Timeline for how to construct a spherical dodecahedron?
Current License: CC BY-SA 2.5
15 events
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| May 24, 2010 at 6:48 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 24, 2010 at 3:45 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 23, 2010 at 16:31 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 23, 2010 at 15:59 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 22, 2010 at 1:42 | comment | added | Will Jagy | Sam, I give the trick after your comment to my answer. If, on the sphere, there is a problem with distances becoming too large, one can simply bisect an overlong $r$ several times and "transport" the length $r/2^k.$ | |
| May 21, 2010 at 22:28 | comment | added | Sam Nead | Is the compass with memory allowed? Or is there a clever trick to duplicate lengths on the sphere? | |
| May 21, 2010 at 21:53 | comment | added | S. Carnahan♦ | Another way to think of it: Each edge of a cube embedded in a dodecahedron spans two nonadjacent vertices in a pentagonal face. Each face of the cube gets two additional vertices that are separated from suitable corners by the side length. | |
| May 21, 2010 at 18:55 | comment | added | Pietro Majer | @David: from the cube to the dodecahedron: the vertices of the cube are 8 out of 20 of the dodecahedron; to find the other 12: let the vertices of a face of the cube be A B C D (cyclically ordered). Consider the intersection of the circle with center A passing through B, and the circle with center D passing through C. You'll get two new vertices of the dodecahedron, and then you'll get all of them. | |
| May 21, 2010 at 18:49 | comment | added | erdos | Pietro, for constructing the great circles we can use the spherical ruler. For me it seems that the great circle acts as a line on a spherical surface. However, the remembering pair of compasses is allowed. | |
| May 21, 2010 at 18:38 | comment | added | Pietro Majer | thanks to you for this nice question. To be precise, I see clearly how to draw a dodecahedron out of a cube, and how to draw a cube and a tetrahedron starting from three mutually orthogonal great circles (that is, an octahedron). But making three orthogonal great circles seems to require a compass "with memory", that is, able to make circles with the same radii when pointed at different points. Is this allowed? | |
| May 21, 2010 at 18:06 | comment | added | David E Speyer | (1) Draw any great circle (using your spherical ruler). (2) Construct a perpendicular to that circle, as in the Euclidean case. You now have two antipodal points, joined by 4 half-great-circles. (3) Bisect the half-great-circles, as in the Euclidean case. (4) Draw the great circle through the midpoints of the half-great-circles. At the point, you have an octahedron. (5) For each of the 8 faces of the octahedron, find the centroid (as in the Euclidean case). These 8 points are the vertices of a cube. | |
| May 21, 2010 at 18:03 | comment | added | David E Speyer | I don't follow how you are going from the cube to the dodecahedron. However, it is easy enough to build a cube: | |
| May 21, 2010 at 17:47 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 21, 2010 at 17:35 | history | edited | Pietro Majer | CC BY-SA 2.5 |
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| May 21, 2010 at 17:25 | history | answered | Pietro Majer | CC BY-SA 2.5 |