Timeline for Two queries on triangles, the sides of which have rational lengths
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2019 at 9:53 | history | edited | R.P. |
edited tags
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| Jul 4, 2019 at 16:07 | answer | added | R.P. | timeline score: 6 | |
| Jul 4, 2019 at 15:49 | comment | added | R.P. | @WillSawin Ah yes, I wasn't looking at it in the right way. Thanks! | |
| Jul 4, 2019 at 12:12 | comment | added | Will Sawin | @RP_ It seems like a quadruple cover to me. $A$ is not a variable, it's fixed. If you are viewing $x,y,P$ as projective coordinates, tehre are four lifts of them to affine coordinates that satisfy the equation, because it's a quartic equation. | |
| Jul 4, 2019 at 8:41 | comment | added | R.P. | @WillSawin I may be way off, but from Chris Wuthrich's equation it would seem to me that this might be a degree 2 del Pezzo, no? Since it is a double cover of $\mathbb{P}^2$ ramified over the quartic $C:(x+y-P/2)(P/2-x)(P/2-y)P =0$. I don't know for sure since $C$ is reducible, but then at least the surface is a degeneration of a family of del Pezzos, hence I guess it must be rational. | |
| Jul 4, 2019 at 7:25 | answer | added | user131781 | timeline score: 0 | |
| Jul 3, 2019 at 0:57 | comment | added | Will Sawin | It seems from counting $\mathbb F_q$-points that this K3 surface should have high rank, so maybe it will have many curves that can be used to find $\mathbb Q$-points. | |
| Jul 3, 2019 at 0:48 | comment | added | Will Sawin | It seems that this equation defines a quartic surface with six nodes at infinity. I guess the resolution is a K3 surface. | |
| Jul 2, 2019 at 19:01 | history | became hot network question | |||
| Jul 2, 2019 at 18:54 | history | edited | Martin Sleziak |
the tag (triangles) seems suitable here
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| Jul 2, 2019 at 18:53 | history | edited | LSpice | CC BY-SA 4.0 |
Deleted signature
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| Jul 2, 2019 at 17:18 | history | edited | YCor | CC BY-SA 4.0 |
clarified, and removed unnecessary emphasizing
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| Jul 2, 2019 at 17:15 | comment | added | YCor | I guess that "general" was meant in the sense "arbitrary". | |
| Jul 2, 2019 at 17:06 | history | edited | YCor |
edited tags
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| Jul 2, 2019 at 14:53 | review | Suggested edits | |||
| Jul 2, 2019 at 16:01 | |||||
| Jul 2, 2019 at 13:26 | answer | added | Chris Wuthrich | timeline score: 13 | |
| Jul 2, 2019 at 13:04 | comment | added | Daniel McLaury | Brahmagupta gave a parametrization of heronian triangles similar to Euclid's parametrization of right triangles. Have you tried playing with that? | |
| Jul 2, 2019 at 12:48 | comment | added | Manfred Weis | Just a remark: I would suggest to provide a definition of what GENERAL triangles are in the context of this question because it can mean different things to different people. | |
| Jul 2, 2019 at 12:31 | history | edited | Manfred Weis | CC BY-SA 4.0 |
fixes a typo in the title and the apparently wrong placement of the NOT relating to rational numbers being the area of a rational right triangle
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| Jul 2, 2019 at 11:00 | review | First posts | |||
| Jul 2, 2019 at 11:52 | |||||
| Jul 2, 2019 at 10:55 | history | asked | R. Nandakumar | CC BY-SA 4.0 |