My Math Forum Finding A Point Which Is Equidistant From Given Points

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 July 8th, 2018, 05:26 AM #1 Member   Joined: Sep 2017 From: Saudi Arabia Posts: 37 Thanks: 1 Finding A Point Which Is Equidistant From Given Points Hello my dears, I have 13 distinct points with known coordinates. Is it possible to find a point which is equidistant from the 13 known points?
 July 8th, 2018, 06:18 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,791 Thanks: 970 Sorry, but you get NO help until you send Maschke his prize: http://mymathforum.com/lounge/342377-competition-2.html Thanks from Maschke
July 8th, 2018, 06:24 AM   #3
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Quote:
 Originally Posted by Denis Sorry, but you get NO help until you send Maschke his prize: http://mymathforum.com/lounge/342377-competition-2.html
I said:
"The member who collects 10 POINTS, I will send a prize to him via a shipping company."

July 8th, 2018, 07:12 AM   #4
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Quote:
 Originally Posted by Hussain2629 Hello my dears, I have 13 distinct points with known coordinates. Is it possible to find a point which is equidistant from the 13 known points?
Yes.

Can you give me my point over PM?

-Dan

July 8th, 2018, 08:43 AM   #5
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Quote:
 Originally Posted by Hussain2629 I said: "The member who collects 10 POINTS, I will send a prize to him via a shipping company."
BUT it is illegal to offer a prize without saying what the prize is:
poor Maschke could possibly have spent countless hours getting
to 10 points....then to receive a pack of chewing gum as reward

July 8th, 2018, 12:16 PM   #6
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Quote:
 Originally Posted by Hussain2629 Is it possible to find a point which is equidistant from the 13 known points?
Not in general, no.

At least not in two or three dimensional space. If you have enough dimensions it's always going to be possible I should think.

July 8th, 2018, 01:38 PM   #7
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Quote:
 Originally Posted by v8archie Not in general, no. At least not in two or three dimensional space. If you have enough dimensions it's always going to be possible I should think.
Right. Just as 2 points determine a 0-sphere (a pair of points) in 1D, and 3 points determine a 1-sphere (a circle) in 2D, and 4 points determine a 2-sphere in 3D, 13 points would determine an 11-sphere living in 12D space.

https://en.wikipedia.org/wiki/N-sphere

July 8th, 2018, 01:41 PM   #8
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Quote:
 Originally Posted by v8archie Not in general, no. At least not in two or three dimensional space. If you have enough dimensions it's always going to be possible I should think.
That's pretty much what I had come up with. And even if I'm wrong about the number of dimensions used I would imagine we could construct a metric which should fix any problems left over from the dimensionality argument.

Admittedly the OP wouldn't be considering such.

-Dan

 July 8th, 2018, 01:47 PM #9 Senior Member   Joined: Aug 2012 Posts: 2,134 Thanks: 621 Actually you can do this in the plane if you can pick the metric. You can even do it on a line. The discrete metric is $d(x,y) = 1$ if $x \neq y$ and $d(x,y) = 0$ if $x = y$. Then given any 13 points in the plane or on a line, pick any point distinct from all of them. Then its distance from each of the 13 points is 1. Thanks from topsquark and v8archie
 October 2nd, 2018, 08:45 AM #10 Member   Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3 Considering you are a very down-to-earth person, I can only imagine you picked 13 geographical points on the surface of the planet and therefore the equidistant point is: "the centre of the earth". Thanks from Denis Last edited by Arisktotle; October 2nd, 2018 at 09:25 AM.

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