Timeline for Which line is more likely to intersect the circle?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19 at 15:48 | history | edited | Theguyabovemeislying | CC BY-SA 4.0 | added 75 characters in body |
| Aug 19 at 15:17 | history | edited | Theguyabovemeislying | CC BY-SA 4.0 | added 1887 characters in body |
| Aug 19 at 15:02 | comment | added | Theguyabovemeislying | Yeah youre right let me just edit the answer | |
| Aug 19 at 14:47 | comment | added | Dan | You "removed the concern over 1-1 correspondences" and then you said "whatever consideration we previously had for correspondence between lines on A and B is radically changed". Why should it matter if it was radically changed, if we weren't concerned about it? Perhaps you could edit your answer to clarify, instead of clarifying in comments. | |
| Aug 19 at 14:26 | comment | added | Theguyabovemeislying | Upon sleeping on it, I’m confident that my argument doesn’t suggest equal probability if you move the circle to the right (nor, obviously, if you move to the left). | |
| Aug 19 at 2:32 | comment | added | Theguyabovemeislying | It’s a cool question but I think you’re dismissing arguments too readily. For what it’s worth I’m reasonably confident my argument is sound. | |
| Aug 19 at 2:14 | comment | added | Dan | I downvoted because the answer (and counter-arguments) seem quite hand-wavy to me. But I appreciate your interest in the question. | |
| Aug 19 at 2:07 | comment | added | Theguyabovemeislying | Okay, I’m going to sleep on this but I think the resolution is that once any non 0 measure length of A (more than 1 point) is inside B, the ‘denominator’ changes for A in a way it doesnt for B. This is because for every single one of those points enclosed by B, you can pick two different points on B that will yield any direction on the plane. That is, whatever consideration we previously had for correspondence between lines on A and B is radically changed. Also, why the downvote? | |
| Aug 19 at 1:52 | comment | added | Dan | When the red circle is coincident with the green circle, the probabilities are (obviously) equal. Then if we start moving the red circle to the left, the probability that $\overleftrightarrow{AB}$ intersects the black circle decreases then increases, hitting equality with $\overleftrightarrow{BC}$ precisely when the red circle is tangent to the green circle. | |
| Aug 19 at 1:45 | comment | added | Theguyabovemeislying | Well no probably not true if you move further out… you can’t sweep the plane anymore for one. But as long as there’s a point of contact, you can. How sure are you of these simulations? Perhaps every red circle that’s still in contact with the green circle and colinear keeps the same probability | |
| Aug 19 at 1:38 | comment | added | Theguyabovemeislying | Also I’m curious about these other values. Have you run simulations and found different probabilities? Clearly the probability is equal if both circles overlap completely, and it looks like that holds at 1 shift over. I’d conjecture even that it might hold for any integer number of shifts over | |
| Aug 19 at 1:25 | comment | added | Theguyabovemeislying | Hmm… let me think about this. My instinct is to say that my argument would no longer apply - I think its very much dependent on the single point of contact and the relative positioning of A and B | |
| Aug 19 at 1:11 | comment | added | Dan | I seem to have misunderstood what you mean by shadow, but anyway there is another problem: Suppose we move the red circle a little bit (say half a radius) to the right. Your argument about shadows would still imply that the two probabilities are equal, but simulations suggest otherwise. | |
| Aug 19 at 0:45 | comment | added | Theguyabovemeislying | Wait sorry I’m confused - I’m pretty slow right now, it’s been a full day. What’s the objection? | |
| Aug 19 at 0:43 | comment | added | Theguyabovemeislying | In particular, you can pick two points on the circle B so that the line they produce is arbitrarily close to the horizontal tangent without hitting T. For example, let C be the top point on B and let the other point be anything counterclockwise of it but arbitrarily close. | |
| Aug 19 at 0:41 | comment | added | Dan | But here is your definition of 'shadow': "Consider the ‘shadow’ of the target circle - the region of the plane where a line passing through the middle circle must hit the target circle." | |
| Aug 19 at 0:36 | comment | added | Theguyabovemeislying | I think you may be misinterpreting what I mean by shadow. It’s very easy to get rays from B past the vertical tangent without hitting T. | |
| Aug 19 at 0:31 | comment | added | Dan | The shadow of B is much larger than the shadow of A. The shadow of B is bounded by the vertical line tangent to the green and black circles. The shadow of A is bounded by the two lines that are tangent to the red and black circles and make an angle of $30^{\circ}$ to the horizontal. | |
| Aug 18 at 13:58 | comment | added | Theguyabovemeislying | You could also probably skip a lot of the caution by just reducing to those two tangents that matter and saying that they dont count because theyre small, and every other line is in a 1-1 correspondence. Im not 100% sure how that handles the other tangents, but im sure someone motivated enough could get around to it | |
| Aug 18 at 13:48 | history | edited | Theguyabovemeislying | CC BY-SA 4.0 | added 4234 characters in body |
| Aug 18 at 5:10 | history | edited | bobble♦ | CC BY-SA 4.0 | orthography |
| Aug 18 at 3:05 | comment | added | Theguyabovemeislying | Will explain tomorrow very tired | |
| Aug 18 at 3:04 | comment | added | Theguyabovemeislying | Urgh, yeah i think because only the tangent lines arent replicable and only 1 (2) tangents to B pass through both other circles | |
| Aug 18 at 3:03 | comment | added | Dan | I don't understand the part after "i found an intuitive solution i think". | |
| Aug 18 at 2:44 | history | answered | Theguyabovemeislying | CC BY-SA 4.0 |