Re: what is the max intersections of a trinagle and a 3d cube?


"Dave Eberly" <dNOSPAMeberly@xxxxxxxxxxxxxxx> wrote in message
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"CliffMacgillivray" <nospam@xxxxxxxxxxxxxx> wrote in message
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I think this is related to a clipping problem...
Suppose a triangle intersects a 3d cube to create some polygon P.
What is the max/min number of vertices P can have?

Suppose the plane of the triangle intersects the cube so that
the intersection is a hexagon. Now sketch a picture of the
hexagon and the triangle. You can draw a triangle that
intersects all 6 edges of the hexagon and that contains
3 of the hexagon's vertices. The resulting intersection is
a polygon with 9 vertices.


this is one possible answer, assuming the OP was asking about a cube/plane
intersection (or that the triangle is larger than the cube).


then again, otherwise (clipping the triangle by the cube), one probably
can't have more vertices than this.

so, I guess, assuming any point of the triangle is inside the cube, the min
number of points is 3 (one needs at least a triangle to be a valid
intersection).

most cases, I suspect the likely result is quadrilateral or pentagonal.

more can occure (so, up to the possible max of 9), but I suspect this is
likely only in more obscure situations.

or something...


--
Dave Eberly
http://www.geometrictools.com




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