On Shortest Paths Between Two Convex Polyhedra

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On Shortest Paths Between Two Convex Polyhedra
Avikam Baltsan
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Thus Cg splits e into at most two remaining portions, each of which is visible from every point on g to which e is not (totally) obscure.
Proof: Let /t/, (resp, fiJj) denote the faces of A (resp. B) incident to g (resp. E), and let 11^, 11/ (resp. TliJlj) be the two planes containing /t/; (resp. FiJj) . By convexity of A, rijt (resp. FI/) is a supporting plane of A. Let Hi- (resp. Hf) denote the closed half space bounded by Fljt (resp. IT/) whose interior is disjoint from A, and let H^ (resp, H
...f) denote the complementary open space.
Suppose for the moment that the faces of B were transparent, so that e could be seen behind them. Three cases can arise: (a) e lies entirely in H^dHf. In this case e is obscure to g.
(b) e lies entirely in H^UHf. In this case e is visible from g.
(c) e crosses one or both planes 11^, 17, . In this case e is split into at most three subsegments, one of which {eg = er\H^nHf) is obscure to g, and the other one or two (e — eDH^nHf) are visible from g.
So far we have assumed that the faces of B are transparent, but obviously they are not.


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