hyperbolic geometry

the sufficiency condition

We will suppose that ΔABC is such that F = F(cosh(a),cosh(b),cosh(c)) > 0,
and write Δ = Δ(cosh(a),cosh(b),cosh(c)). Then the circumradius equation
is FR² = Δ, and, as F > 0, it has a positive solution R.

If the circumcircle exists, then its radius must satisfy cosh(r) = R.
Now, cosh(x) > 1 for x ≠ 0, so we will have no circumcircle if R ≤ 1.
In fact we prove something more powerful and useful.

size lemma
With ΔABC as above, and R the positive solution of the equation,
we have R = cosh(r), where 2r ≥ max{a,b,c}.
proof
We have R²= Δ²/F. We may as well assume that a ≥ b,c.
2R²-1 - cosh(a) = (2Δ²-(1+cosh(a))F)/F = (2(Δ²-F)+(1-cosh(a))F)/F
By the difference formula Δ²-F =2(cosh(a)-1)(cosh(b)-1)(cosh(c)-1),
so that 2R²-1-cosh(a) = (cosh(a)-1)(4(cosh(b)-1)(cosh(c)-1)-F)/F.
After some tidying up, this gives
2R²-1-cosh(a) = (cosh(a)-1)(cosh(a)+1-cosh(b)-cosh(c))²/F.
Now, cosh(a) > 1, F > 0 and the other factor is a square, so that
2R²-1-cosh(a) ≥ 0.
It follows that 2R² ≥ 1+cosh(a) > 2, so R > 1.
Now we can write R = cosh(r) with r > 0.
Note that cosh(2r) = 2cosh²(r)-1 = 2R²-1, so our inequality gives
cosh(2r) ≥ cosh(a), so 2r ≥ a.
Since we assumed a ≥ b,c, we have the result.
Our object is a sufficiency proof, but we note that the lemma indicates
the existence of exceptional configurations where the computed value
of r is equal to half the length of a side. We will pursue this in another
digression page.

The lemma brings us close to our sufficiency proof.
Assuming that F > 0, we have a solutrion of the circumradius equation
R = cosh(r). Consider the hyperbolic circles K(A,r), K(B,r) and K(C,r),
where K(P,r) denotes the hyperbolic circle with hyperbolic centre P and
hyperbolic radius r. By the size lemma, 2r ≥ a, so K(B,r) and K(C,r) do
intersect at least once. In general they will meet twice, once on each
side of the hyperbolic line AB. Similarly the other pairs meet. Thus we
have up to six points of intersection. Now note that K(P,r) will be the
circumcircle if and only if each of A,B,C is on K(P,r). This means that
P must lie on the three circles K(A,r), K(B,r), K(C,r). We need to show
that one of the intersections of K(A,r) and K(B,r) is actually on K(C,r).
sufficient condition for a hyperbolic circumcircle
If the hyperbolic triangle ΔABC satisfies F(cosh(a),cosh(b),cosh(c)) > 0,
then it has a hyperbolic circumcircle.

proof
The given condition guarantees the existence of a solution R of the
circumcircle equation. The size lemma shows that R = cosh(r), and
the remarks above show that K(A,r) and K(B,r) meet. Suppose that
they meet at P. We have d(P,A) = r = d(P,B). Let d(P,C) = t.
By the basic lemma, t is such that
G(cosh(r),cosh(r),cosh(t),cosh(a),cosh(b),cosh(c)) = 0.
Since the two circles meet in two points (perhaps coincident),
and we have observed that the equation is quadratic in cosh(t),
the roots must correspond to points of the hyperbolic plane.
Now, r arises from the circumradius equation, which is really
G(cosh(r),cosh(r),cosh(r),cosh(a),cosh(b),cosh(c)) = 0,
t = r is one of the solutions of the first equation.
Thus one of the intersections of our circles is also on K(C,r),
as required.

main page

size lemma With ΔABC as above, and R the positive solution of the equation, we have R = cosh(r), where 2r ≥ max{a,b,c}. proof We have R²= Δ²/F. We may as well assume that a ≥ b,c. 2R²-1 - cosh(a) = (2Δ²-(1+cosh(a))F)/F = (2(Δ²-F)+(1-cosh(a))F)/F By the difference formula Δ²-F =2(cosh(a)-1)(cosh(b)-1)(cosh(c)-1), so that 2R²-1-cosh(a) = (cosh(a)-1)(4(cosh(b)-1)(cosh(c)-1)-F)/F. After some tidying up, this gives 2R²-1-cosh(a) = (cosh(a)-1)(cosh(a)+1-cosh(b)-cosh(c))²/F. Now, cosh(a) > 1, F > 0 and the other factor is a square, so that 2R²-1-cosh(a) ≥ 0. It follows that 2R² ≥ 1+cosh(a) > 2, so R > 1. Now we can write R = cosh(r) with r > 0. Note that cosh(2r) = 2cosh²(r)-1 = 2R²-1, so our inequality gives cosh(2r) ≥ cosh(a), so 2r ≥ a. Since we assumed a ≥ b,c, we have the result. Our object is a sufficiency proof, but we note that the lemma indicates the existence of exceptional configurations where the computed value of r is equal to half the length of a side. We will pursue this in another digression page.
The lemma brings us close to our sufficiency proof. Assuming that F > 0, we have a solutrion of the circumradius equation R = cosh(r). Consider the hyperbolic circles K(A,r), K(B,r) and K(C,r), where K(P,r) denotes the hyperbolic circle with hyperbolic centre P and hyperbolic radius r. By the size lemma, 2r ≥ a, so K(B,r) and K(C,r) do intersect at least once. In general they will meet twice, once on each side of the hyperbolic line AB. Similarly the other pairs meet. Thus we have up to six points of intersection. Now note that K(P,r) will be the circumcircle if and only if each of A,B,C is on K(P,r). This means that P must lie on the three circles K(A,r), K(B,r), K(C,r). We need to show that one of the intersections of K(A,r) and K(B,r) is actually on K(C,r). sufficient condition for a hyperbolic circumcircle If the hyperbolic triangle ΔABC satisfies F(cosh(a),cosh(b),cosh(c)) > 0, then it has a hyperbolic circumcircle. proof The given condition guarantees the existence of a solution R of the circumcircle equation. The size lemma shows that R = cosh(r), and the remarks above show that K(A,r) and K(B,r) meet. Suppose that they meet at P. We have d(P,A) = r = d(P,B). Let d(P,C) = t. By the basic lemma, t is such that G(cosh(r),cosh(r),cosh(t),cosh(a),cosh(b),cosh(c)) = 0. Since the two circles meet in two points (perhaps coincident), and we have observed that the equation is quadratic in cosh(t), the roots must correspond to points of the hyperbolic plane. Now, r arises from the circumradius equation, which is really G(cosh(r),cosh(r),cosh(r),cosh(a),cosh(b),cosh(c)) = 0, t = r is one of the solutions of the first equation. Thus one of the intersections of our circles is also on K(C,r), as required.