Volume of spheres inscribed in a cone.

Question

There are five perfectly spherical scoops of ice cream with various radii placed inside a waffle cone. Each scoop of ice cream is in contact with the adjacent scoop of ice cream. Also, each scoop of ice cream comes in contact all around the waffle cone wall. If the radius of the smallest ice cream scoop is 8 and the radius of the largest ice cream scoop is 18, find the volume of the middle ice cream scoop.

My intuition is that the radius should be the geometric mean of 18 and 8 (which is 12), but I don't know why. Is it really the geometric mean and why/why not?

Also, this isn't homework.

Answer 1

Extend the cone to its tip. Consider the following $2$ shapes:

Shape $1$: The cone, from the tip up to the place where it meets the second scoop from the bottom, together with the bottom scoop and the second scoop.

Shape $2$: The cone, from the tip up to the place where it meets the third scoop from the bottom, together with the second scoop from the bottom, and the third scoop from the bottom.

Shapes $1$ and $2$ are similar. Let $r$ be the scaling factor that gets us from the bottom scoop of Shape $1$ to the top scoop of Shape $1$. Then scaling again by the factor $r$ gets us to the top scoop of Shape $2$. And scaling twice more by the factor $r$ gets us to the top scoop in the diagram of the OP.

Thus $\frac{18}{8}=r^4$. It follows that $r^2=\frac{3}{2}$, and therefore the middle scoop has radius $(8)(3/2)$. Now that we have the radius, the volume is easy to write down.

Remark: We could do basically the same argument by drawing a bunch of similar triangles. But that would take longer, and anyway I wanted to give a pure scaling argument.

The argument above shows that the radius is indeed the geometric mean of $12$ and $18$. Precisely the same argument shows that if the bottom scoop has radius $a$ and the top scoop has radius $b$, then the middle scoop has radius $\sqrt{ab}$.

The same argument shows that if the bottom scoop as surface area $A$, and the top scoop has surface are $B$, then the middle scoop has surface area $\sqrt{AB}$.

The same argument shows that if the cone is not right-circular, again the middle scoop has radius $\sqrt{ab}$. The same applies to waffle cones that have the shape of an upside-down pyramid.

Answer 2

Since I'd never seen the similarity argument, I decided to work one out.

Following the "walls" of the cone to its apex, we can construct similar triangles using one wall and the symmetry axis of the cone. We will call the distance from the apex to the surface of the smallest sphere $ \ Y \ $ , and the radii of the three smallest spheres $ \ r_1 \ , \ r_2 \ , \ $ and $ \ r_3 \ $ . (It will be sufficient to work with just the first three spheres, since the argument is easily extended to any number of spheres in a "stack".)

Since the cone is the result of revolution of a straight line, the "slope" of the wall relative to the symmetry axis is constant (this is also the tangent value for half of the "opening angle" of the cone). This permits us to describe similar triangles, for which this tangent value is

$$ \frac{r_1}{Y \ + \ r_1} \ = \ \frac{r_2}{Y \ + \ 2r_1 \ + r_2} \ = \ \frac{r_3}{Y \ + \ 2r_1 \ + 2r_2 \ + \ r_3} \ \ . $$

From pairing the first two ratios, we have

$$ r_1 Y \ + \ 2 \ r^2_1 \ + \ r_1 r_2 \ = \ r_2 \ Y \ + \ r_1 r_2 \ \ \Rightarrow \ \ Y \ (r_2 - r_1) \ = \ 2 \ r_1^2 \ \ . $$

Pairing the second two ratios, and using our result for $ \ Y \ $ , we then obtain

$$ r_2 \ Y \ + \ 2 \ r_1 r_2 \ + \ 2 \ r_2^2 \ + \ r_2 r_3 \ = \ r_3 \ Y \ + \ 2 \ r_1 r_3 \ + \ r_2 r_3 $$

$$ \Rightarrow \ \ r_2 \ \left( \frac{2 \ r_1^2}{r_2 \ - \ r_1} \right) \ + \ 2 \ r_1 r_2 \ + \ 2 \ r_2^2 \ = \ r_3 \ \left( \frac{2 \ r_1^2}{r_2 \ - \ r_1} \right) \ + \ 2 \ r_1 r_3 $$

$$ \Rightarrow \ \ 2 \ r_1^2r_2 \ + \ 2 \ r_1 r_2^2 \ - \ 2 \ r_1^2r_2 \ + \ 2 \ r_2^3 \ - \ 2 \ r_1 r_2^2 \ = \ 2 \ r_1^2r_3 \ \ + \ 2 \ r_1 r_2 r_3 \ - \ 2 \ r_1^2 r_3 $$

[multiplying through by $ \ r_2 \ - \ r_1 \ $ and canceling like terms]

$$ \Rightarrow \ \ 2 \ r_2^3 \ = \ 2 \ r_1 r_2 r_3 \ \ \Rightarrow \ \ r_2^2 \ = \ r_1 r_3 \ \ . $$

So our intuition concerning a scaling argument is correct, and the geometric mean relation between radii (and so of surface areas and volumes) of contiguous spheres follows from the walls of the cone having constant slope and the spheres being in direct contact with one another.