MathDB
congruent tetrahedra

Source: Polish MO Recond Round 1978 p5

September 9, 2024
3D geometrygeometrytetrahedron

Problem Statement

Prove that there is no inclined plane such that any tetrahedron placed arbitrarily with a certain face on the plane will not fall over.
It means the following:
Given a plane π \pi and a line l l not perpendicular to it. Prove that there is a tetrahedron T T such that for each of its faces S S there is in the plane π \pi a triangle ABC ABC congruent to S S and there is a point D D such that the tetrahedron ABCD ABCD congruent to T T and the line parallel to l l passing through the center of gravity of the tetrahedron ABCD ABCD does not intersect the triangle ABC ABC .
Note. The center of gravity of a tetrahedron is the intersection point of the segments connecting the centers of gravity of the faces of this tetrahedron with the opposite vertices (it is known that such a point always exists).