Is it true that, if a closed, strictly convex curve has exactly four vertices (extrema of curvature), then any circle has at most four points of intersection with it?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ what is the question? $\endgroup$– Francesco PolizziCommented Oct 27, 2018 at 18:49
-
$\begingroup$ I understand that the Moon's trajectory is strictly convex with respect to the Sun, yet it intersects the Earth's orbit about 26 times. I'm not sure what a "vertex" of a curve is. $\endgroup$– Gerry MyersonCommented Oct 27, 2018 at 23:40
-
2$\begingroup$ @GerryMyerson: a vertex of a smooth curve is an extremum point of the curvature. This question is related to the classical "four vertex theorem": every smooth closed convex curve has at least four vertices. $\endgroup$– Ivan IzmestievCommented Oct 28, 2018 at 9:45
-
$\begingroup$ Thanks, @Ivan. That makes my Moon comment totally irrelevant. $\endgroup$– Gerry MyersonCommented Oct 28, 2018 at 11:12
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Yes, this is true. More generally, if a smooth closed strictly convex curve intersects some circle in $2n$ points, then it has at least $2n$ vertices. This is stated in Blaschke's book "Kreis und Kugel" at the end of the Appendix on vertices of curves. For a proof, see
Jackson, S. B., Vertices of plane curves, Bull. Am. Math. Soc. 50, 564-578 (1944). ZBL0060.34909.