Hard geometry probs

for 1. try find the ratio of the side lengths using the pythogorean theorem, and also properties of equilateral triangles

for 2 there's a formula for volume of a tetrahedron knowing only  all six side lengths - it appears in murdereous maths, and just plug the sidelengths in - inefficient to do that tho

for 3 it's the easiest find the side lengths ratio - because the pyramid has 3 45-45-90 triangles, and cube it, multiply by 4, and substract from original cube

Problem 1:

Since ABCD is a regular tetrahedron with volume 18, each of its four triangular faces has the same area. Let this area be K.

Step 1: Finding the Height of Tetrahedron ABCD

The volume of a tetrahedron is given by:

Volume = (1/3) * Base Area * Height

In this case, the base area is K (area of one face), and the volume is 18. We need to find the height (h) of the tetrahedron.

18 = (1/3) * K * h

Solving for h:

h = (18 * 3) / K = 54 / K

Step 2: Relating Tetrahedron ABCD to Pyramid EFGH

Each face of pyramid EFGH is half the size of a face of tetrahedron ABCD. This is because each face of EFGH is formed by connecting the midpoints of the edges of a face in ABCD.

Therefore, the base area of pyramid EFGH is K/2.

Step 3: Finding the Height of Pyramid EFGH

Since each face of pyramid EFGH is the centroid of the corresponding face in ABCD, the height of EFGH (h') is two-thirds of the height of ABCD (h).

h' = (2/3) * h = (2/3) * (54 / K) = 36 / K

Step 4: Finding the Volume of Pyramid EFGH

Using the formula for the volume of a pyramid:

Volume = (1/3) * Base Area * Height

The base area of EFGH is K/2, and the height is 36/K.

Volume of EFGH = (1/3) * (K/2) * (36 / K) = 6

Therefore, the volume of pyramid EFGH is 6.

Problem 3

Imagine we make one such cut through a corner of the cube. Let s be the length of the cut, as shown below.

Since all the edges have the same length, the four triangles meeting at the point of the pyramid must all be equilateral.

Thus the sides of the square at the base of the pyramid all have length s. By the Pythagorean Theorem, s2+s2=62, so s=62​/2=32​.

The volume of each small pyramid is then 31​(32​)2⋅32​=9, so the volume removed from the cube is 8⋅9=72.

The volume of the original cube is 63=216, so the volume of the resulting polyhedron is 216 − 72 = 144​.