Geodesics on $S^2$ Correct description of this problem is here: Geodesics on $S^2$ with specific Riemannian metric
I was asked to show the following:

Given a Riemannian metric on $S^2$, let $x \in S^2$ be the north pole, and let $V(\theta) \in T_xS^2$ be unit tangent vectors at $x$. Then there exists a constant $t > 0$ such that $\exp_x(tV(\theta))$ is the south pole for all $\theta$. 

The conclusion in this statement is really not intuitive for me, and I am not sure how to approach this problem. Could someone give a hint on this?
 A: Hints: What are the geodesics of a round unit sphere?
Start at a point $x$ of a round unit $2$-sphere and walk at unit speed. How long does it take to reach the antipode $-x$?

By contrast, consider an arbitrary metric on a $2$-sphere. As a physical analogy, imagine the surface of a squash or a gourd, where the center of the stem end is the north pole and the south pole is the center of the blossom end. It's not obvious that a geodesic starting at the north pole ever reaches the south pole. Certainly, different geodesics between the poles have different lengths.
A: Definition. Let $(M,g)$ be a Riemannian manifold and $p, q\in M$ aretwo points. The pair $(p,q)$ is called a wiedersein pair if every geodesic through $p$ also passes through $q$. 
Given your comments to Andrew Hwang's answer, your precise question is the following:
Assuming that the north pole - the south pole pair $(p,q)$ in $S^2$ is a wiedersein pair, show that all simple geodesic arcs from $p$ to $q$ have the same length. 
To prove this, consider the 1-parameter family of geodesics (parameterized by arclength) $\gamma_\theta(t)$ with $\gamma_\theta(0)=p$, $\gamma_\theta'(0)=V(\theta)$ as in your question. Let $L(\theta)$ denote the least value of $t$ such that $\gamma_\theta(L(\theta))=q$. Then $L(\theta)$ is the length of $\gamma_\theta|[0,L(\theta)]$. Recall that every geodesic  between two points $p, q$ (in any Riemannian manifold) is a critical point of the length function restricted to the space of all smooth curves connecting $p$ and $q$. Therefore, the derivative $L'(\theta)$ is identically zero. Hence, since the circle is connected, $L$ is constant. qed 
