Timeline for Is the circle in the square best at avoiding random lines?
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13, 2017 at 3:43 | comment | added | fedja | @Eckhard The square with the vertices at the midpoints of the sides of the original square is, obviously, optimal :lol:. But if you bring the area down a bit, that version may get more interesting. | |
| Aug 10, 2017 at 5:37 | vote | accept | Aaron Meyerowitz | ||
| S Aug 9, 2017 at 10:16 | history | bounty ended | CommunityBot | ||
| S Aug 9, 2017 at 10:16 | history | notice removed | CommunityBot | ||
| S Aug 1, 2017 at 9:01 | history | bounty started | Aaron Meyerowitz | ||
| S Aug 1, 2017 at 9:01 | history | notice added | Aaron Meyerowitz | Improve details | |
| Aug 1, 2017 at 8:53 | answer | added | Aaron Meyerowitz | timeline score: 2 | |
| Jul 30, 2017 at 14:23 | comment | added | Eckhard | What about the dual problem of the convex set of area 1/2 that maximises the probability of being hit by a random line segment? | |
| Jul 24, 2017 at 22:59 | answer | added | fedja | timeline score: 7 | |
| Jul 24, 2017 at 8:02 | answer | added | Will Sawin | timeline score: 16 | |
| Jul 24, 2017 at 7:31 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 8 characters in body
|
| Jul 24, 2017 at 7:29 | comment | added | Aaron Meyerowitz | @TimothyChow you are right. I'll fix it. | |
| Jul 24, 2017 at 6:17 | comment | added | Will Sawin | @fedja By a symmetry argument, I believe the differential equation has to have the form $\ddot{y} =f_{\lambda}(x)$ which should be solvable in reasonably explicit form. | |
| Jul 24, 2017 at 2:53 | comment | added | Timothy Chow | I'm slightly confused...is $p(\Delta)$ supposed to be the probability that a random line does not intersect $\Delta$? Currently it is stated that it is the probability that a random line does intersect $\Delta$. | |
| Jul 24, 2017 at 0:32 | comment | added | fedja | @WillSawin Are you sure it is solvable? Anyway, so far the best upper bound I can get in the initial setting with a (relatively) decent proof is $1/2$. | |
| Jul 23, 2017 at 19:49 | comment | added | Will Sawin | @fedja Perhaps I or someone else should solve the differential equation for that problem, which might shed some light. | |
| Jul 23, 2017 at 19:22 | comment | added | fedja | @WillSawin Indeed, you are right. :-) That makes our life even more complicated than I thought. | |
| Jul 23, 2017 at 18:56 | comment | added | Will Sawin | @fedja I don't think the right half is optimal, it can be improved by making it slightly convex. If you just replace it with the pentagon with top vertices $(0,a), (1/2, 1-a), (1,a)$ then we get all lines between $(1,r)$ and $(1,s)$ as long as $r \geq a$, $s \geq a$, $r+s \geq 2-2a$ which is one-half times $(1-a)^2 - (2-4a)^2/2$ which is optimized at $a=3/7$. | |
| Jul 23, 2017 at 18:40 | comment | added | fedja | @WillSawin I can prove it :-) Unfortunately, the proof uses one peculiar coincidence (in addition to the traditional Cauchy-Schwarz) that is not there in the original setup. It is funny that the optimal shape for the complementary problem (about the lines joining two opposite sides) is most likely the right half of the square though I do not have an argument to support it. If so, we would have an upper bound of $\frac 38=0.375$ for the original question. | |
| Jul 23, 2017 at 18:23 | comment | added | Will Sawin | @fedja How do you know that's optimal for the other problem? | |
| Jul 23, 2017 at 15:46 | comment | added | fedja | The area bounded by four hyperbolas, one of which (provided that the left bottom corner of the square is the origin) is $xy=\frac a4$ where $a$ solves the equation $a(1+\log\frac 1a)=\frac 12$ beats the circle slightly. However, I do not think it is optimal either (though it is optimal for a different problem: maximize the probability that a random interval with the endpoints on two adjacent sides of the square misses the region). | |
| Jul 23, 2017 at 14:16 | comment | added | Yoav Kallus | It is perhaps worth mentioning that if lines are chosen according to a measure that is uniform with respect to orientation and, conditioned on orientation, uniform with respect to translation, then the circle is optimal as a consequence of Crofton's formula: en.wikipedia.org/wiki/Crofton_formula | |
| Jul 23, 2017 at 12:52 | comment | added | Gerry Myerson | I've seen some pretty long lines at The Circle in the Square. en.wikipedia.org/wiki/Circle_in_the_Square_Theatre | |
| Jul 23, 2017 at 9:58 | history | asked | Aaron Meyerowitz | CC BY-SA 3.0 |