The mutilated tetrahedron

[bonanova]

A solid in the shape of a regular tetrahedron had uniform density and a mass of 1 kg. It was mutilated by the removal of its vertices, each made by a planar cut, parallel to its opposite face. The solid now has eight faces, whose areas are 1, 2, 3, 4, 5, 6, 7 and 8 square units, in some order. What is the area of its original faces? What is the mass of the resulting solid?

[bonanova]

Perhaps begin by deciding how to control the area of (at least three of) the resulting faces when the vertices are truncated by planar cuts. It's a nice result, by itself.

[plasmid]

This assumes that the area left over from each of the initial faces is greater than the area of any face formed by a cut vertex,

Call the original faces A, B, C, and D. Before any cuts were made, we of course know that areas A = B = C = D = initial area. If you make a cut from the vertex opposite of face A, it will remove some area E from each of B, C, and D. Likewise for each of the other sides: the cut opposite B will remove area F from each of {A, C, D}, the cut opposite C will remove area G from {A, B, D}, and the cut opposite D will remove area H from {A, B, C}. So the areas of the initial faces after the cutting will be

1) A = initial - F - G - H
2) B = initial - E - G - H
3) C = initial - E - F - H
4) D = initial - E - F - G

The cuts will each remove a small tetrahedron from the initial large tetrahedron, so the areas of the new faces that are generated by the cuts will be equal to the area that they remove from the adjacent faces. So the areas of the new sides of the tetrahedron formed by cutting are simply
5) E
6) F
7) G
8) H

Assign the values 1-4 to areas E-H, and you get
A = initial - 2 - 3 - 4 = initial - 9
B = initial - 1 - 3 - 4 = initial - 8
C = initial - 1 - 2 - 4 = initial - 7
D = initial - 1 - 2 - 3 = initial - 6

Since the values A-D must be 5-8, the initial area of the faces must have been 14.
For an equilateral triangle we have
area = sidelength² * sqrt(3)/4
sidelength = sqrt[area * 4/sqrt(3)]
and the volume of a tetrahedron is
volume = edgelength³ / (6 * sqrt(2))
So the initial tetrahedron would have volume
sqrt[14 * 4/sqrt(3)]³ / (6*sqrt(2)) ~= 21.6658

The volumes of each of the four cut vertices would be
sqrt[1 * 4/sqrt(3)]³ / (6*sqrt(2)) ~= 0.4136
sqrt[2 * 4/sqrt(3)]³ / (6*sqrt(2)) ~= 1.1698
sqrt[3 * 4/sqrt(3)]³ / (6*sqrt(2)) ~= 2.1491
sqrt[4 * 4/sqrt(3)]³ / (6*sqrt(2)) ~= 3.3088

So the final volume after cutting would be 14.6245, or 67.5% of the original volume, for a mass of 675 grams from the initial 1 kg tetrahedron.

[bonanova]

This assumes that the area left over from each of the initial faces is greater than the area of any face formed by a cut vertex,

 

It probably does, but I had not thought about that point.

If we create all the 5 6 7 8 sides by truncation, would that chew up the original solid?

Could any one or two of 5 6 7 8 be a side created by truncation?

 

Upon reflection the OP should also have asked the amount by the surface area was decreased.

Since the area created at each vertex is also removed from its three associated faces,

The answer is, trivially, 2 x (1+2+3+4) = 20.

[plasmid]

You should be able to make a cut face be any proportional size compared to an initial face, but whether (and how) you could make all of the cut faces be larger than the remainder of the initial faces is not clear.

For what it's worth: if you cut off the vertices such that each plane of cutting is halfway between a vertex and its opposite face, that would create an octohedron where the remainder of the original faces are indistinguishable from the newly cut faces. If you take an octohedron and start shaving away one of its faces, I think that face might get larger before it gets smaller again, but I haven't done the math.