Given two points $y,z$ on the circle, find $x$ on the shortest arc between $y$ and $z$ such that $|x-y|^2+|x-z|^2$ is minimized. The Langrange conditions give $-y-z+\lambda x=0$ which yields $x=\frac{y+z}{|y+z|}$. Note that the minimizer satisfies $|x-y|=|x-z|$.
I want to generalize this result to a manifold embedded in a high-dimensional sphere, $M\subset S^d\subset R^{d+1}$ for some large $d$ (e.g. the Stiefel manifold or the Grassmannian). Given $y,z\in M$ (with $|y|=|z|=k$ in the Frobenius norm), let $T=\{x\in R^{d+1}|\,|x-y|=|x-z|\}$. Under what conditions on $M$ can we conclude that there is a local solution to $\min_{x\in M}|x-y|^2+|x-z|^2$ which belongs to $T$? It is true when $M$ is a circle (see above). When $M$ is $S^n$ it is true for $x=\frac{y+z}{|y+z|}$, but not the other points of $T$.
Is it true when $M$ is the Stiefel manifold $V_{p,n}=\{X\in R^{n\times p}|X^TX=I_p\}$? Using the Lagrange conditions I get $-Y-Z+X\Lambda=0$. Then $\Lambda=-X^T(Y+Z)$. Assuming $X,Y$ and $Z$ are sufficiently close to each other, we get that $\Lambda$ is nonsingular. Then $X=(Y+Z)[X^T(Y+Z)]^{-1}$. I can't seem to get further. Perhaps there is some geometric argument?