Let $A_1 A_2 A_3 A_4$ be a regular tetrahedron. Let $P_1$ be the center of face $A_2 A_3 A_4,$ and define vertices $P_2,$ $P_3,$ and $P_4$ the same way. Find the ratio of the volume of tetrahedron $A_1 A_2 A_3 A_4$ to the volume of tetrahedron $P_1 P_2 P_3 P_4.$

  27 : 1

Step-by-step explanation:

The faces of a regular tetrahedron are equilateral triangles. The incenter, circumcenter, and centroid are all the same point, located 1/3 of the distance from the edge to the opposite vertex of the face. The vertical height of the point that is 1/3 the slant height from the base is 1/3 of the height of the tetrahedron.

Then the "inscribed" tetrahedron has 1/3 the height of the original. The ratio of volumes is the cube of the ratio of linear dimensions, so the ratio of the larger volume to the smaller is ...

  3³ : 1³ = 27 : 1

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