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

[CliffMacgillivray]

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?

[Dave Eberly]

"CliffMacgillivray" <nos...@fake-email.net> wrote in message
news:JAFHh.299$m%.230@newsfe19.lga...

>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.

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

[cr88192]

"Dave Eberly" <dNOSPA...@usemydomain.com> wrote in message
news:bRIHh.27446$7g3....@newsfe14.phx...

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...

[CliffMacGillivray]

Thanks to all who replied.
I also ran this problem past some friends and
I think that we are all in agreement about 9 as the max.
Now, I'm curious, is there any proof out there in the literature that this
is definitely the max? I certainly can't think of how more could be
possible, then again, I had previously convinced myself that the max
was 6. ;)

[Dave Eberly]

"CliffMacGillivray" <nos...@fake-email.net> wrote in message
news:ZpMHh.7485$C_....@newsfe22.lga...

Generally, what is the maximum number of vertices of the
intersection of a triangle and an N-sided convex polygon?

The essence of a proof:

Start with the N vertices of the polygon. If a triangle edge
intersects a polygon edge at a single point that is interior on
both edges, you gain one vertex. One triangle-edge endpoint
is outside the polygon. If the other is outside, then you gain
a second vertex. In both cases, you lose one vertex from the
polygon. So you gain 1 or 2 and lose 1. Worst case is that the
net gain is 1 vertex per triangle edge, so the intersection
polygon has at most N+3 vertices.

I leave the details of making this a rigorous proof to you.