We say that the ordered list (A(1),..,A(n)) of distinct points of the disk defines a
hyperbolic polygon if the hyperbolic segments A(1)A(2),A(2)A(3),..,A(n-1)A(n),
A(n)A(0) meet only in the points A(i). When the condition is satisfied, we refer to
the polygon as the polygon A(1)..A(n). The A(i) are its vertices. The segments
are its sides. Clearly, only the cyclic ordering of the vertices is significant, the
list (A(2),..,A(n),A(0)) defines the same polygon. To avoid degeneracy, we shall
assume that the vertices do not all lie on a hyperbolic line. The the polygon has
a non-empty interior and exterior. The latter extends to the disk boundary.
Suppose that we have a hyperbolic polygon A(0)..A(n). The generalized triangle
To express this condition algebraically, we use the function
B(l(1),..,l(n)) = Πi(S - 2l(i)), where S = Σjl(j)
This section began as an attempt to provide a simple proof of the sufficiency of
Even in euclidean geometry, there does not appear to be a simple proof for n > 4.
Having failed in our task, we now make the problem harder by imposing additional
As in euclidean geometry, we say that a polygon P is convex if, for A,B in or on P,
Again, this is tricky, even in euclidean geometry. One solution in the euclidean case
In the hyperbolic case, it is too much to ask for a cyclic polygon, i.e. for a polygon
Let us say that a hyperbolic polygon is i-linear if its vertices lie on an i-line which is
We shall see that such a polygon is convex. We end up by trying the yet harder
Now we do succeed. Theorems A and B follow as special cases.
The proof takes us on quite a long journey through hyperbolic and neutral geometry.
main hyperbolic geometry page