Three mutually tangent spheres of
radius 2 cm rest on a horizontal plane. A sphere of radius 3 cm rests on them.
What is the distance from the plane to the top of the larger sphere? Your
answer should be in either simplest radical form or as a decimal rounded to the
nearest 1000^{th}.

SOLUTION

Figure 1 is an approximate side
view showing two of the smaller spheres with the larger sphere on top. The line length we’re solving for is the
total of the three segments shown in dark blue, light green, and pink.

The dark blue segment equals the
radius of the smaller sphere.

The pink segment equals the radius
of the larger sphere.

The length of the light green
segment is the tricky part of this problem.

The light green segment shown in
this view is only an approximation because the larger sphere is sitting in the
indentation of the three smaller balls.
Hence, the red triangle is really leaning at an angle (see Figure 3)
with its base between the two small spheres and its apex leaning back to a line
perpendicular to the tabletop and running through the center of the three
smaller spheres.

So, in Figure 3 let’s see what we know.

The red line represents a side view of the red triangle in Figure 1. However, it is not equal to the length of a side of the triangle. In Figure 3, the red line equals the altitude of the red triangle in Figure 1. The red triangle is an isosceles triangle. Its base equals the length of the distance between the centers of the two small spheres and the two equal sides measure the length of the small triangle plus the length of the large triangle.

From this information, I leave it to you to calculate the length of the altitude.

The light blue line in Figure 3 is represented in Figure 2. Figure 2 is a top-down view of the three small spheres. The green triangle is an equilateral triangle joining their centers. The three green internal lines are the perpendicular bisectors of the three angles.

How can we calculate the length of the light blue line in this triangle? Here’s what we know. The base of this triangle equals the radius of the small sphere. Since the green triangle is an equilateral triangle, each angle is 60 degrees. The hypotenuse is the perpendicular bisector of that angle, so the small angle opposite the light blue line is 30 degrees, which means the line opposite equals one half the hypotenuse. If the length of the light blue line is X, then we can set up this equation.

X^{2} + 2^{2} = (2X)^{2}

I leave it to you to solve for X, which is the length of the light blue line.

Finally, in Figure three, you can now solve for the length of the light green line since you should now know the length of the other two lines.

And that’s all there is to it.

~Zeyda

The SPHERES Answer to the nearest 1000^{th}
is 9.726 cm.