Poincaré Kreisscheibe
Author: Prof. Dr. Dörte Haftendorn
Date: 2012+26 Juni 2013

Bilder

Ziehe E bis die kleinen Kreise zusammenfallen

Hyperbolische Ebene

Finde die Gerade durch zwei Punkte

Lösung

Hinführung

Viereck

Ha 2012

Werkzeug hypger[pkt,pkt,c]
verwenden

A = Point(xAxis)
B = Point(xAxis)
c = Circle(A, B)
C = (-1.62, 0.96)
D = (-1.16, -0.56)
a = LineBisector(C, D)
E = Point(a)
b, d = Tangent(E, c)
e = Polar(E, c)
F = Intersect(e, b)
G = Intersect(e, d)
f = Circle(E, F)
g = Circle(E, D)
C' = Mirror(C, c)
D' = Mirror(D, c)
h = false
i = LineBisector(D, D')
H = Intersect(a, i)
k = Circle(H, D)
j = false
I, J = Intersect(c, k)
p = CircumcircleArc(I, D, J)
l = false
K = (0.6, -0.32)
L = (-0.64, 1.36)
q, M, N, O = hypger(K, L, c)
r, P, Q, R = hypger(D, K, c)
s, S, T, U = hypger(C, D, c)
t, V, W, Z = hypger(C, L, c)
m = false
n = LineBisector(C', C)
C_1 = Point(yAxis)

Tool: hypger
Punkt, Punkt, Hyp-Kreis  ergibt Geradenbogen
Input: C, D, c
Output: p, J, I, H

a = LineBisector(C, D)
D' = Mirror(D, c)
i = LineBisector(D, D')
H = Intersect(a, i)
k = Circle(H, D)
I, J = Intersect(c, k)
p = CircumcircleArc(I, D, J)