the arcs theorem
Suppose that K is a horocycle or hypercircle, and that A,B are points on K.
Then, for any C on K other than A and B,
(1) E^{2}(A,B,C) is independent of C, and
(2) E(A,B,C) < 0 if and only if C lies on the finite arc AB.
proof
(1) We have 1  E^{2}(A,B,C) = J(A,B,C)s^{2}(AB).
For A,B,C lie on K, J(A,B,C) = j(K), i.e. is fixed.
It follows that, if we fix A and B, then E^{2}(A,B,C) is constant.
Since we have a horocycle or hypercycle, j(K) ≤ 0, so E^{2}(A,B,C) ≥ 1.
Indeed, if K is a horocycle, then j(K) = 0, so E^{2}(A,B,C)=1.
(2) If we vary C along a particular arc, then J(A,B,C) is continuous.
But its values are the E(A,B,C) where E^{2}(A,B,C) is fixed, and nonzero.
It follows that E(A,B,C) is constant along each arc.
For an infinite arc, we can choose C so that d(A,C) is larger than d(A,B).
Then s(AC) > s(AB). It follows that E(A,B,C) < 0 for this C, and hence
for all C on the arc.
For a finite arc, we use the symmetry lemma to pick the point C where
the hyperbolic bisector of AB cuts K. Since we have a horocycle or a
hypercycle, we have H(s(AB),s(BC),s(CA)) ≤ 0. It follows that one of
the arguments is at least as large as the sum of the others. For this C,
s(AC) = s(BC), so we must have s(AB) ≥ s(AC)+s(BC). Then we have
s^{2}(AB) ≥ (s(AC)+s(BC))^{2} > s^{2}(AC)+s^{2}(BC), so E(A,B,C) < 0. By the
continuity argument, this holds for all C on the finite arc AB.
Note We have not established the continuity of J(A,B,C) with respect to C.
The formula is quite complicated, and certainly has poles when C = A or B.
Rather than get involved with these matters, we can give a proof which is
purely geometric.
Alternative proof of (2)
Let H be the hyperbolic circle on AB as diameter. Then H cuts K at A and B,
and hence, as ilines, nowhere else. Thus, H divides K into its three arcs.
As H cannot contain points arbitrarily close to the boundary C, the finite arc
of K must be that lying inside H.
Thus, if C is on th e finite arc, C is inside H.
By the hyperbolic semicircle theorem, s^{2}(AC)+s^{2}(BC) < s^{2}(AB).
Then E(A,B,C) < 0, as required.
If C lies on an infinite arc defined by A,B, then
either A lies on the finite arc BC, so E(B,C,A) < 0,
or B lies on the finite arc AC, so E(A,C,B) < 0.
By the definition, at most one of E(A,B,C), E(B,C,A), E(A,C,B) < 0.
It follows that, for C on an infinite arc, E(A,B,C) > 0.
Since E^{2}(A,B,C) is constant, it follows that E(A,B,C) is constant
on each arc.
