Trigonometry of an hyperbolic Omega Triangle

Question

I was puzzeling with the trigonometry of an Omega Triangle ( a triangle / biangle in hyperbolic geometry where one of the vertices is an Ideal point see https://en.wikipedia.org/wiki/Hyperbolic_triangle#Ideal_vertices )

And using the Poincare disk model ( https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model )

Translated the Omega triangle to the following problem in Euclidean geometry:

(the angle $\angle AOB $ in one of the angles of the Omega triangle the other angle is $\angle CAO - 90 ^{\circ} $ )

Given:

• the unit circle (circle centered around $ (0 . 0)$ and radius 1 )

• a point $A = (a,0) $ with $ 0 < a < 1 $

• a point $ B$ on the arc of the unit circle with $0^\circ < \angle AOB < 90^\circ $

• A circle $c$ with centre $C$ such that $c$ is orthogonal to the unit circle and passes trough $A$ and $B$

Question:

• What is $ \angle CAO$ as function of $ \angle AOB $ and $ a$ ?

Answer 1

the cosine of angle $CAO=$

$$\frac{\left(a^2-1\right)\sin \theta }{1+a^2-2a \cos \theta }$$

where $\theta$ is angle $AOB$

Outline of a proof usin trig

Let $OB=1, OA=a, AB=b, BC=CA=c$

angles in degrees: $CAO=x, AOB=\theta, OBA=\beta,CAB=ABC=90-\beta$

Law of Sines: $\frac{\sin \beta }{a}=\frac{\sin \theta }{b}$

Calculating $\text{OC}^2$ in two different ways we get

$$a^2+c^2-2a c \cos x=c^2+1$$

and therefore $2a c \cos x=a^2-1$

On the other hand

$$\frac{a \sin \theta }{b}=\sin \beta =\cos (90-\beta )=\frac{b}{2c}$$

$$2a c \sin \theta =b^2$$

$$\cos x = \frac{\left(a^2-1\right)\sin \theta }{b^2}=\frac{\left(a^2-1\right)\sin \theta }{1+a^2-2a \cos \theta }$$

Some interesting special cases: $$a=\cos \theta , x=\theta +\pi /2$$ $$a=2 \cos \theta , x = 3\theta -\pi /2$$ $$\theta =\pi /2, x=\arccos \left(\frac{a^2-1}{a^2+1}\right)$$

another special case

$$\cos \theta = \frac{2a}{a^2+1},x=\pi $$

Answer 2

HINT.-Here an answer put in equations (perhaps there may be another easier one). Do not solve the system nor calculate the angle between the straight lines $CA$ and $OA$ (I leave this as an exercise for the OP).

In order to know angle $\angle {CAO}$ one can find out the two points $B=(x_1,y_1)$ and $C=(x_2,y_2)$ so there are four unknowns, therefore one needs four (independent) equations.

First equation.- $B$ is in the unit circle: $$x_1^2+y_1^2=1\qquad (1)$$ Second equation.-$C$ is in the line segment bisector of $\overline{AB}$. The midpoint of $\overline{AB}$ being $M=(\frac {x_1+a}{2},\frac {y_1}{2})$ and the slope of $\overline{AB}$ being $\frac{y_1}{x_1-a}$ we have $$\frac{2y_2-y_1}{2x_2-x_1}=\frac{-x_1+a}{y_1}\iff 2y_1y_2+2(x_1-a)x_2+ax_1=1\qquad (2)$$ Third equation.- $|\overline{CB}|=|\overline{CA}|$ (radius of the circle $c$). $$(x_2-x_1)^2+(y_2-y_1)^2= (x_2-a)^2+y_2^2\iff 2(a-x_1)x_2-2y_1y_2+1=-2ax_2+a^2\qquad (3)$$ Fourth equation.-Line $CB$ is tangent to the unit circle. $$\frac{y_1-y_2}{x_1-x_2}=\frac{-x_1}{y_1}\iff x_1x_2+y_1y_2=1\qquad (4)$$