The construction below achieves $2^{c n}$ for constant $c$ (e.g., $c=\frac{1}{3}$). But it is flawed in that I have two triangles sharing a vertex and paths passing through that vertex. It is natural to insist that the triangles be disjoint. I think the same basic construction can accomplish a $2^{c n}$ bound by (a) separating the touching triangles, while (b) arranging a detour that makes a turn at the center of the X exactly the same length as shooting straight through on a diagonal. But I have not carried through the construction in detail.
     Many Shortest Paths http://cs.smith.edu/%7Eorourke/MathOverflow/ShortestPathsMany.jpg

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Update. Here is an exponential construction using disjoint triangles, where the marked lengths satisfy $a=b+c$. Length $d$ is shared by all paths. Here there are $2^2=4$ paths, and $2^{1+(n-1)/5}$ for $n$ triangles.
     ![Many Shortest Paths][2]

The construction below achieves $2^{c n}$ for constant $c$ (e.g., $c=\frac{1}{3}$). But it is flawed in that I have two triangles sharing a vertex and paths passing through that vertex. It is natural to insist that the triangles be disjoint. I think the same basic construction can accomplish a $2^{c n}$ bound by (a) separating the touching triangles, while (b) arranging a detour that makes a turn at the center of the X exactly the same length as shooting straight through on a diagonal. But I have not carried through the construction in detail.

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     ![Many Shortest Paths][1]

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Update. Here is an exponential construction using disjoint triangles, where the marked lengths satisfy $a=b+c$. Length $d$ is shared by all paths. Here there are $2^2=4$ paths, and $2^{1+(n-1)/5}$ for $n$ triangles.

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      [![Many Shortest Paths][2]][2]

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The construction below achieves $2^{c n}$ for constant $c$ (e.g., $c=\frac{1}{3}$). But it is flawed in that I have two triangles sharing a vertex and paths passing through that vertex. It is natural to insist that the triangles be disjoint. I think the same basic construction can accomplish a $2^{c n}$ bound by (a) separating the touching triangles, while (b) arranging a detour that makes a turn at the center of the X exactly the same length as shooting straight through on a diagonal. But I have not carried through the construction in detail.
     Many Shortest Paths http://cs.smith.edu/%7Eorourke/MathOverflow/ShortestPathsMany.jpg

The construction below achieves $2^{c n}$ for constant $c$ (e.g., $c=\frac{1}{3}$). But it is flawed in that I have two triangles sharing a vertex and paths passing through that vertex. It is natural to insist that the triangles be disjoint. I think the same basic construction can accomplish a $2^{c n}$ bound by (a) separating the touching triangles, while (b) arranging a detour that makes a turn at the center of the X exactly the same length as shooting straight through on a diagonal. But I have not carried through the construction in detail.
     Many Shortest Paths http://cs.smith.edu/%7Eorourke/MathOverflow/ShortestPathsMany.jpg

---

Update. Here is an exponential construction using disjoint triangles, where the marked lengths satisfy $a=b+c$. Length $d$ is shared by all paths. Here there are $2^2=4$ paths, and $2^{1+(n-1)/5}$ for $n$ triangles.
     ![Many Shortest Paths][2]