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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?

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. 
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 threepoint 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  
Senior Member Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389  Quote:
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? 

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