In experiments with circles, we looked at the possibility of drawing a hyperbolic circle through three points A,B,C in the disk. The CabriJava applet should have indicated that this is not always possible. This is easy to prove. Recall that a hyperbolic circle is alsoa euclidean circle lying wholly within the disk. If A,B and C lie on a euclidean line, thenno such circle exists. Equally, if A,B and C lie on a hyperbolic line, then this is either part of a euclidean line, or part of a euclidean circle K orthogonal to the boundary . In theCformer case, there is no circle, as before. In the latter case, K is the only circle throughthe points, and it cuts twice. Thus we do have cases with no hyperbolic circumcircle.
C
Here we identify the region
It is difficult to identify the region within hyperbolic geometry. This is probably innevitable
Suppose that A and B are distinct points in Choose a circle K with euclidean centre A, and denote inverses with respect by dashes.We observe that any line or circle through A and B inverts to a line through B'. We alsoobserve that inverts to c circle C, centre Z'. C' does not contain B' since B is inside C'.
CThe hyperbolic line AB is an arc orthogonal to , it maps to a line through B', orthogonal Cto . This must be the line B'Z'. This is illustrated
in the figures below.
C'
Now concentrate on the right hand figure. There are two lines though B' which touch Let the points of contact be X' and Y'. A line through B' which does not meet must lieC'entirely in the yellow shaded regions. A little thought shows that these are the images of the yellow regions in the left hand figure. Thus, the points C for which ΔABC does have a hyperbolic circumcircle are precisely those in these regions.
The boundaries are arcs of the euclidean circles through A and B which touch the inverse images of X' and Y'. In fact, we can describe X and Y in hyperbolic terms.
Let L through X and Y, orthogonal to and to the hyperbolic line AB. The first shows that the arc XY is a hyperbolic line. InversionCin this line maps each of the green circles to themselves - look at the corresponding map in the right hand figure. The inversion therefore interchanges the intersections, A and B. It follows that the hyperbolic line XY is the hyperbolic bisector of the hyperbolic segment AB. Thus, we could dispense with the right hand firure. The boundaries we require are arcs of the euclidean circles through A,B,X, and through A,B,Y, where X and Y are the boundary points of the hyperbolic bisector of AB. Indeed, they are the arcs which touch the boundary. This leads us to define to a new concept:
Note that, if figure in hyperbolicgeometry. However, since M neither lies within nor is orthogonal to it, the figure is notDa hyperbolic line or circle.
each meets in a boundary point for the hyperbolic bisector of AB.
C
The first is easy - the image of This means that horocycles are hyperbolic objects. The proof of the second is contained in the above discussion. With this language, we can now state the result we obtained above as the
There is a CabriJava applet which illustrates the result.
We can also tackle the problem algebraically, characterizing the triangles which do A somewhat more detailed approach, with additional results appears in circumcircles |