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This sort of addresses the question in your last paragraph, but it is actually a bit tangential. Hopefully it is still of interest. It is well known that every planar graph has an embedding such t …

Regarding III, the Alexandrov-Urysohn Theorem gives sufficient conditions. Any zero-dimensional, separable, nowhere compact, and completely metrizable space is homeomorphic to $J$.

If the length divided by the width is rational, then yes. Just partition the rectangle into congruent squares and cut each square along a diagonal.

For hyperbolic geometry, you can find the answers to your questions on Wikipedia. An interesting thing is that there is an absolute upper bound on the area of a hyperbolic triangle, even though lengt …

There is a wide body of work on this in connection with the classic De Bruijn–Erdős theorem. De Bruijn–Erdős Theorem. Every set of $n$ points in the plane (not all lying on the same line) deter …

Terry Tao has a nice blog post on a 'cheap version' of the Kabatjanskii-Levenstein bound mentioned in Lucia's answer, using the so-called 'tensor product trick'.

Note that the answer is yes in 2 dimensions, since any polygon can be triangulated (without adding additional vertices). Thus, every point in the interior sees at least 3 vertices of $P$. One can …

Yes. See this paper of Negami. The main result is that for any fixed knot (or link) of type $k$, there is a constant $R(k)$ such that every straight line embedding of $K_{R(k)}$ in $\mathbb{R}^3$ con …

Consider the complete bipartite graph $G$ with bipartition $(A,B)$, and let the weight of an edge $ab$ be $d(a,b)$. Then $d(A,B)$ is simply the weight of a minimum weight perfect matching of $G$. F …

As Boris Bukh points out, three points suffice, but I'd like to point out that your question is related to this MO question. Here is a summary of the information in the previous question. For the …

I think the answer is no for both questions. Let $T$ be the unique tree on $2n$ vertices with two adjacent vertices $u$ and $v$ of degree $n$. Let $e=uv$. Let the weight of $e$ be $n^2$ and all oth …

Yes, there are non-square rectangles that admit a perfect squaring. The smallest number of squares in a perfect squaring of a rectangle is 9. On the other hand the smallest number of squares in a per …

Pilipczuk and Siebertz proved that every planar graph has such a partition with an even stronger property. Namely, each part $V_i$ is a geodesic path, and the graph obtained by contracting each part …

This is an answer to your last question. As far as I know, it is still open whether there exists a dense subset $S$ of the plane with all pairwise distances rational. Such a set $S$ would imply a po …

The answer to the third question is no. This is a rather counter-intuitive discovery of Micha Perles from the sixties. See this paper of Ziegler, for a simpler construction and other pertinent infor …