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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Posted 02 June 2014 - 09:52 AM
Draw a circle. Inscribe a square ABCD, at best AC being horizontal. Move the points A and C some degrees towards D, delete the arc AC. Move D some 4 units from A. You get a kind of hot air baloon or an ice cream cone.
#define RAC = Remaining Arc of the Circle
- 3 points on the RAC: 4th point must be on the circle, at least 90 degrees missing;
- 3 points on the cone: 4th point cannot be on the cone (lines not parallel) nor on the RAC;
- 2 points on the cone and 2 points on the RAC: draw two parallel lines, resulting segments can never have the same length
It is hard to inscribe a square into a regular pentagon. I conjecture that many polygons will fit.
Posted 02 June 2014 - 01:00 PM
Posted 02 June 2014 - 03:37 PM
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