Absane: You are correct. I did not understand the concept indicated by your diagram. Clever idea and it does suggest a framework for the solution. This concept might not lead to a usable method of solving the problem, but it is better than the total lack of ides in my mind. Expressing the sum of the white areas as a function of coordinates of three points looks like a formidable task. That would only be a part of the solution.
I do not think so but it is not important as huge number of different ones must be considered. Simple proof that "equilateral triangle and centered at the circle's center" is not the most typical: Imagine circle divided into 6 equal sector "pies" 1, 2, ...6. Let first point fall in sector 1. Lets agree to consider that points 2 & 3 form something resembling your "equilateral triangle and centered at the circle's center",ETCCC, if one is in sector 3 abd the other is in sector 5. This is very generous as the distance between two of them could eaily be half as large as the longer side of their triangle (probably is, but I will not use this or try to prove it.) When second point is selected, if it is in sector 3 or 5 you still have chance at ETCCC, but if it falls in 1,2,4 or 6 you have failed already to get your ETCCC. Thus you have 1/3 of a change at selection of point two but lets assume you were lucky and it went into sector 3. Now the third point MUST FALL in sector 5 but it can fall in any of sectors 1,2,3,4,or,6 for you to still fail (even with the luck you had on the fall of the second point) I.e. this is less than a 17% chance. Over all (without luck) the approximately ETCCC is expected approximately ~0.33x0.17 < 6% probability to get this "generously defined" ETCCC!! Now we can be a little stiffer on the requirements for the ETCCC. For example divide the circle in 9 sectors and very reasonably require that an "ETCCC" has points in 1, 4, & 7. I.e. must have two sectors between each, still with no requirements on their separation distances (yet we call them "equilateral"! I.e. the 9 sector division of the circle only requires their angular spacing be at least 2(360/9) = 80 degrees, not the 120 degrees of an exact ETCCC. That is, if all their angular separations are greater than 80 but less than 160 this "requirement" accepts the angular separtion as appromately the 120 of an ETCCC - still very generous, I think, especailly as there is still zero requirements on their distances from each other!!!! I am not trying to be argumentive, but to show that anything that can reasonably called and "equilateral triangle" is not to be expected. I even doubt that the typical triangle contains the center of the circle with more than 50% probability, regardless of it shape. SUMMARY: Far from being typical, the ETCCC is actually rare. Your useful drawing shows the triangle with more area than the three-perimeter section, but I suspect this is also atypical. I.e. I bet the typical triangle has more area in the three perimeter red areas than in the triangle. But again, none of this is important. Even if you triangle is not typical, it is very useful. My solution outlined in post 17, does not make any use of the figure. I did not “turn the crank” to get the numerical answer to question as a percent, and do not plan to do so. (My only strong interest was to learn how to get the answer.) Perhaps Pete will - he has done amazing amounts of work on other problems. - I am much too lazy to “turn the crank.”
What's the status on this problem? I haven't even tried the problem yet but when I get around to it, I'll try attacking it once and for all.
I'm working on: 1) Finding the area of a segment formed by the chord connecting two points, given the distances of the two points from the centre, and the angle between them, and 2) Finding the probability density function of that area, given the density functions we already know.
If you will use Monte Carlo, then I have the method of solution clearly (I think) described in my post 17. It is just a question of "turning the crank" now. My trick was to reduce the problem to another. Briefly, in four steps, here is computer program for the solution: (1) Randomly pick 4 quadrilateral points and name them (in cyclic order) a,b,c & d. Form lines ac & bd and solve for their intersection point, P which is at (Xp, Yp). (2) Then define horizontal "half line" from P called "L" with equation: y = Yp but x>Xp. (3) Now solve for intersections of L with each of the 4 line segments* boundaries of quadrilateral. If there is only one such solutions, then P is inside quadrilateral and it is convex. If there are 0 or 2 such solutions, then it is concave. If the solution is exactly a corner of the quadrilateral, then return to step (1). - (This will never happen in practice. i.e. a set of zero measure.) (4) Score this Monte Carlo trial accordingly and start next, if "exit loop" I also gave in post 17 is not satisfied. Then you have the probability to better than 1% error. Very probably accurate to three significant figures with "exit loop" test I gave in post 17. ------------------------------- *In post 17 I also explained how to avoid at least testing more than 2 boundary lines (often all 4 if quadrilateral is concave) by simple ordering of the 4 values of Ya, Yb, Yc, & Yd. -I.e. if Yp is greater or smaller than all 4Ys, then quadrilateral is concave. this however is more to code and only makes program faster, which probably is unimportant if three place accuracy is OK.
