[Math] Maximum number of vertices in intersection of triangle with box

geometrypolygonspolyhedra

Suppose we have a triangle and a box in 3D. The intersection of the triangle with the (solid) box will be a polygon with some number of vertices (possibly zero). The vertex count will vary according to the precise shapes and positions of the triangle and box.

How can we establish the maximum number of vertices that this polygon could have, for any possible triangle and box?

Intuitively, an upper bound is 9 vertices because each plane of the box can add at most one vertex to the polygon, and the box has 6 planes. But I'm not sure if this is a sharp upper bound; the problem's difficult enough to visualize that I can't convince myself either way. Is there a geometry theorem that answers this?

Best Answer

An answer emerged from a couple of the comments, observing that

  1. The problem can be reduced to finding the intersection of the box with the triangle's plane, which produces a polygon, followed by the 2D problem of intersecting that polygon with the triangle.
  2. It's possible to cut a box with a plane in such a way that the intersection is a hexagon.

One can easily see that intersecting a triangle with a hexagon can give a nonagon, since each of the triangle's three edges can cut off one vertex of the hexagon. So the upper bound of 9 vertices is fulfilled in such a case.

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