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using only a spherical ruler (to construct great lines) and a pair of compasses, how can you construct a regular dodecahedron on the surface of the sphere? thank you very much.

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Sorry, but I am a bit confused by your question: what is one allowed to do with a compass and what is one allowed to do with two compasses (that can't be done with one)? –  Willie Wong May 21 '10 at 20:57
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The traditional description was a "pair of compasses" as in a pair of pants en.wikipedia.org/wiki/Compass_(drafting) –  Will Jagy May 21 '10 at 21:05
    
oh silly me. That's what I get for having learned straight-edge-compass geometry and engineering drafting in a different language. Thanks Will. –  Willie Wong May 22 '10 at 0:51
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You must be very careful about the language your parents teach you. –  Will Jagy May 23 '10 at 18:26
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3 Answers

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the details of the construction are described in various people's comments below. One needs a "spherical ruler" (able to draw the great circle passing for two non-antipodal given points) and a compass (C) (able to draw a circle with given center passing for a given point). One first draws three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.

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I don't follow how you are going from the cube to the dodecahedron. However, it is easy enough to build a cube: –  David Speyer May 21 '10 at 18:03
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(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. –  David Speyer May 21 '10 at 18:06
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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? –  Pietro Majer May 21 '10 at 18:38
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@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. –  Pietro Majer May 21 '10 at 18:55
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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 $r/2^k.$ –  Will Jagy May 22 '10 at 1:42
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I'm not certain if this quite fits the ruler and compasses rules, but perhaps it can be modified: Carolyn A. Yackel has a paper in the proceedings of the Bridges 2009 conference, called "Marking a Physical Sphere with a Projected Platonic Solid".

She gives a method to construct the regular dodecahedron following the "rules" of temari (a Japanese art form involving thread wrapped around balls in various symmetries), which involves measuring with paper tape and using traditional paper folding techniques.

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Dear Sir, thanks for your answer. I have read in her article that she began the construction by dividing the great line to tenths. Abul Al Waffa in his book ("On parts of geometry needeed by craftsmen") describes two methods for the tesselation of sphere with regular pentagons. His first method also needs us to divide the great circle to equal parts. Yet the scripts remained are wrong as we can hardly speak about congruent spherical triangles. His second method is illegal in a way that he does euclidean constructions inside the sphere. Will I see you in the Bridges conference this year? ... –  erdos May 21 '10 at 17:26
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I don't know temari, but it is known that origami allows one to trisect any given angle, thus making that set of techniques strictly more powerful than compass and straightedge. –  Will Jagy May 21 '10 at 19:31
    
A propos origami vs compass-straight-edge kahuna.merrimack.edu/~thull/omfiles/geoconst.html ... but I wasn't aware that temari allows full set of origami rules. The little I've seen corresponds somewhat to the sphere ruler construction rule: the paper strip is used to mark great circles, and one is allowed to arbitrarily divide the circumference (by folding the sheet of paper; though in practice most people only use divisions of $2^n 3^m$, since it is kind of hard to fold a piece of paper into fifths).... –  Willie Wong May 22 '10 at 1:05
    
...continuing... now off the top of my head I can't remember whether length trisection is an allowable construct on the sphere, I suspect not as trisection of length is equivalent to trisection of angle. If that is the case then the temari rules would allow more operations. –  Willie Wong May 22 '10 at 1:07
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Not an answer to the question as stated but relevant:

I don't seem to have a reference, but the quickest description of possible constructions on, say, the unit sphere, is by the angles made by intersecting curves. The constructible angles are the same as the constructible angles in the Euclidean plane. Thus the constructible lengths are those arclengths $\alpha$ for which $\cos \alpha$ or $ \sin \alpha$ or $\tan \alpha$ (the conditions are equivalent) are in the "constructible field," the smallest extension of the rationals in which the square root of any positive element is still in the field. One might also wish to require $\alpha \leq \pi.$

Actually, let me make that a request. If anybody knows of a reference on the constructible lengths and angles on the surface of the sphere, please let me know.

This is strictly analogous to (and presumably far, far older than) the situation in the hyperbolic plane, I will try to make a working link:

http://zakuski.math.utsa.edu/~jagy/papers/Intelligencer_1995.pdf

See also Marvin Jay Greenberg, "Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries," The American Mathematical Monthly (an M.A.A. journal), Volume 117, number 3, March 2010, pages 198-219. I have a pdf of that as well if anyone cannot find it.

There is a bit of a story. The results on constructibility in $H^2$ were in a string of papers in Russian and Ukrainian in the 1930's and 1940's. I found, and used, the simple conclusions. I later sent my paper to Greenberg, so that material is in the M.A.A. paper mentioned and in the fourth edition of his book. Meanwhile, Robin Hartshorne (yes, that Hartshorne) heard of this result from Marvin and came up with his own proof using the Hilbert Field of Ends formalism, expressing regret that such a pretty result did not make it into his own book on the subject, "Geometry: Euclid and Beyond."

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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? –  Sam Nead May 21 '10 at 22:35
    
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. –  Will Jagy May 22 '10 at 1:25
    
to continue, this is the same trick as in the regular plane or the non-Euclidean plane –  Will Jagy May 22 '10 at 5:06
    
Got it. Very nice. Thank you. –  Sam Nead May 23 '10 at 17:48
    
I wonder, if you only have a memoryless compass, but not the straightedge, doesn't the hyperbolic plane and the sphere look the same? –  Zsbán Ambrus May 23 '10 at 20:00
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