Given two planar systems X'=F(X) and X'=G(X) (so F and G are both $C^1$). Assume the dot product of F(X) and G(X) is always zero on $R^2$. Now if F has a closed orbit, prove that G has a zero.
My attempt:
The first thing pops up in my mind is a circle (the closed orbit of F). On this circle, G(X) is perpendicular to F(X), and then there must be an equilibrium point inside the circle. However I cannot generalize my idea.
P.S.: this should be an easy exercise after learning the Poincare-Bendixson theorem.
If $G$ has an equilibrium on the closed orbit of $x'=F(x)$, then you are done. Otherwise, along the closed orbit of $x'=F(x)$ the vector field $G$ points always inside or always outside, in view of the orthogonality of the two vector fields. So, by the Poincaré-Bendixson theorem, there is a zero of $G$ inside the closed orbit.