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 April 7th, 2018, 03:41 PM #1 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 444 Thanks: 29 Math Focus: Number theory 4 coplanar points nearly equidistant 2 or 3 separate points may be equidistant on a plane. How close can 4 points come to obeying this qualification (minimal deviation from a circle?) April 8th, 2018, 07:44 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 How is closeness defined? Thanks from romsek April 8th, 2018, 11:09 AM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 An interesting problem. Symmetry would seem to dictate that the points form a rhombus. Using a rhombus with side length 1 I calculated the squared distances between each point, averaged them, and then took the sum of the difference between the individual squared distances and the average, as a function of the angle of the rhombus. I then minimized this using standard calculus. The answer was a square. There are other choices of what we are trying to minimize. Thanks from Loren April 8th, 2018, 11:35 AM #4 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 444 Thanks: 29 Math Focus: Number theory skipjack, I suppose statistically, like finding the minimal standard deviations of distances from circles with three-point combinations to the fourth point. So take all circles containing three points where the distance to the fourth is minimized and test which of these standard deviations are minimal. Square sounds good, romsek. Maybe the above is more general for including asymmetries. April 8th, 2018, 12:24 PM   #5
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 Originally Posted by Loren skipjack, I suppose statistically, like finding the minimal standard deviations of distances from circles with three-point combinations to the fourth point. So take all circles containing three points where the distance to the fourth is minimized and test which of these standard deviations are minimal. Square sounds good, romsek. Maybe the above is more general for including asymmetries.
by circles do you mean the equilateral triangle formed by the 3 equidistant points?

I'm not sure how circles enter into things. Would you then find the distance to the fourth point from the center of these 3 points? April 8th, 2018, 12:40 PM #6 Senior Member   Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 Ok, optimizing the placement of a 4th point given 3 equidistant points so that the total distance of the 3 points to the 4th is minimized results in the 4th point being placed at the center of the original triangle. Given the symmetry of the problem this is entirely expected. April 8th, 2018, 01:58 PM #7 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 444 Thanks: 29 Math Focus: Number theory Imagine any four points on a plane. It is impossible for all of them to be equidistant to each other. My approach is to construct every circle that includes each combination of three (out of the four points), then compare their centers of curvatures with the position of the fourth points. The ideal weighting of the center of curvature with the fourth point may yield the best case of arbitrary four points overall. Like you said, the greater symmetries might yield a square, or perhaps a rhombus. Or, romsek: "Ok, optimizing the placement of a 4th point given 3 equidistant points so that the total distance of the 3 points to the 4th is minimized results in the 4th point being placed at the center of the original triangle." But how does one choose which might be the actual fourth point given an arbitrary circle? Tags coplanar, equidistant, points Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post EvanJ Geometry 2 September 6th, 2017 05:46 PM LasLasso Linear Algebra 1 February 11th, 2015 01:42 PM elifast Algebra 6 September 18th, 2012 06:35 PM Rutzer Number Theory 1 April 10th, 2011 02:49 PM elifast Applied Math 0 December 31st, 1969 04:00 PM

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