4 spheres in a tetrahedron

• Jan 20th 2010, 07:35 AM
4 spheres in a tetrahedron
Three spheres, A, B and C, each 1 inch in diameter, are arranged on a horizontal flat surface so that they touch each other and their centers form an equilateral triangle. A fourth sphere D, also 1 inch in diameter, is placed on top of the other three so as to form a regular tetrahedral configuration (it touches each of the other three spheres).

(a) What is the distance between the centers of the spheres?
(b) How far above the flat surface is the center of sphere A?
(c) How far (vertical distance) above the center of sphere A is the center of sphere D?

Well I drawed the 4 spheres in the arrangement of a tetrahedron in a cube to help me out. for (a) I said the distance is just half the diagonal of the cube i.e.1 inch for (b) its just the radius of sphere A i.e. 0.5 inch.
For (c) I said its the radius of sphere A plus the radius of sphere D which would be 1 inch as well. But it seems much more complicated than this, can anyone check for errors?
• Jan 20th 2010, 07:48 PM
Pulock2009
• Jan 20th 2010, 07:51 PM
Pulock2009
b)half inch because the distance is nothing but the radius which is half the diameter
c)1 inch. (3/2 inch -1/2 inch)
• Jan 21st 2010, 12:15 AM
Opalg
Quote:

For (c), you already know (from (a)) that the centres of the four spheres form a regular tetrahedron with side 1 inch. You then need to know that the height of such a tetrahedron above its base is $\displaystyle \sqrt3/6$ inches (see here for a good explanation of that).