I was not sufficiently interested in the numerical answer to write program I out lined in my last post to compute it. I only found it interesting to see if I could invent a way to solve the problem (that did not use darts Please Register or Log in to view the hidden image! ). I think I did. At least no one has suggested I did not. If you want to both test my method and write program I would be happy to know if I did find a working method and what the answer is. If you want to do this and have any questions about my method, just PM me.
Well, I am (was?) looking for the exact answer to the problem. I remember when I looked up the answer online, the same method I proposed was used (that is, calculate the average area of all triangles in the circle).
Do you have a link to any method of doing that? As there are an infinite number of possible triangles (I think you mean quadralaterials) possible I do not see how ones does what you state produces an exact answer.
Area of triangles works. The quadrilateral is convex if and only if the fourth point lies outside the triangle formed by the first three points. So you can find the complementary probability: concave quadrilaterals. Given the first three points, the fourth point will make the quadrilateral concave with probability (Area of triangle)/(Area of circle). All you have to do is find a suitable formula for the area of a triangle of three points and integrate it over the distribution of those other three points.
http://www.sciforums.com/showthread.php?p=1263484#post1263484 I will work on finding the solution again.
Yes, that and Absane are correct. So long since I last thought about this I forgot that. thanks. PS it is really the area of triangle and three "pies" - see Absane's very useful drawing at his link.
Yes, true - that makes the integration a lot more complex. It's easy to change the probability distribution of points to polar coordinates (make the probability proportional to r), but then how can you write the triangle+pie area in terms of \(r_1, \theta_1, r_2, \theta_2, r_3, \theta_3\)?
So easy, that I did it, in earlier post this thread, if Memory serves me correctly. but the general expression for Absanes "red area" is a bitch to workout and probably a room full of bitches to multiply integrate over the randomnly chosen variables. (I did not get even close to the first part - the general area expression. I made a few false posts thinking I was on my way to it though.) As I found method I could actually do, I quit. All methods will have a finite error, I am practically sure; certainly any with either finite number of randomn cases or even an infinite number of cases if the red area integration is not analytic. So from my POV, my solutions is just as good in principle. - I.e. just tell me what error is OK and I can (in principle) get answer to that accuracy by my method.
This is a famous problem, Sylvester's "Four Point Problem". There is an exact answer. The problem was posed in 1865, solved in 1867. It appears to be the heart of a lot of work in computational geometry. The "trick" to solving this problem in a few minutes is to remember attending a seminar that discussed this or a related problem.
hard for me to do this "trick" so I googled: Sylvester's Four Point Problem and went to the first 10 hits, but none gave clear, simple answer as to method. (it does seem to be a sub set of more complex problem.) Do you have a link to reasonably easy explanation of how exact solution to the simple question* is found? ---------------------- *What is probablility that four random points inside a circle form convex quadralaterial? Thanks for giving name/ history.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122451 Read a about half way down. Right after the math formula someone provides, a member shares a link to the solution.
Great thanks already even thought I have not reach that point: Someone there (Yauning lion) noted: Actually it does indeed reduce to finding the average area of a triangle. Iff ABCD is not a convex quadrilateral, exactly one of the triangles formed by 3 of the points contains the fourth one. Hence the answer is 1-4P(triangle ABC contains D)." Thus we do not need to consider Absane's red "pies" after all!!! (His drawing blinded me to considering the area of the four different possible triangles as fracion of the circle and averaging for that case's probability of convex.) I see that "jmerry," like me, noted that the location intersection of AC and BD, wrt to the points, is also an approach. people there had problems (I am sure I would) downloaing the link tht gives the solution (something to due about TEX formating.) so Idid not try to download it, but your "just after jmerry's triple integral formulae" is good wait to get to link for anyone who does. It seems to be a long download also.
