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. I've also been racking my brain trying to come up with a way to "see" this without any algebra, but nothing has yet come to mind. This is a very cool fact. 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 wonder if this has any meaning for a circularly polarized wave hitting a corner reflector... Now that many tests allow the use of calculators, I wonder if they're ready to start allowing access to Macsyma/Mathematica/Maple/... on tests? At 06:22 PM 7/3/03 -0400, Dan Hoey wrote:
Meanwhile, I'm intrigued by the straw problem. So far, the best I've done is to express the length as sqrt(tt+uu+vv) where (t,u,v) satisfy equations
(x-t)^2 (tt+uu+vv) = 4 r^2 (uu+vv) (y-u)^2 (tt+uu+vv) = 4 r^2 (tt+vv) (z-v)^2 (tt+uu+vv) = 4 r^2 (tt+uu)
in the box dimensions (x,y,z) and the straw radius r. Adding them up, I'm startled (possibly not for the first time) that 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. Is there an easy geometrical proof?
But I'm pretty much flummoxed about solving the whole system. Maybe I should have gone to high school in New York.
Dan Hoey@AIC.NRL.Navy.Mil