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. 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. 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. So what _is_ the geometric significance of a circle sqrt(2) bigger than the incircle? At 04:14 AM 7/7/03 -0400, Dan Hoey wrote:

About my remark that

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any way you lean a disc of radius r on the floor against two walls, the center will be sqrt(2)r from the corner.

Henry Baker <hbaker1@pipeline.com> says:

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You are absolutely right! I checked this using a symbolic algebra package on the plane to Boston today. From one of my previous forays, I already had a good method for finding the incenter of a circle in 3D, so I found the incenter of the triangle (x,0,0), (0,y,0), (0,0,z). I then found the distance^2 to the origin. I then divided by the radius^2 of the incircle, and voila'! The ratio is exactly _2_ ! Taking the square root, we get a ratio of sqrt(2) for the ratio of the distance to the radius, as you point out.

Well, that is even more startling, because it's a solution to a different problem! Your calculation is of a disc supported by three perpendicular lines; mine is of a disc supported by three perpendicular planes. Does this mean what I suspect it means--that if you support the disc on three perpendicular planes, the lines from the points of tangency to the origin must be mutually perpendicular? I'm becoming motivated to make sure I get this solved, just to verify that conjecture.

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A man in the street could understand this -- nail one end of a string into a corner of a room. Take a large circular table with a small hole in the center, and put the string through the center. Now push the table into the corner, and pull the string taut. You can move the table around -- up to and including flat against one wall, and you won't break the string, and the string won't become slack either!

I thought of a gadget something like that, only with a rigid strut holding the disc center to the corner. Then you don't have to watch it to know it doesn't go loose. Of course, neither of these tethers lets you turn the disc over, as you can in theory. (If the above plane-support=line=support conjecture is true, then you could also mount a tripod of perpendicular struts from the corner so that they meet the disc at the points of tangency.)

I wonder if this linkage has any useful qualities as a somewhat universal joint? Say one that provides heavy-duty support (from the disc resting on the planes) and light-duty tension (from the strut). The strut can be connected with ball-and-socket joints on both ends, since it isn't subjected to great force. I don't know if there's a better way to support the disc than with sliding edges. If the planes are backed off, a carriage could be inserted that slides on the plane and accepts the disc in a groove. Optionally, a hole through the carriage could maintain alignment with the tripod leg, if that conjecture works out. This would be interesting just to watch.

Meanwhile, I'm still trying to solve the original problem (which is the key to all the other stuff.) I looked at the two-dimensional problem to see if I could get anywhere. I think I've seen rectangles inscribed in rectangles before, but I'm not sure where, so I worked on it from scratch. Anyway, for an a x b rectangle is inscribed in a c x d rectangle, I get

0 = (a^2 - b^2)^2 - (ad - be)^2 - (ae - bd)^2

which means solving a quartic (unless I've overlooked some simplication.)

(I find it interesting that this equation can also be written

0 = (a^2 - b^2)^2 - (a^2 + b^2) (d^2 + e^2) + 4 abde

which is only quadratic in the reparameterization of the rectangles from (width,length) to (hypotenuse squared,area). And then the two kinds of parameters are nicely segregated from each other, though the a^2b^2 term is of ambiguous kind.)

I still haven't worked out a way of using this to solve the three-dimensional problem.

Dan Hoey@AIC.NRL.Navy.Mil