A unit cube is revolved around its body diagonal. What is the maximum distance between two points in the resulting solid?
The distance to beat is the main diagonal from (0,0,0) to (1,1,1) which is sqrt(3) and right from the start I suspect this is the answer.
Want to know how far is the perpendicular distance from that diagonal line to any of the other vertices; then double it.
(Could do an axis modification so the opposite vertices are at (0,0,0) and (0,0,sqrt(3)) and look at the x,y coordinates of one of the other vertices. I don't recall how to do a 3D axis rotation, so I'll try the other way.)
3D line of the main diagonal: x=y=z
A vertex is (0,0,1), closest point on diagonal line is (a,a,a)
minimize dist = √((a0)^2 + (a0)^2 + (a1)^2)
dist^2 = 3a^2  2a + 1
min if 6a  2 = 0, a = 1/3.
dist = √((1/3)^2 + (1/3)^2 + (2/3)^2)
dist = √(6/9) = √6/3
2*dist = 2√6/3
So if we now imagine the cube rotated so the main diagonal is vertical, the widest horizontal distance is:
2√6/3 = 1.6329931618554518 which is less than √3. In this rotated configuration, where opposite vertices are at (0,0,0) to (√3,√3,√3) there are 2 wide cross sections, at z heights of √3/3 and 2√3/3.
If we take a frontal plane through this figure, call the axes u and z, there will be extreme points at (u,z) coordinates of:
(√6/3, √3/3) and (√6/3, 2√3/3).
That distance is √( (2√6/3)^2 + (√3/3)^2 )
= √( (24/9) + (3/9) )
= √(27/9) = √3 which should not surprise me because those two edges of the solid are exactly the result of 2 opposite vertices rotating.
So: √3

Posted by Larry
on 20240810 17:52:55 