This result sums up all of our work on hyperbolic polygons. We shall only
outline the proof
since each step uses earlier results and ideas.
We shall use the following notation :
- B(x(1),..x(n)) = Πi(Σjx(j)-2x(i)),
- s(x) = sinh(½x),
We shall say that H is an (l(1),..,(l(n)) hyperbolic polygon if it has sides
whose lengths are, in order, l(1),..l(n), and no two sides intersect.
The hyperbolic polygon theorem
Suppose that n ≥ 3 and that l(1),..,l(n) are positive real numbers.
- There exists an (l(1),..,l(n)) hyperbolic polygon if and only if B(l(1),..,l(n)) > 0,
- If B(l(1),..,l(n)) > 0, then there is a convex (l(1),..,l(n)) hyperbolic polygon,
- If B(l(1),..,l(n)) > 0, then there is an (l(1),..,l(n)) hyperbolic polygon whose
vertices lie on an i-line K. Moreover,
- K is unique up to hyperbolic conjugacy,
- K is a hyperbolic circle if B(s(l(1)),..,s(l(n))) > 0,
- K is a horocycle if B(s(l(1)),..,s(l(n))) = 0,
- K is a hypercircle if B(s(l(1)),..,s(l(n))) < 0.
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proof
- The necessity is clear from the triangle inequality.
The sufficiency follows at once from part (2).
- This follows from part (3) and remarks about convexity.
- Now earlier results demonstrate the existence of at least one such polygon.
That proof began by dividing the problem into two cases, depending on how large
an angle was subtended at the centre of a hyperbolic circle of diameter max(l(i))
by successive chords of length l(j), j ≠ i.
Indeed, when the angle was greater than π, the proof showed the existence of
a unique hyperbolic circle circumscribing an (l(1),..,l(n)) polygon. In this case,
we know from the mapping result that B(s(l(1),..,s(l(n))) > 0.
In the other case, we need to use our monotonicity theorem whose detailed
proof demonstrates the existence of a unique i-line K (once we map a point to O).
Thus, the i-line is unique up to conjugacy. The proof also shows that we will get
a hyperbolic circle, horocycle or hypercircle according as B(s(l(1)),..,s(l(n))) is
positive, zero or negative.
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