Timeline for how to construct a spherical dodecahedron?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| May 23, 2010 at 21:01 | comment | added | Sam Nead | (At least, I think that is correct!) | |
| May 23, 2010 at 21:00 | comment | added | Sam Nead | @Zsban - I don't think so. In the Euclidean plan, with a memoryless compass, you can produce a "triangular tiling" as follows: pick any pair of points a,b and produce all of radius d(a,b) that result. So we have the circle with center a going through b, the circle with center b going through a, and so on. In the sphere and in the hyperbolic plane you only get tilings for a measure zero set of choices for a and b. Additionally, the way in which the tiling fails records the curvature. | |
| May 23, 2010 at 20:00 | comment | added | Zsbán Ambrus | I wonder, if you only have a memoryless compass, but not the straightedge, doesn't the hyperbolic plane and the sphere look the same? | |
| May 22, 2010 at 5:06 | comment | added | Will Jagy | to continue, this is the same trick as in the regular plane or the non-Euclidean plane | |
| May 22, 2010 at 1:25 | comment | added | Will Jagy | For "collapsible" or "memoryless" compass: Given a center $A$ and a circle around it of radius $r,$ to draw the circle around $B$ of same radius. Draw the line $l$ between $A$ and $B,$ keep repeating distance $r$ segments beginning with $A$ until you pass $B,$ at some point $C$ on the same line $l,$ with the point $D$ just before $B.$ Bisect the segment $BC,$ call this point $E.$ With center $E,$ draw the circle with radius $DE,$ arriving at a new point $F,$ where $F$ is on $l$ and on the same side of $B$ as $C.$ But the length of segment $BF$ is $r.$ Anyway, email me if I've misunderstood. | |
| May 21, 2010 at 22:35 | comment | added | Sam Nead | I looked at your paper (but have not yet looked at the references). One point of confusion for me - how does one "duplicate distances" in the spherical or hyperbolic plane? In the euclidean plane I can do this by building a parallelogram. I don't see how to do the duplication in the other geometries? Is there a (clever?) trick I missing? | |
| May 21, 2010 at 20:44 | history | edited | Will Jagy | CC BY-SA 2.5 |
pretty
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| May 21, 2010 at 20:22 | history | edited | Will Jagy | CC BY-SA 2.5 |
marvin article
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| May 21, 2010 at 20:02 | history | answered | Will Jagy | CC BY-SA 2.5 |