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If you have an arbitrary system of linear equations and add to it one equation, the dimension of the space of solutions can decrease by at most 1. Do take any $d$ equations of your hyperplanes, order …

This paper is contained in the 1-st volume of Aleksandrov's Selected works. This has an English translation: Alexandrov, A. D. Reshetnyak, Yu. G. (ed.) Selected works. Part 1: Selected scientific pape …

Your broken line closes iff the triangles $(v_1,v_2,v_3)$ and $(v_3,v_4,v_5)$ are equal, and $[v_1,v_3]$ is their common side. So there are two conditions: one is that $m_1=m_3$, by the hyperbolic law …

Denote the diameter by $d$ and distance by $|x-y|$. Then there are $y,z$ such that $d=|y-z|$ and we have by triangle inequality for every $x$: $$d=|y-z|\leq |y-x|+|x-z|\leq 2f(x),$$ so we obtain your …

I think this may help: Lemma. For any two convex sets, and any vector $x$, the function $F(t)={\mathrm{Vol}}\left(\mathrm{conv}\left(K\cup(L+xt)\right)\right)$ is convex, as a function of real variab …

All that is needed is an estimate of the length of the piece of $\gamma$ in the disk $|z|\leq r$. This can be done as follows. Remove a bounded arc of $\gamma$ so that two unbounded pieces remain, an …

The necessary and sufficient conditions are: $(1/n)\sum x_j=(1/3)(a+b+c)$, and all $x_j$ lie inside the closed triangle $(a,b,c)$. Proof of sufficiency. For every affine function, we have equality. L …

This has nothing to do with circles (or polygons). If one convex curve is inside another, then the length of the inner curve is smaller.