
Can you solve it? The four points, two distances problem - mkl
https://www.theguardian.com/science/2019/oct/21/can-you-solve-it-the-four-points-two-distances-problem
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ColinWright
Nice to see this get a little traction ... I've posted the puzzle here a
couple of times in the past, including posting a link to this article, and
it's never got traction in the past.

It's such a lovely puzzle, and the _meta-_ puzzle is fascinating. Once you
have a few configurations, how can you _know_ that you have them all.

It's like accounting for corner cases in code and user-interface design and
implementation. How do you _know_ you've accounted for everything that might
happen, especially when dealing with the unwashed masses.

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user_rob
Yo - I'm not totally stupid - used circles and a table approach to show that
there are only 6. Just took the time to write and draw in notebook.

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ColinWright
Cool - I'd love to see your approach. Could you write it up and post a link?
Or feel free to email me, my contact details are in my profile. Alternatively,
the article here has a link to my blog, and there's a comment box.

Or just here: p4d2_HN@solipsys.co.uk

I'm always interested in seeing a better way to do things.

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AstralStorm
I wonder if multidimensional solutions are ok, or if we have to keep it
planar. And if any metric is acceptable.

The answer is a really big set. All balls including really silly graphs.

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ColinWright
The original problem I think intends the points to be on the plane. Feel free
to find solutions in higher dimensions - the trivial answer is that there are
infinitely many. The less trivial answer is to find all the _families_ of
solutions.

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mywacaday
Have to admit I would have missed the kite, literally did not think outside
the box.

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janpot
an equilateral triangle with a dot in the center?

Edit:

Note to self: read articles before commenting

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mkl
That is one of them, yes. The challenge is to find _all_ possible
configurations.

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kangnkodos
I found a number of them.

Here's how I would go about systematically trying to find them one by one. I
suppose this method could be built up to a proof.

First consider the shape of the outside points: point - no straight line
segment - no, after a little work triangle - maybe quadrialteral - yes
pentagon and up - no

For the case of 4 outside points, consider the possible lengths. aaaa - Yes.
Square. Other angles - maybe aaab, b is longer - Maybe aaab, b is shorter -
Maybe aabb - Maybe abab - Maybe

For the case of 3 outside points, consider the possible lengths. aaa - maybe
aab, b is shorter - maybe aab, b is longer - maybe

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AstralStorm
This is not even close.

Obviously you missed a crossed circle and every tesselation of regular polygon
has at least one of those. And that's just a start. Metric spaces are a huge
area of mathematics.

1) Question did not mention distinct points. Two joined equal length segments
with overlapping points qualify.

2) Question did not mention metric used, you assumed euclidean distance.

3) There are many more solutions in multidimensional spaces even just for
euclidean metric. Question did not specify 2D euclidean (but explanation sort
of assumed that) nor even planar. Have fun finding these in hyperbolic
geometry - current model of reality due to General Relativity. Or in quantized
space.

4) Do infinite distances count as distinct? Zero distances are more obvious,
but what about zero limits with different signs?

Mainly etc. Never ask a general question about geometry. ;)

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sonofgod
1) is answered in the article

Clarifications: The four points must be distinct. That is, no point is allowed
to be superimposed on another point.

