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Hilbert space satisfies the following condition: if two triangles $\triangle ABC$, $\triangle A_1B_1C_1$ have equal sides lengths: $|AB|=|A_1B_1|$, $|BC|=|B_1C_1|$, $|AC|=|A_1C_1|$ they also have equal medians lengths: $|AM|=|A_1M_1|$, where $M=(B+C)/2$, $M_1=(B_1+C_1)/2$. Are there other Banach spaces with this property? (If yes, there are 2-dimensional examples, and the next question is complete classification. But I doubt.)

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Firstly, notice that you may apply your condition to the triangles $ABM$ and $A_1B_1M_1$ and iterate this process. So, e.g.. for every points $X$ and $X_1$ on $BC$ and $B_1C_2$ with $BX/BC=B_1X_1/B_1C_1=k/2^n$ with integer $k,n$ we obtain $AX=A_1X_1$. Since the binary rationals are dense on $[0,1]$, the same holds for every ratio. Similarly, we may apply this to other sides and hence see that the affine transform mapping $ABC$ to $A_1B_1C_1$ is metric-preserving on the whole triangle $ABC$. THis already should suffice.

Indeed, consider an equilateral triangle $ABC$; by an affine transform we may assume that it is equilateral also in the Euclidean metric. Take a triangle $A'B'C'$ inscribed into $ABC$, with $A'$, $B'$, $C'$ dividing the sides of $ABC$ at the same ratio. Then $A'B'C'$ is also regular (in both metrics), and under the affine transform mapping $ABC$ to $A'B'C'$ the metric scales at the coefficient $\mu$ depending only on the ratio at which the vertices of $A'B'C'$ divide the sides of $ABC$.

If the sides of $A'B'C'$ are rotated at $\pi/n$ with respect to the sides of $ABC$ (angles are also Euclidean) then we may iterate inscribing such triangles $n$ times getting a triangle homothetic to $ABC$ with a known ratio. Since the metric on this triangle is $\mu^n$ times the initial one, we obtain that $\mu$ is the same for both metrics. Thus the ratios of segments rotated at $\pi/n$ is the same in our metric and Euclidean one. This is what we want.

For passing to an arbitrary space, see Anton Petrunin's answer...

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In dimension 2, your condition implies existence of rotation by arbitrary small angle. Simply apply your condition recurcevely to a sequence of points on the unit circle $\dots,A_0,A_1,A_2,\dots$ such that the midpoint of $[A_{i-1}A_{i+1}]$ lies on $[OA_i)$. The later implies that the metric is Euclidean.

Now in your space any 2-dimesional subspace is Euclidean. In particular the parallelogram identity holds. The later implies that this is a Hibert space.

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  • $\begingroup$ I am confused. What are triangles to which we apply it? $\endgroup$ – Fedor Petrov Sep 30 '15 at 13:26
  • $\begingroup$ @FedorPetrov $\triangle A_{i-1}OA_{i+1}$. $\endgroup$ – Anton Petrunin Sep 30 '15 at 20:47
  • $\begingroup$ But why $A_{i-1}A_{i+1}$ does not depend on $i$? $\endgroup$ – Fedor Petrov Sep 30 '15 at 21:19
  • $\begingroup$ @FedorPetrov, I was solving a more general problem (did not read the formulation carefully). I thought your condition holds only if $AB=AC$. This is likely the reason of your confusion. These equalities are easy to see, but it seems that you are satisfied with the other solution :) $\endgroup$ – Anton Petrunin Sep 30 '15 at 22:09

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