This question might be related to the four-legged stool. Or not.

Draw a simple (no crossings) closed (not necessarily convex) curve on a sheet of paper.

Can you always find four points on the curve that form the four vertices of a square?

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Guest Message by DevFuse

Started by bonanova, Jun 01 2014 06:51 AM

3 replies to this topic

Posted 01 June 2014 - 06:51 AM

This question might be related to the four-legged stool. Or not.

Draw a simple (no crossings) closed (not necessarily convex) curve on a sheet of paper.

Can you always find four points on the curve that form the four vertices of a square?

- Bertrand Russell

Posted 02 June 2014 - 09:52 AM

Spoiler for

Posted 02 June 2014 - 01:00 PM

Informally, the proof is to take a square and let 2 opposite corner A and C run around the loop. Doing that, you rotate the square while trying to fit the 2 other corners B and D within the loop. If you start with A and C on a horizontal line, and B and D can fit within the loop, then after 1/4 turn, A and C are on a vertical line, where BD originally was and now B and D are outside of the loop.

So, at some point (for some angle between AC and the horizontal), you go from "B and D both fit in the loop" to "B and D fit outside the loop". At that point, it should be possible to fit both B and D on the loop.

There are plenty of loose ends, but it makes me feel such a square must exist.

Posted 02 June 2014 - 03:37 PM

It's probably better to define a rhombus for each direction:

Spoiler for

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