Timeline for Finding closest point to a set of circles
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Mar 21, 2014 at 18:50 | comment | added | Gerhard Paseman | I'm sorry to see that Douglas. Perhaps it is for the best. After further searching, I am guessing that you answered the wrong question. Hopefully the right question (and questioner) will come along. Gerhard "Sometimes Answers The Wrong Question" Paseman, 2014.03.21 | |
| Mar 21, 2014 at 18:27 | comment | added | Douglas Zare | I deleted my answer. I regret trying to help people who don't appreciate actual answers. | |
| Mar 21, 2014 at 18:21 | comment | added | Douglas Zare | @Benjamin Dickmaan: Why would that be better than using the quadratic formula to find the intersection of circles? People keep suggesting that the quadratic formula is slow, uses calculus, should be avoided, etc. I don't understand. | |
| Mar 21, 2014 at 18:16 | comment | added | Benjamin Dickman | With regard to Fermat's point etc, perhaps you can rig up a physical model to estimate where this point would be. See the related MO question mathoverflow.net/questions/104714/… | |
| Mar 21, 2014 at 17:58 | comment | added | Douglas Zare | @user3098199: Do you actually prefer taking a long time to get an iterative approximation over using the quadratic formula and Newton's method in 1 dimension? | |
| Mar 21, 2014 at 10:09 | vote | accept | user3098199 | ||
| Mar 21, 2014 at 7:38 | answer | added | user25199 | timeline score: 1 | |
| Mar 21, 2014 at 4:19 | comment | added | user39815 | This is minimum steiner tree problem. | |
| Mar 21, 2014 at 0:49 | answer | added | Joseph O'Rourke | timeline score: 2 | |
| Mar 20, 2014 at 10:08 | comment | added | user3098199 | @DouglasZare: Can you think of simple way to solve this, assuming the case fermat point won't work? An approximation will also do. | |
| Mar 20, 2014 at 10:01 | comment | added | Douglas Zare | I never said that. I explicitly said otherwise, that the solution is sometimes on one of the circles. This was left out by the "answers" on math.stackexchange. In addition, the radii of the circles affects which solutions are valid which are not on any circle. | |
| Mar 20, 2014 at 9:58 | comment | added | user3098199 | @DouglasZare Then are you saying the radius of the circles do not matter? I am sorry, but from what you are saying, it means only the coordinates of the circles matter, and not the radius. However, we can easily debate that change in radius can easily impact the closest point. | |
| Mar 20, 2014 at 9:54 | comment | added | Douglas Zare | No, there are times when the Fermat point is optimal even though the circles intersect. There are times when the circles don't intersect, but the Fermat point is not optimal. | |
| Mar 20, 2014 at 9:42 | comment | added | user3098199 | @Carl/Douglas: Most of the times the circles could intersect, which means the Fermat point may not be what I'm looking for. Fermat point would make sense only when the intersection does not happen between circles. | |
| Mar 20, 2014 at 9:28 | comment | added | Douglas Zare | Yes, one possible solution is the Fermat point. en.wikipedia.org/wiki/Fermat_point There are a lot of other cases. Sometimes the solution is on one of the circles. Sometimes it is a sort of anti-Fermat point. | |
| Mar 20, 2014 at 9:25 | comment | added | user25199 | I didn't look at this in detail, but would try the following, for the real problem: As in soap bubbles, I suspect it is where the lines to the circle centres meet at 120 degrees (this can be confirmed/refuted by differentiation). Find the equation of the locus of points at which the centres of two circles subtend 120 degrees. Then the desired point is the intersection of these three loci. | |
| Mar 20, 2014 at 7:58 | review | First posts | |||
| Mar 20, 2014 at 8:53 | |||||
| Mar 20, 2014 at 7:41 | history | asked | user3098199 | CC BY-SA 3.0 |