʻûdî , Hanover, Orient-Buchhandlung Heinz Lafaire, 1927, p. 722 ff. (Section: About the construction of the side of the regular heptagon inscribed in the circle ), digitized (PDF) Jan P. Hogendijk, University of Utrecht, Department of Mathematics; accessed on October 14, 2019.
7. ↑ 3 Zeroing the function graph. Wolfram Alpha, accessed July 13, 2020 . 
8. ↑ ^{a b} Carl Schoy: The trigonometric teachings of the Persian astronomer Abu'l-Raiḥân Muḥ. Ibn Aḥmad al-Bîrûnî based on Al-qânûn al-masʻûdî , Hanover, Orient-Buchhandlung Heinz Lafaire, 1927, p. 721 (section: About the construction of the side of the regular heptagon inscribed in the circle ), digitized (PDF) on Jan P Hogendijk, University of Utrecht, Department of Mathematics; accessed on October 14, 2019.

Remarks

1. ↑ In the literature one often finds 836 AD as the year of birth
2. ↑ If so, the length of the line is equal to the root of the so-called silver constant. The root or , according to Wolfram Alpha, finally leads to the function. The third zero of the graph of this function supplies the length of the line ${\displaystyle a=1,}$ ${\displaystyle {\overline {AM}}=1,801937735\ldots }$ ${\displaystyle S.}$ ${\displaystyle S}$ ${\displaystyle {\sqrt {S}}={\sqrt {2+2cos\left({\tfrac {2\pi }{7}}\right)}}}$ ${\displaystyle f(x)=x^{3}-x^{2}-2x+1.}$ ${\displaystyle \left(1{,}801937735\ldots \mid 0\right)}$ ${\displaystyle {\overline {AM}}.}$