Timeline for Can an arbitrary collection of circles of total area 1/2 fit into a circle of area 1?
Current License: CC BY-SA 2.5
9 events
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| May 21, 2010 at 8:56 | history | edited | Roland Bacher | CC BY-SA 2.5 |
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| May 20, 2010 at 20:49 | comment | added | fedja |
If all radii are small enough, the greedy algorithm (arrange disks in the decreasing order and pack them so that each next circle touches either at least two of the previously packed one or at least one of the previously packed one and the boundary circle) does the job, the reason being that if a disk D of radius $r$ touches two bigger disks $F,G$, then the area of $2D\setminus (2F\cup 2G)$ is not greater than t times the area of $D$ with t < 2, so this case is trivial. The troublesome disks are those with radii greater than 1/20 or so. If you can pack those, you can pack the rest.
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| May 20, 2010 at 11:29 | comment | added | Victor Protsak | I don't buy "one can use a packing argument showing that a solution always exists if the largest radius is small enough" claim. The best you can reasonably hope for is a simplified solution if the ratio of the largest to the smallest radius is assumed bounded by an a priori constant, which is a rather strong condition. | |
| May 20, 2010 at 11:14 | comment | added | Wadim Zudilin | @roland-bacher: Thanks for your kind words above. Yes, this trick should have a combinatorial interpretation. It looks like if $4R^2/9\le r_1^2+\dots+r_n^2\le R^2/2$, then one can place circles of radii $r_1,\dots,r_n$ inside the circle of radius $R$ in such a way that there is a room inside the large circle for one circle of radius $\sqrt{R^2-2(r_1^2+\dots+r_n^2)}$. You show this for $n=1$ and last couple of hours I spent on $n=2$: it works but the solution is too complicated. | |
| May 20, 2010 at 8:36 | history | undeleted | Roland Bacher | ||
| May 20, 2010 at 8:36 | history | edited | Roland Bacher | CC BY-SA 2.5 |
added 1193 characters in body; added 85 characters in body
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| May 12, 2010 at 6:43 | history | deleted | Roland Bacher | ||
| May 11, 2010 at 8:00 | history | edited | Roland Bacher | CC BY-SA 2.5 |
Answer to the wrong question
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| May 11, 2010 at 7:52 | history | answered | Roland Bacher | CC BY-SA 2.5 |