Posted June 1, 2014 This question might be related to the 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? 0 Share this post Link to post Share on other sites

0 Posted June 2, 2014 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. 0 Share this post Link to post Share on other sites

0 Posted June 2, 2014 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. 0 Share this post Link to post Share on other sites

0 Posted June 2, 2014 It's probably better to define a rhombus for each direction: 1) pick a point A on the curve and project a line in the chosen direction. 2) For each point of intersection, C, find the midpoint of AC and project a line perpendicular to the chosen direction. 3) Each pair of intersections of the perpendicular with the curve, BD, defines a kite, ABCD. 4) I'm not 100% convinced, but I think you can then pick a continuous sequence of kites where A, C, B and D all move continuously around the curve (each point potentially reversing direction multiple times) so that the line AC sweeps across the entire width of the shape. At one side of the shape, AC will be left of the midpoint of BD; at the other, it'll be right of the midpoint. Since you have a continuous sequence, at some point AC must be the perpendicular bisector of BD and ABCD be a rhombus. Once you have a set of rhombi for each direction, you can again find continuous sequences. If you can find one that returns to the original shape after a net 90-degree rotation, you're done. So the holes are: A) proving you can span the width of the shape with a continuous sequence of (not completely degenerate) kites B) proving you can find a continuous sequence of rhombi that gives a 90 degree rotation There's an additional assumption that the curve is smooth for the continuity arguments to work. Of course, there's also the degenerate solution where you pick the same point 4 times and produce the size-0 square. 0 Share this post Link to post Share on other sites

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This question might be related to the 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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