## Tetrahedrons

My puzzle coffer is low. I've a few submissions to read through. So here are a few quick problems for you. On a regular tetrahedron of side-length 1:
1. From the midpoint of one edge, two lines are drawn along the surface to the two edge-midpoints of one side. What is the angle between these lines?
2. From the midpoint of one edge, two lines are drawn along the surface to the edge-midpoints of different sides. What is the angle between these lines?
3. What is the farthest surface distance possible between any two points?
4. What is the shortest distance between two non-adjacent edges?
5. Remove one tetrahedron from each corner, each with side-length of 1/2. For the figure created:
1. What is it called?
2. Connect two opposite vertices via the shortest surface line. What is the angle formed by this line?
3. Each of the removed tetrahedra were 1/8 the volume of the large one. How many more 1/8 tetrahedra can be cut from the remaining figure, in no more than one piece each?
If you know of other good questions to add to this list, send them. I'll add them as their received.

Source: Original, though influenced by many past readings.

Solutions were received from Joseph DeVincentis:
1. 1. 60 degrees. The three such lines on one face form an equilateral triangle.
2. 2. Either 60 or 90 degrees. (The lines can be either two edges of one of the new triangular faces revealed step 5, or part of a square which bisects that figure (and the original tetrahedron).
3. 3. This is tricky. I'm going to say it's the distance from one vertex to the center of the opposite face, which is 2/sqrt(3) or about 1.15, but I'm not entirely certain there's not a better solution using, say, 2 points each located halfway between a vertex and the center of a face.
4. 4. Since there's only one edge not adjacent to any given edge, there's no real choice here. The distance is the altitude of a triangle two of whose edges are face-altitudes, and the third edge is an edge of the tetrahedron. So, it's the other leg of a right triangle with short leg 1/2 and hypotenuse sqrt(3)/2, or sqrt(2)/2.
5. 5.
1. a. It is a regular octahedron.
2. b. The surface line consists of two face-altitudes, and half of the angle involved is the angle in a right triangle of the same dimensions we just solved in part 4, whatever that comes out to.
3. c. I think you can cut just one, possibly sharing a face with the octahedron. Because the square pyramids which make up an octahedron are flatter than a tetrahedron, we can't put one in each of two opposite corners; they'd overlap.

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