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This problem trains for: SAT-I, GMAT, AMC-8, AMC-10.

Three spheres of diameter 6 units must be packed in a rectangular box (parallelepiped) with dimensions: 6 x 10 x 11.5 cubic units. What is the approximate volume (in cubic units) of filler pellets that must be used in order to fill the box?

Is there a way in which we can place the spheres so as to make them fit in the box?

It is obvious that the spheres will not fit if they are placed as if each was encased in a cube. However, it is possible to 'pack the spheres' as in the figure:

To calculate the dimensions of the smallest box the spheres would fit in, we 'enclose' the group of spheres with a rectangle that would be the bottom of the box. The height of the box (not represented in the figure since the figure is a projection onto the bottom plane) would be equal to 6 units.

What are the length and width of such a box?

We use the fact that, if two circles/spheres are tangent (exterior or interior) then the point of contact is on the same line (is collinear) as the two centers.

Therefore, by connecting the centers of the three spheres through the points of contact, we obtain an equilateral triangle with side 6 units.

The height of this triangle is (use Pythagorean theorem, or the 30° -60° -90° triangle):

One side of this rectangle is :

and the other side is:

Therefore, this rectangle fits entirely in the box with dimensions 12 x 11.5 square units.

Now that we are sure that the spheres fit in the given box, we can find the volume of filler pellets by subtracting from the volume of the box, the sum of the volumes of the three spheres.