Timeline for Bijection from the plane to itself that sends circles to squares
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2017 at 11:46 | comment | added | Vincent | I'd like to use the sudden interest in this question to generate some overflow interest into a different but similar question on MSE: math.stackexchange.com/q/2221362/101420 from a few days ago. It is not mine, but I am still curious to see an answer. | |
| Apr 11, 2017 at 10:52 | review | Close votes | |||
| Apr 11, 2017 at 13:19 | |||||
| Apr 10, 2017 at 19:36 | vote | accept | Tom Solberg | ||
| Apr 10, 2017 at 19:35 | answer | added | Will Brian | timeline score: 66 | |
| Apr 10, 2017 at 19:34 | comment | added | Denis T | As you tagged this as AG, I guess you may be interested in some works of Vladlen Timorin, who wrote a few papers on classification of $\mathbb {RP}^n$ selfmaps taking lines into plane curves of some degree. See arxiv.org/abs/math/0212098 and later. | |
| Apr 10, 2017 at 19:33 | comment | added | Neal | Maybe some interesting related questions are: For which $\mathcal{S}$ does this proposition hold, where $\mathcal{S}$ is the set of boundaries of some class of convex bodies? Is number of intersection points the only obstruction? | |
| Apr 10, 2017 at 19:29 | answer | added | amakelov | timeline score: 8 | |
| Apr 10, 2017 at 19:27 | comment | added | Denis T | I guess any two circles have 1, 2 or infinitely many common points, whereas squares can intersect in 4 points. | |
| Apr 10, 2017 at 19:27 | comment | added | amakelov | You can have two squares that intersect at 8 points (for example - take two congruent squares on top of each other and rotate by $45^\circ$ around their common center. Under the inverse of such a bijection, your squares would have to go to two distinct circles with 8 common points, which is impossible. | |
| Apr 10, 2017 at 19:17 | history | asked | Tom Solberg | CC BY-SA 3.0 |