Angle in the Bernoullian lemniscate

Angle on Bernoulli's lemniscate

An angular relationship in the Bernoullian lemniscate goes back to the mathematician Gerhard Christoph Hermann Vechtmann , which , according to the Italian mathematician Gino Loria , is to be regarded as very remarkable . Vechtmann presented this in his dissertation in 1843.

Representation of the angular relationship

It can be specified as follows:

A Bernoullian lemniscate with the two defining focal points and and the center is given in the Euclidean plane . ${\displaystyle {\mathfrak {L}}}$ ${\displaystyle F_{1}}$ ${\displaystyle F_{2}}$ ${\displaystyle O}$

A point is given that is not on the connecting line through and through . ${\displaystyle P\in {\mathfrak {L}}}$ ${\displaystyle g}$ ${\displaystyle F_{1}}$ ${\displaystyle F_{2}}$

The normal to in the point intersects in the point . ${\displaystyle n_{P}}$ ${\displaystyle {\mathfrak {L}}}$ ${\displaystyle P}$ ${\displaystyle g}$ ${\displaystyle R}$

Then:

The point in the triangle adjacent exterior angle is three times the size of the center located inside angle . ${\displaystyle R}$ ${\displaystyle POR}$ ${\displaystyle O}$

Remarks

Fig. 1: Construction of the tangent through a point Connect the point with and determine with the help of the distance at the vertex The normal to the lemniscate cuts into The final tangent is a vertical . ${\displaystyle t}$ ${\displaystyle P}$
${\displaystyle P}$ ${\displaystyle O}$ ${\displaystyle \angle {\;2\alpha }}$ ${\displaystyle |AB|}$ ${\displaystyle P.}$ ${\displaystyle n}$ ${\displaystyle {\mathfrak {L}}}$ ${\displaystyle {\overline {OF_{2}}}}$ ${\displaystyle R.}$ ${\displaystyle t}$ ${\displaystyle n}$

Fig. 2: Bernoulli's lemniscate
As the example shows, for a given angular width, the angular width at the vertex is unequal . ${\displaystyle 3\alpha }$ ${\displaystyle P}$ ${\displaystyle 2\alpha }$

• According to the set of external angles, the aforementioned angular relationship is synonymous with the fact that the associated internal angle at the point is twice as large as said central angle . ${\displaystyle P}$

• According to Gino Loria, the angle relation is remarkable (Fig. 1), as it not only provides an easy construction method for the normal in any point of the lemniscate (and therefore also for the tangent ), but also proves that the problem of the tripartite division of the angle The main thing is identical to drawing a normal or a tangent of a given direction to a lemniscate .

• Even if at first it seems that Bernoulli's lemniscate is suitable for the three-part division of any angle, this is not the case (Fig. 2). For a given angular width , the angular width at the apex is unequal and thus the direction of the angle leg is also determined. This means you would a perpendicular to the angle leg through build, would this the lemniscate twice cut ; once in and once z. B. in one point . As already described in the previous paragraph, there is no constructive possibility of a lemniscate "... to draw a tangent from a given direction." The vertex with the angular width cannot therefore be represented at a given angular width . ${\displaystyle 3\alpha }$ ${\displaystyle P}$ ${\displaystyle 2\alpha }$ ${\displaystyle {\overline {PC}}}$ ${\displaystyle {\overline {PC}}}$ ${\displaystyle P}$ ${\displaystyle {\mathfrak {L}}}$ ${\displaystyle P}$ ${\displaystyle P_{2}}$ ${\displaystyle P}$ ${\displaystyle 2\alpha }$ ${\displaystyle 3\alpha \neq 90^{\circ }}$

Proof according to Loria

The proof given by Loria is essentially based on the two equations of the lemniscate and on the addition theorems for multiple angles of sine and cosine and goes as follows:

The normal form of the lemniscate is assumed to be given, in which the straight line coincides with the abscissa axis and the center with the coordinate origin . ${\displaystyle g}$

The defining equation of in Cartesian coordinates can then be written as ${\displaystyle {\mathfrak {L}}}$

(I) ${\displaystyle (x^{2}+y^{2})^{2}-2a^{2}(x^{2}-y^{2})=0}$

and those in polar coordinates in the form

(II) ${\displaystyle r^{2}=2a^{2}\cos(2\alpha )}$

with as the polar angle and as the distance to the coordinate origin. ${\displaystyle \alpha }$ ${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$

