I have three pairs of points in 3D space. These may or may not be coplanar. I want to find a point such that it is equidistant from each pair of points. I know that may or may not be possible depending on the positions of the points. What I want is the best average point, which I can take safely as the centre and draw a sphere from there whose radius is the maximum distance of this point from any of the six points, then I want all the points to remain inside the sphere.
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1$\begingroup$ Your question would probably be more appropriate at MO's sister site math.stackexchange.com. This site is for research-level questions, in mathematics itself. 'Research-level' means, roughly, questions that might be discussed between two professors, or between graduate students working on PhD's. $\endgroup$– HJRWCommented Oct 29, 2010 at 14:20
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$\begingroup$ If you mean that research-level means questions can be asked only by research students or professors then please note that I do have a Masters in Mathematics. Right now I am involved in computer graphics, and this question relates to that. Thanks for the reference to the site, I appeciate. $\endgroup$– Alok GandhiCommented Oct 29, 2010 at 14:36
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$\begingroup$ Alok - no, I mean that there are plenty of very interesting questions with some mathematical content that are not suitable for MO. I think this is one of them. $\endgroup$– HJRWCommented Oct 29, 2010 at 19:05
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$\begingroup$ @Henry : I agree absolutely, I went to the site and it is quite interesting as well as engaging. Thanks again. $\endgroup$– Alok GandhiCommented Oct 29, 2010 at 19:23
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2 Answers
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Distance from a pair of points instead of from each one of a pair of points does not seem to be well defined. However, you could simply take the centroid of all six points (add the coordinates and divide by 6 in each axis), then compute the distance to each of the six points, and use the maximum for the radius of the sphere. It is not clear how you want us to use the fact that the points come in pairs instead of six single points.
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$\begingroup$ As I read it, "equidistant from each pair of points" means that if the pairs are p_{1,1}, p_{1,2}, p_{2,1}, p_{2,2}, p_{3,1}, p_{3,2} then we want |p - p_{i,1}| = |p - p_{i,2}| for each i. $\endgroup$ Commented Oct 29, 2010 at 14:35
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$\begingroup$ Thanks, that partially do answer my question. In fact I have tried the centroid approach before and it seems to work but I was not sure it is the most suitable candidate. As for the pair of points, what I actually have in each pair are points which are maximum and minimum from the origin(0,0,0) in x,y and z axis. For example, the points two points in x - pair will have their x coordinate, minimum and maximum, the y and z coord of these two points are arbitrary. $\endgroup$ Commented Oct 29, 2010 at 14:44
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$\begingroup$ If Qiaochu's interpretaion is right, each of these equidistance conditions defines a plane, and generically three planes meet in a unique point. I'm not sure how this squares with Alok's final sentence about the "best" point and spheres which seem to be looking for something like the point whose greatest distance from the given points is minimized. $\endgroup$ Commented Oct 29, 2010 at 14:50
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$\begingroup$ I think what Robin says"and generically three planes meet in a unique point." is exactly the solution I want. $\endgroup$ Commented Oct 29, 2010 at 15:28
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$\begingroup$ @Alok I'm still not clear on the pairs of points. Is one pair $(x_{min},y,z) \text{and} (x_{max},y,z)$ with the same $y$ and $z$? The sense would be that you really want the whole segment to be in the sphere. That is OK as the sphere is convex. Could you give an example of three pairs? $\endgroup$ Commented Oct 29, 2010 at 17:01
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You want mini ball.
