Benutzer:Adam Bernau/Arbeitsseite

degenerated angles

In spherical geometry, Lexell's theorem states that, for any two distinct non-antipodal points A, B ∈ S², the locus of points C ∈ S² along which the spherical triangle with vertices A, B, C has given fixed area, is a circular arc with endpoints at the antipodes of A and B. The theorem is named after Anders Johan Lexell.

History

Lexell tried to find the spherical analog of the first proposition, 6th book of Euclid's Elements

"Triangles and parallelograms which are of the same height are to one another as their bases."

He formulated his thought in the following way: "This proposition came to my mind on the occasion of my research into the curves that unite the vertices of spherical triangles that share the same baseline and have equal area." . The original statement from 1784 reads as follows:

The locus of points along which the spherical triangle △ABC has given fixed area, is part of circle.

Jakob Steiner was discontent with the construction of the circle in the proof of Lexell. 1827 he published and proved the modern version of the theorem. .

Proofs

Two proofs will be presented here. Overall the first one follows Steiner and is close to the basic idea of the construction. The second one maps the problem to the euclidian plane via a Möbiustransformation. Afterwards one can complete the proof using elementary geometry. However, both together should give a deeper understanding of spherical geometry.

First Proof (Steiner)

Proposition: Two spherical triangles △ABC and △ABC´ that share the same baseline AB have a common circumcircle Σ if and only if the interior angles satisfy α + β - γ = α´ + β´ - γ´.

A spherical quadrilateral with a circumcircle

Proof. Let △ABC be a spherical triangle and M be the center of its circumcircle Σ. The triangles △ABM, △BCM and △CAM are isosceles and therefore

$\alpha =\varphi +\eta \qquad \beta =\psi +\varphi \qquad \gamma =\eta +\psi$

where

$\varphi :=\angle BAM\qquad \psi :=\angle CBM\qquad \eta :=\angle ACM$

Expressing 𝜑 in terms of α, β and γ yields

$\varphi =\alpha -\eta =\alpha -\gamma +\psi =\alpha -\gamma +\beta -\varphi ={\frac {\alpha +\beta -\gamma }{2}}$

Note that 𝜑 is invariant under the movement of C on Σ. Thus α + β - γ is a constant.

degenerated angles

Conversely suppose △ABC and △ABC´ satisfy α + β - γ = α´ + β´ - γ´ and the point C´ does not lie on the circumcircle Σ of △ABC. Concider the degenerated cases: Send C on Σ to A and observe the limit of the angles. Then β shrinks to zero and α becomes the enclosed angle between the line tangent to Σ at A and the line segment AB. Since the area of the degenerated triangle is zero we obtain γ = π - α. Equally we can send C´ on the circumscribed circle Σ´ of △ABC´ to A. But by construction

$\Sigma \cap \Sigma '=\{A,B\}$

and therefore

$\alpha \neq \alpha '\qquad \Rightarrow \qquad 2\alpha -\pi \neq 2\alpha '-\pi$

This is in contradiction to the assumtion α + β - γ = α´ + β´ - γ´.

A spherical quadrilateral with a circumcircle

Proof. (of the theorem) Consider the upper hemisphere H and two distinct non-antipodal points A, B ∈ ∂H at the boundary of H. Let C ∈ H be an arbitrary point and A', B' be the antipodal points of A and B. Moreover let α', β´ and γ´ be the interior angles of the triangle △A´B´C. Expressing the interior angles of △ABC in terms of α´, β´ and γ´ yields:

$\alpha =\pi -\alpha '\qquad \beta =\pi -\beta '\qquad \gamma =\gamma '$

The formula for the area of a spherical triangle provides

${\begin{aligned}\mathrm {area} (\triangle ABC)&=\alpha +\beta +\gamma -\pi =\pi -\alpha '+\pi -\beta '+\gamma '-\pi \\&=\pi -(\alpha '+\beta '-\gamma ')\end{aligned}}$

The requirement of fixed area implies that α´ + β´ - γ´ is constant. Applying the proposition completes the proof.

Second Proof

A spherical quadrilateral with a circumcircle

Proof. Given a spherical triangle △ABC and the antipodes A´, B´ of A and B. Furthermore let Σ be the circle defined by the points A´, B´, C and 𝜑 be the enclosed angle between Σ and the line segment B´A. Focus on the degenerated case: Send C on Σ to B´. This process limits in the bigon BB´, where the line BC limits in the great circle L passing through B´, B and intersecting B´A with angle 𝜑. Since 𝜑 is uniquely deteremined by the circle Σ, it suffices to prove that the aera of the bigon equals the area of the triangle, i.e.

$2\varphi =\alpha +\beta +\gamma -\pi$

A spherical quadrilateral with a circumcircle

To see this, we apply an inversion centered at A´ to the configuration. Recall that this map preserves angles. All circles become lines, except L and the great circle passing through the points B, C and B´. From now on only angles are involved. So we can delete L, because 𝜑 is deteremined by Σ. Since B and B´ were antipodal points, the line segment BB´ is the diameter of the circle determined by B´CB. Due to Thales' theorem the enclosed angle between the line tangent to the circle at C and CB is equal to 𝜑. The same holds true for the enclosed angle between the line tangent to the circle at B and BC. Adding all interior angles of the euclidian triangle △ABC yields

$\pi =\alpha +\beta -\varphi +\gamma -\varphi \qquad \Leftrightarrow \qquad 2\varphi =\alpha +\beta +\gamma -\pi$