# The ratio of the ball's volume to the volume of the cone

I've been struggling to solve this problem and I'm still unable to get the right answer. I keep getting a ratio that is bigger than one and I'm not sure what I'm doing wrong.

The question is the following: The largest possible ball is placed inside a right circular cone. What is the ratio of the ball's volume to the cone's volume, when the cone's radius is $r$ and the slant height is $2r$?

• What do you get for the radius of the ball? – almagest Jan 27 '18 at 17:12
• I got 3r/(2+√3) – Pöytä Laatikko Jan 27 '18 at 17:14
• That looks wrong to me. How did you get it? – almagest Jan 27 '18 at 17:16
• I drew a triangle inside the cone and used similarity to calculate the radius of the ball: – Pöytä Laatikko Jan 27 '18 at 17:18
• a/h=(h-a)/2r , where a is the radius of the ball and h is the height of the cone (I used Pythagoras theorem and got r*√3). – Pöytä Laatikko Jan 27 '18 at 17:23

A section of the cone is an equilateral triangle and the inradius of an equilateral triangle equals $\frac{1}{2\sqrt{3}}$ times its side length. It follows that the volume of the cone is $\frac{\pi r^2}{3}\cdot\sqrt{3} r=\frac{\pi}{\sqrt{3}}r^3$ and the volume of the largest inscribed sphere is $\frac{4\pi}{3}\left(\frac{r}{\sqrt{3}}\right)^3 = \frac{4\pi}{9\sqrt{3}}r^3.$ The wanted ratio is so $\color{red}{\frac{4}{9}}$.

• How did you get the radius of the sphere? – Pöytä Laatikko Jan 27 '18 at 17:28
• The radius of the sphere is the inradius of an equilateral triangle with side length $2r$. – Jack D'Aurizio Jan 27 '18 at 17:29
• My method was the following: I drew a triangle inside the cone and used similarity to calculate the radius of the ball. a/h=(h-a)/2r , where a is the radius of the ball and h is the height of the cone (I used Pythagoras theorem and got r*√3). With these calculations I got 3r/(2+√3) as the radius of the sphere. What did I do incorrectly? – Pöytä Laatikko Jan 27 '18 at 17:40
• You used a similarity between which triangles? The centroid and the incenter of an equilateral triangle are the same point, so the inradius is just one third of any height. – Jack D'Aurizio Jan 27 '18 at 17:43
• I cut the cone in half and drew a line from the senter point of the sphere to the side of the cone. This formed a new triangle which is similar to the big one (the half cone). – Pöytä Laatikko Jan 27 '18 at 17:50

One can't argue with Jack's solution, but it would be more instructive to show how the solution can be carried out for an arbitrary right-circular cone. In fact, we show below how the ratio of the cone to the sphere is dependent upon a single variable, here shown in terms of the cone vertex half-angle. First of all, I set it up in the plane, as if to be a volume of revolution. The figure below shows the layout. The trick here is to know that the incenter of the triangle shows where to place largest inscribed circle. You can look it in Wikipedia to find that the distance from point A to point I is given by

$$d(A,I)=c\frac{\sin \gamma}{\cos \gamma}=c\tan \gamma$$

Then we can say that radius of the circle/sphere is

$$r=h-d(A,I)=c~(\cos \alpha-\tan \gamma)$$

and we also note that

$$h=c\cos \alpha\\ R=c\sin \alpha$$

Now we can put together the ratio of the cone to the sphere as follows:

\begin{align} \frac{V_{cone}}{V_{sphere}} &=\frac{\frac13\pi R^2h}{\frac43 \pi r^3}\\ &=\frac{\frac13\pi ~c^2\sin^2\alpha ~c\cos\alpha}{\frac43 \pi c^3 (\cos\alpha-\tan\gamma)^3}\\ &=\frac14 \frac{\sin^2\alpha ~\cos\alpha}{(\cos\alpha-\tan\gamma)^3}\\ \end{align}

Finally, we note that $\gamma=\pi/4-\alpha/2$, we can write

\begin{align} \frac{V_{cone}}{V_{sphere}} &=\frac14 \frac{\sin^2\alpha ~\cos\alpha}{\left[\cos\alpha-\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\right]^3}\\ \end{align}

I have verified this result numerically for arbitrary vertex angles.