3D Geometry Contest Math Problem

Question

The problem is as follows:

Six solid regular tetrahedra are placed on a flat surface so that their bases form a regular hexagon H with side length 1, and so that the vertices are not lying in the plane of H (the "Top" vertices) are themselves coplanar. A spherical ball of radius $r$ is placed so that its center is directly above the center of the hexagon. The sphere rests on the tetrahedra so that it is tangent to one edge from each tetrahedron. If the ball's center is coplanar with the top vertices of the tetrahedra, compute $r$.

I know that the height of the "top" vertices is $\sqrt{\frac{2}{3}}$ but i'm not sure how to calculate the point where the sphere centered in the center of the hexagon at height $\sqrt{\frac{2}{3}}$ touches the tetrahedrons edges (the edges have to face inward). How would I find this point?

Answer 1

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Answer 2

As far as Asimov's question, it can be restated in 2D as: For an equilateral triangle of side = 1, how can you prove that the distance from the centroid to any vertex is sqrt(1/3)?

Divide the triangle symmetrically by drawing one of its medians (from a vertex to the midpoint of opposite edge). Euclid proved that the centroid (intersection of 3 medians) lies at the 2/3 distance along each median. From trig. of 30-60-90 triangle of hypotenuse = 1, we know that the longer leg is sqrt(3/4). So just take 2/3 of sqrt(3/4).