The regular tetrahedron inscribed in the hemisphere.

Question

Let there be four points in the space region $$x ^ 2 + y ^ 2 + z ^ 2 \leqslant1,z\geqslant 0,$$ and the distance between any two points is $d$, then find the maximum value of $d$ .

I know it is equivalent to finding the length of the regular tetrahedron inscribed in the unit hemi-sphere. And I thinkt he tetrahedron must be "centered" in the hemisphere by intuition, so start off by placing the bottom equilateral on the circle, then calculate the tetrahedron's height. If the height is larger than 1, then find the dilation factor needed to size the tetrahedron down so the tip touches the top-most point of the hemisphere.

And the tetrahedron could also be flipped, with its tip touching (0,0,0). I am not as sure what to do here, but it likely involves taking cross sections.

to get a regular tetrahedron, every other vertex of a regular cube. Cube edge 2 gives: $(1,1,1); ; ; (1,-1,-1); ; ; (-1,1,-1); ; ; (-1,-1,1); ; ;$ — Will Jagy, Jul 08 '21 at 01:00

score 2 · Answer 1 · answered Jul 08 '21 at 11:31

There are three possible candidates:

3 points on the plane $z=0$ and 1 point on a sphere
2 points on the plane 2 points on a sphere
1 point on the plane 3 points on a sphere.

It's easy to see how (1) beats (3) by flipping tetrahedron vertically. It's also possible to show that in case of (2) you can always rotate tetrahedron around the side that lies on a plane to get the same tetrahedron with 3 points on the plane.

Finally, we conclude that the largest tetrahedron is case (1). Can you finish the rest?

yes, I find another way to solve it, just use the height . – Mr.He Jul 09 '21 at 00:14 — Mr.He, Jul 09 '21 at 00:14

The regular tetrahedron inscribed in the hemisphere.

1 Answers1