Q:

Property of compact, convex sets in $\mathbb{R}^3$

How to solve the following:

Let $K\subset \mathbb{R}^3$ be a convex, compact set with smooth boundary $C=\partial K$ and let $\vec{u}$ be any vector. Show that there exist points $x\neq y$, $x,y\in C$ such that vector $\vec{xy}$ is colinear with $\vec{u}$ and tangent planes $T_{x}C$, $T_{y}C$ are parallel.

Any hint would be helpful, because I don't have any idea how to start.
Thanks in advance.

A:

It seems I find solution. Let $\phi: C \rightarrow C$ is the involution, which maps point x to its opposite point y by vector u.
Let's consider the following vector field on C:
$$v(x) = pr_{T_x} [u, n(x) + n(\phi(x))]$$ where n(x) is the normal vector for C in point x with length one. By the Hairy ball theorem it's exists the point x for which $v(x)=0$. This is the required point.
To proof this we need to check that if $v(x) = pr_{T_x} [u, n(x) + n(\phi(x))]=0$ then $n(x) + n(\phi(x)) = 0$. 
Let $a = n(x)$, $b = n(\phi(x))$. 
Suppose that $pr_{T_x}[u, a+b] = 0$. Then vector $v = [u, a+b]$ is parallel to vector a, and if v is nonzero then vector a is orthogonal to vector u, because v is orthogonal to u. But if vector a is orthogonal to u then u belongs to the tangent space $T_x$ of surface C. In this case a = b and $pr_{T_x}[u, a+b] \neq 0$. We have contradiction. Thus $v = [u, a+b] = 0$, so $a + b = 0$.      

