Timeline for Maximum possible number of similar three-colored triangles
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2015 at 14:51 | vote | accept | Morteza | ||
| Sep 11, 2014 at 20:09 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced deprecated tag 'geometry'
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| Sep 11, 2014 at 10:53 | answer | added | Ilya Bogdanov | timeline score: 2 | |
| Aug 14, 2014 at 6:02 | comment | added | Morteza | @Gerry Myerson Good example for 3D, but Of course planarity is important, if not you can find infinitely many three colored triangles in the 3 dimensional space. ($1\times1\times\infty$) | |
| Aug 12, 2014 at 23:18 | comment | added | Gerry Myerson | I suppose for the $2\times2\times2$ you insist on planarity, else the vertices of a regular octahedron will do. | |
| Aug 12, 2014 at 23:06 | answer | added | Joseph O'Rourke | timeline score: 2 | |
| Aug 12, 2014 at 19:06 | comment | added | Lev Borisov | Regarding the 2x2x2 case: It might be a mess to do by hand, but there is a way of coding this. Specifically, have the computer go through all possible angle permutations of 8 triangles (there are $6^7$ possibilities, since you can fix the angles of the first one). You can use symmetry of the problem to reduce this a bit further, if needed. This gives you equations among the squares of the sides $|X_iY_j|^2,|Y_iZ_j|^2,|Z_jX_i|^2$. Add to these the equations of planarity (mathworld.wolfram.com/Cayley-MengerDeterminant.html) and see if the only solution is all zeros by Groebner bases. | |
| Aug 12, 2014 at 16:54 | history | edited | Morteza | CC BY-SA 3.0 |
edited title
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| Aug 12, 2014 at 16:44 | history | edited | Morteza | CC BY-SA 3.0 |
added 25 characters in body
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| Aug 12, 2014 at 16:35 | history | asked | Morteza | CC BY-SA 3.0 |