Timeline for Locus of equal area hyperbolic triangles
Current License: CC BY-SA 2.5
8 events
when toggle format | what | by | license | comment | |
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May 13, 2010 at 0:11 | comment | added | Grant Lakeland | 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. | |
May 13, 2010 at 0:01 | comment | added | Will Jagy | 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. | |
May 12, 2010 at 20:43 | comment | added | Grant Lakeland | @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. | |
May 12, 2010 at 17:27 | comment | added | Grant Lakeland | @Will: They seem to be equidistant curves only. A related question would be: Does the geodesic with the same endpoints have any significance? | |
May 12, 2010 at 17:26 | vote | accept | Grant Lakeland | ||
May 12, 2010 at 12:12 | answer | added | Sam Nead | timeline score: 9 | |
May 12, 2010 at 2:24 | comment | added | j.c. | Very nice! I'm now wondering idly what can be said about higher dimensions and spherical geometry... | |
May 11, 2010 at 23:12 | history | asked | Grant Lakeland | CC BY-SA 2.5 |