Proof that a compact surface has a tangent plane orthogonal to position

Given a compact surface $S$, is it true that $S$ has a tangent plane that is orthogonal to the position vector for at least one of its points?

I believe that the statement is true, but I'm having trouble writing a formal proof. My idea is that, evaluating the position and tangent vectors at a starting point and moving it along the surface until we get back to the start, the dot product must be zero somewhere. I don't know if this is completely right or even how to write this down properly, would appreciate some help.

Solution 1:

Hint

Suppose that the origin $O \notin S$ and consider the map defined on $S$ by $\Vert x \Vert^2$. What happens for a point $a$ such that $f$ is minimum at $a$?