Let $(M^n,g)$ be a connected Riemannian manifold with boundary, $n \geq 2$. A chord of $M$ is a non constant geodesic $c : [a,b] \to M$ with endpoints $c(a), c(b) \in \partial M$ but $c(t) \in \operatorname{int} M$ for $a < t < b$. Let
$$\ell = \ell(M,g) = \inf \{ \operatorname{length}(c) : c \text{ is a chord of } M \}.$$
Assuming that $M$ is compact, it it true that $\ell(M,g) > 0$? Is it achieved by a chord of $M$?
Now let
$$SC(M,g) = \{ \gamma : [0, \ell] \to M \text{ chord of } M \text{ of length } \ell(M,g) \}.$$
(SC stands for "shortest chords"). Does there exist $\gamma \in SC(M,g)$ which meets $\partial M$ orthogonally at the endpoints?
For instance, let $(M,g)$ be the upper hemisphere of the unit round sphere in $\mathbb{R}^3$. Then $\ell = \pi$ and every chord has the same length. In particular, through every point at the boundary there passes exactly one shortest chord which is orthogonal to the boundary.
It is false that $\ell(M, g) > 0$ under these hypotheses. To see this, consider $(\mathbb{D}, g_{\mathbb{D}})$ to be the closed unit disk in the plane with the Euclidean metric. Taking nearly-vertical straight lines through the point $(1,0)$ gives a counter-example. The salient difference between this example and the half-sphere is that the boundary of $\mathbb{D}$ is concave, while the boundary of the half-sphere is a geodesic.