For reasons of symmetry , it is sufficient to show the theorem for that part of the lemniscate which is in the first quadrant , i.e. for and , and it is still sufficient to prove the asserted equation only for  , i.e. excluding the high point there at where the tangent to the lemniscate runs parallel and the associated normal perpendicular to the abscissa axis. Because for this exceptional case the equation follows for reasons of continuity . ${\displaystyle x\geq y>0}$ ${\displaystyle r>0\;,\;0<\alpha <{\frac {\pi }{4}}}$ ${\displaystyle \alpha \neq {\frac {\pi }{6}}}$

It is now said outer angle with . ${\displaystyle \psi }$

By taking into account that, on the one hand, the center angle and the polar angle of the point in the first quadrant coincide in the representation in polar coordinates and, on the other hand, the real tangent function in the dotted interval is injective , one sees that only the equation ${\displaystyle ]0\;,\;\pi [\setminus \{{\frac {\pi }{2}}\}}$

${\displaystyle \tan(\psi )=\tan(3\alpha )}$

is to be shown.

The proof of this equation now takes place in several calculation steps:

First one obtains by means of implicit differentiation from (I)

${\displaystyle (x^{2}+y^{2})(x+yy^{'})-a^{2}(x-yy^{'})=0}$

and it

${\displaystyle y^{'}={\frac {a^{2}-(x^{2}+y^{2})}{a^{2}+(x^{2}+y^{2})}}\cdot {\frac {x}{y}}}$  .

Now is

${\displaystyle \tan(\psi )=-{\frac {1}{y^{'}}}}$

and because of and then the equation results ${\displaystyle r^{2}=x^{2}+y^{2}}$ ${\displaystyle {\frac {y}{x}}={\frac {\sin \alpha }{\cos \alpha }}}$

${\displaystyle \tan(\psi )=-{\frac {a^{2}+r^{2}}{a^{2}-r^{2}}}\cdot {\frac {\sin \alpha }{\cos \alpha }}}$  .

and because of (II) further

${\displaystyle \tan(\psi )=-{\frac {a^{2}+2a^{2}\cos(2\alpha )}{a^{2}-2a^{2}\cos(2\alpha )}}\cdot {\frac {\sin \alpha }{\cos \alpha }}=-{\frac {1+2\cos(2\alpha )}{1-2\cos(2\alpha )}}\cdot {\frac {\sin \alpha }{\cos \alpha }}={\frac {1+2\cos(2\alpha )}{2\cos(2\alpha )-1}}\cdot {\frac {\sin \alpha }{\cos \alpha }}}$  .

Since you have at the same time , follow on ${\displaystyle \cos(2\alpha )=1-2{\sin }^{2}(\alpha )=2{\cos }^{2}(\alpha )-1}$

${\displaystyle \tan(\psi )={\frac {3-4{\sin }^{2}(\alpha )}{4{\cos }^{2}(\alpha )-3}}\cdot {\frac {\sin \alpha }{\cos \alpha }}={\frac {3\sin \alpha -4{\sin }^{3}(\alpha )}{4{\cos }^{3}(\alpha )-3\cos \alpha }}}$  .

Finally, then because of the multiple angle equations mentioned

${\displaystyle \tan(\psi )={\frac {\sin(3\alpha )}{\cos(3\alpha )}}=\tan(3\alpha )}$

and everything is shown.

literature

• GM Fichtenholz : Differential and integral calculus I (=  university books for mathematics . Volume 61 ). 3rd, unchanged edition. VEB Deutscher Verlag der Wissenschaften, Berlin 1968, p. 484-485 ( MR0238635 ).
• Alexander Ostermann, Gerhard Wanner : Geometry by Its History (=  Undergraduate Texts in Mathematics. Readings in Mathematics ). Springer Verlag, Heidelberg / New York / Dordrecht / London 2012, ISBN 978-3-642-29162-3 , pp. 207-208 , doi : 10.1007 / 978-3-642-29163-0 ( MR2918594 Google books.google.de ).
• Gino Loria : Special algebraic and transcendent plane curves: theory and history . First volume: The algebraic curves (=  BG Teubner's collection of textbooks in the field of the mathematical sciences including their applications . V, 1). 2nd Edition. BG Teubner Verlag, Leipzig / Berlin 1910 ( 1902 edition on archive.org ).
• GCH Vechtmann : Diss. Inaug. phil. de curvis lemniscatis . Göttingen 1843 ( books.google.de ).

References and footnotes

1. ↑ ^{a b c} Gino Loria: Special algebraic and transcendent plane curves. Theory and history. Leipzig, printing and publishing house BG Teubner 1902. ( p. 202, on archive.org )
2. ^ ^{A b c} Alexander Ostermann, Gerhard Wanner: Geometry by Its History. Springer, 2012, pp. 207-208