# Locus of equal area hyperbolic triangles

Henry Segerman and I recently considered the following question:

Given a fixed area $A < \pi$ and two fixed points in the upper half-plane model for hyperbolic $2$-space, what is the locus of points which give rise to a hyperbolic triangle of the given area?

We found it a fun exercise in hyperbolic geometry to show that the answer is a Euclidean straight line, or an arc of a Euclidean circle. As this requires only elementary properties of hyperbolic geometry, we strongly suspect it should be known, but have thus far been unable to find a reference for it. Does anyone know whether it's known, and if so, where one can find it?

• Very nice! I'm now wondering idly what can be said about higher dimensions and spherical geometry...
– j.c.
Commented May 12, 2010 at 2:24
• @Will: They seem to be equidistant curves only. A related question would be: Does the geodesic with the same endpoints have any significance? Commented May 12, 2010 at 17:27
• @Will: The endpoints of the locus are actually distinct from the geodesic through the two fixed points, and hence define a separate geodesic (which necessarily does not intersect the first, even on the boundary). If, say, the two fixed points lie on a vertical line, we get one "banana" curve to the right of the line, and its reflection on the other side, but these do not share endpoints. Commented May 12, 2010 at 20:43
• There is one special case, take the two points as $0$ and $\infty$ along the imaginary axis, area $\pi / 2 .$ Then the third point has real and imaginary parts equal. That is, if both your original points are on the "boundary," in this case the equidistant curve meets them. Specific value of the area should not matter, the "angles" at $0$ and $\infty$ are $0,$ so we are fixing the third angle, a diffeent third angle giving a different slope. Commented May 13, 2010 at 0:01
• Yes, and in fact the same is true for any area: once you have one vertex $v$ giving the right area, with $0$ and $\infty$, the locus is the straight line through $0$ and $v$. The proof we constructed actually used this as a warm-up case, and the proof when all vertices are non-ideal is built from this. Commented May 13, 2010 at 0:11