Henry Baker <hbaker1@pipeline.com> wrote:
I think that the right way to think about this is to keep the straw stationary, and wiggle a right-angled corner around on the end of it. The corner itself traces out the surface of a hemisphere of radius r sqrt2 around the center of the end of the straw.
Okay, but how to wiggle it so that the point at (X/2,Y/2,Z/2) (in the coordinates of the corner) lies on the axis of the straw? I haven't even figured out how to map from the (three-parameter) position and orientation of the corner to the vertices of the triangle.
But I disagree that you think that these are two different problems. The plane determined by the end of the straw intersects the corner in a triangle, and I contend that the end of the straw is the incircle of that triangle.
You are correct, of course. I misread your earlier statement, "I found the incenter of the triangle (x,0,0), (0,y,0), (0,0,z)...," as referring to the center of the circle through those three points (the circumcenter of the triangle), rather than the center of the circle inscribed in the triangle. Now that I see it, I admire your transformation very much! Tangent planes to tangent lines, axes to vertices.
Question: Does the base of the hemisphere (described above) traced out by the corner form the circumcircle of the triangle? That doesn't seem right, since the circumcircle isn't always concentric with the incircle.
No, and, "Indeed, it couldn't be right."
So what _is_ the geometric significance of a circle sqrt(2) bigger than the incircle?
(and concentric with the given incircle). Answer: The circumcircle of a square with that incircle. For when the corner is in the plane of the incircle, the triangle has a right angle at the corner (and you get to pick any hypotenuse tangent to the other side of the incircle). Dan Hoey@AIC.NRL.Navy.Mil