Wikipedia:Reference desk/Archives/Mathematics/2023 September 29

= September 29 =

For which natural number n, this is true?
For any closed curve in a plane, there are always n points that from the n vertices of a regular n-gon. 118.170.51.132 (talk) 12:49, 29 September 2023 (UTC)


 * If you are confused about “regular n-gon” for n=1 and n=2, just consider the n roots of unity in the complex plane. 118.170.51.132 (talk) 13:04, 29 September 2023 (UTC)
 * For $$n = 1, 2$$ this is trivial.
 * For $$n = 3$$, this is true at least for Jordan curves (and in fact, for any Jordan curve $$C$$ and triangle $$T$$, there are infinitely many triangles similar to $$T$$ that are inscribed in $$C$$; see Inscribed square problem.)
 * For $$n = 4$$, this is not known for Jordan curves (as it is the inscribed square problem) and so it is not known in general, although it is possible that there is some non-simple closed curve that admits no inscribed square.
 * For $$n = 4k$$, the presence of an inscribed regular $$4k$$-gon implies the existence of an inscribed regular $$4$$-gon (i.e. a square) and so the problem cannot have been proven true for $$n = 4k$$ yet. It is possible that in this case some Jordan or non-simple closed curve does not admit an inscribed regular $$4k$$-gon for $$k > 1$$, although naturally that doesn't itself imply that such a curve admits no inscribed square. GalacticShoe (talk) 14:50, 29 September 2023 (UTC)
 * If you can prove that every closed curve admits a Jordan subcurve, then $$n = 3$$ becomes true for every closed curve, while $$n = 4$$ becomes fully reduced to the inscribed square problem. GalacticShoe (talk) 14:21, 5 October 2023 (UTC)
 * The general form of the equation of a conic section as an algebraic curve is given by $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0,$$ which can be normalized such that $$A^2+B^2+C^2+D^2+E^2+F^2=1,$$ so there are 5 degrees of freedom. This means that the 5 vertices of a regular pentagon determine a conic section, which we know will be a circle. Therefore an ellipse that is not a circle does not contain 5 points that form the vertices of a regular pentagon. The same reasoning also excludes $$n>5.$$ --Lambiam 15:52, 29 September 2023 (UTC)
 * Good point. So an ellipse does not contain any 5 concircular points, so a fortiori it doesn't contain the vertices of a regular pentagon. There are an infinite number of conics that pass through the vertices of a square, a circle, ellipses, hyperbolas, and a degenerate case with two lines. No parabolas though, in fact a parabola does not contain the vertices of any parallelogram. --RDBury (talk) 01:03, 30 September 2023 (UTC)

Dual polyhedron of rotundas
What are the dual polyhedron of the rotundas? 118.170.51.132 (talk) 12:55, 29 September 2023 (UTC)


 * The dual of the $$n$$-th rotunda appears to be a polyhedron composed of $$2n$$ diamonds and $$2n$$ triangles, looking like this from the bottom, side, and top respectively, where you have $$n$$ diamonds packed around a top apex point, $$n$$ diamonds forming a ring below those top diamonds, and then $$2n$$ triangles drawn from points of that middle ring to a bottom apex point. GalacticShoe (talk) 15:53, 29 September 2023 (UTC)