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Is the hyperbolic or spherical analogy of the following Euclidean fact, true?

Two triangles with equal corresponding medians are congruent.

More precisely: Assume that $\Delta ABC$ and $ \Delta A'B'C'$ are two triangles in the hyperbolic space $\mathbb{H}^2$ or elliptic space $\mathbb{S}^2$ such that $$AM_1=A'M_1',\; BM_2=B'M_2',\;CM_3=C'M_3'$$ where $M_i (M_i')$ in $XM_i(X'M_i')$ is the mid point of the edge opposite to the vertex $X(X')$, respectively.

Under this condition, is there an isometry of the corresponding space carrying the first triangle to the second one?

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1 Answer 1

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This is not true for $S^2$. Namely, one can construct two convex non-isometric spherical triangles such that all medians have length $\frac{3\pi}{4}-\varepsilon$.

1) The first triangle is the standard equilateral triangle. Clearly medians of such triangles all have the same length and their length vary from $0$ to $\pi$.

2) The second triangle will be a perturbation of the following one. Take on $S^2$ two opposite points $A$ and $B$ and join them by two geodesics (of length $\pi$) that bound a sector with angle $\frac{3\pi}{4}$. Call one of these geodesics $AB$. And choose $C$ as the mid point of the other geodesic. It is easy to see that for this degenerate triangle all medians have length $\frac{3\pi}{4}$. Now, by perturbing slightly $A$, $B$ and $C$ one can decrease the lengths of all medians by the same amount.

Note. It is important in the second construction that by varying $A$ and $B$ as little as we want, we can achieve that the median starting from $C$ takes any length in $(0,\pi)$. However the length of two other medians stay close to $\frac{3\pi}{4}$.

PS, hyperbolic case. Concerning the hyperbolic case, it looks like the answer is positive. If I would like to prove it, I would do as follows.

i) Consider the map from the convex cone in $\mathbb R_{\ge 0}^3$ consisting of triples $(a,b,c)$ satisfying the (non-strict) triangle inequality to itself: lengths of sides $\to$ lengths of medians.

ii) Realise that this map if proper (preimage of compact is compact). And it is an isomorphism on the boundary on the cone.

iii) The map is a "diffeo" close to the point $(0,0,0)$ - because it is so for Euclidean triangles.

iv) This is the most complicated bit -- show that the map has non-vanishing differential. This is where one needs to make a calculation. But it looks very plausible.

v) If all the above holds then the map is an isomorphism. QED

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