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This question is cross-posted in MO and MSE https://math.stackexchange.com/questions/2276064/about-pairwise-distances-of-some-points-in-a-riemannian-manifold-m-of-rm-se

Assume that there are points $p_i,\ 1\leq i\leq m$ in a Riemannian manifold $(M,d)$ s.t.

(1) all sectional curvatures are equal to or greater than $1$ and

(2) $$ d(p_i,p_j)> \frac{\pi}{2},\ i\neq j $$ Then there is a set of points $\eta_{ij},\ i\neq j$ s.t. $$ d(p_i,\eta_{ij}) < \frac{\pi}{2} <d(p_j,\eta_{ij}) $$ and $$ d(p_l,\eta_{ij} )=\frac{\pi}{2} $$ for all $l$ not in $\{ i,j\}$.

How do we prove this ? Thank you in advance.

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(a) $\angle p_ip_jp_k > \frac{\pi}{2}$

(b) Note that if there exists such $\eta_{ij}$, then $|\{ \eta_{ij} \}|=m(m-1)$.

(c) $m=3$ Case is solved : Consider a geodesic triangle $[p_1p_2p_3]$. Define $\eta_{31}\in [p_2p_3]$ s.t. $d(p_2,\eta_{31} )= \frac{\pi}{2}$. By applying Toponogov theorem to a hinge at $p_2$, $d(p_1,\eta_{31}) > \frac{\pi}{2}$.

(d) My difficulty is to find $\eta_{m(m-1)}$

By an induction, there is $q$ around $p_m$, which may help to find $\eta_{m(m-1)}$, s.t. $$l_i:=d(q,p_i)=\frac{\pi}{2},\ 1\leq i\leq m-3,\ L:=d(q,p_{m-3}) > \frac{\pi}{2} $$

Hence there is a point that increases $l_i$ and diminishes $L$

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We can assume that the set $\{p_1,\dots,p_n\}$ is maximal. Note that the set $$K_{n,n-1}=\{\,x\in M\mid |x-p_i|\ge\tfrac\pi2\ \text{for}\ i<n-1\,\}$$ is convex and it contains a point $z$ such that $|p_i-z|=\tfrac\pi2$ for any $i<n$.

Since $\{p_1,\dots,p_n\}$ is maximal, $|p_n-z|\le \tfrac\pi2$. By comparison, we get $$\measuredangle[z\,^{p_i}_{p_j}]>\tfrac\pi2$$ for all $i\ne j$.

Flow $z$ in the gradient flow for $\mathrm{dist}_{p_{n-1}}$ in $K_{n,n-1}$ for short time. We get a point $z'$ which will satisfy all your conditions for $\eta_{n,n-1}$.

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