existence of convex polygons

We are now in a position to show that, with the obvious necessary
condition B(a(1),..,a(n)) > 0, there is a convex polygon with sides
of lengths a(1),..,a(n). In euclidean geometry, there is a unique
convex cyclic polygon. In hyperbolic geometry, we cannot expect
such a theorem in view of result 4 which shows that there will be a
cyclic polygon if and only if R(s(a(1)),..,s(a(n)) > 0. This is a more
stringent condition. However, we will show that there is always at
least one convex cyclic, horocyclic or hypercyclic polygon.

Suppose that a(1),..a(n) is a sequence of positive numbers such
that B(a(1),..,a(n)) > 0. Note that although the order is significant,
we may choose the starting point so that a(n) ≥ a(i) for all i. Note
that the condition for existence may be stated in the simpler form
a(1)+..+a(n-1) > a(n) as all other factors of B are clearly positive.

We draw a vertical segment AB of length a(n). Any circle K which
passes through A and B has its centre C on the perpendicular
bisector of AB. If X lies on such a circle, then for each i, the circle
with centre D and radius a(i) cuts K as a(n) ≥ a(i). In general it
cuts K twice. We choose the intersecion E nearer to D in the
counter-clockwise direction. Starting with A(1) = A, we choose
points A(2) to A(n) in this way, with |A(i)A(i+1)| = a(i). If we are
to obtain a convex polygon, we must have A(n) = B and A(1),..,A(n)
on the left-hand arc AB.

We begin with the circle on AB as diameter. The CabriJava applets
below show two possible figures.

In the first, some of the A(i) lie beyond B. By dragging C to the left,
we arrive at a figure with A(8) = B.

In ther second, all of the A(i) lie on the left arc AB. By dragging C to the
right, we get a suitable polygon.

The experience of these applets suggests that there is exactly one solution.
The formal proof shows that this is the case.

convexity in hyperbolic geometry