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 inequality shows that the length of any side must be less than the sum of the lengths of the other sides. 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 the above condition. We state this as Theorem A If the positive numbers l(1),..,l(n) satisfy B(l(1),..,l(n)) > 0, then there is a hyperbolic polygon whose sides have lengths l(1),..l(n). 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 conditions on the polygon! As in euclidean geometry, we say that a polygon P is convex if, for A,B in or on P, the segment AB lies in the interior of P. We now try to prove Theorem B If the positive numbers l(1),..,l(n) satisfy B(l(1),..,l(n)) > 0, then there is a convex hyperbolic polygon whose sides have lengths l(1),..l(n). Again, this is tricky, even in euclidean geometry. One solution in the euclidean case is to go still further and ask for a cyclic polygon. In the hyperbolic case, it is too much to ask for a cyclic polygon, i.e. for a polygon whose vertices lie on a hyperbolic circle. In fact, this is too much even for n = 3. Let us say that a hyperbolic polygon is i-linear if its vertices lie on an i-line which is not orthogonal to the boundary i.e. which does not define a hyperbolic line. We shall see that such a polygon is convex. We end up by trying the yet harder Theorem C If the positive numbers l(1),..,l(n) satisfy B(l(1),..,l(n)) > 0, then there is an i-linear hyperbolic polygon whose sides have lengths l(1),..l(n). 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. Some of the steps along the road involve objects and results which are interesting in themselves. To guide the reader through the main proof, and to point out the prettier byways, we provide an overview of the section.

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