Can you solve it? The four points, two distances problem | Hacker News Reader

Can you solve it? The four points, two distances problem

theguardian.com

16 pointsmkl12 comments

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.

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.

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.

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.

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.

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

an equilateral triangle with a dot in the center?

Edit:

Note to self: read articles before commenting

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

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

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. ;)

I deleted my list of spoiler solutions.

But you are correct. I was only thinking in 2D euclidean space. Khaaaaaan!

Your other ideas are very creative, and as far as I understand them, correct (except for the idea of four DISTINCT points), but probably outside the problem. How many solutions would there be on a hyperbolic plane? No idea.

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.

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