Sphere Inscribed in Two Cones

In this proof, I will refer to:

• the volume of the sphere as $v$ ,
• the radius of the sphere as $r$ ,
• the volume of the two cones as $V$ as given in the question,
• the radius of the base of the cones as $s$ (because $r$ is already taken), and
• the height of each cone as $h$ .

First, we will get the dimensions of the sphere in terms of a single variable $s$ .

From the formula for the volume of a cone: $V=2\times\left(\frac{1}{3}\pi s^2 h\right)\\ h=\frac{3V}{2\pi s^2}$ Shown here is a cross-section of part of the sphere and cones: We can easily prove that $\triangle ABD$ is similar to $\triangle ADC$ , because of the right angle and that $\angle CAD$ is common.

$AC=\sqrt{s^2+h^2}$ from the Pythagorean theorem, and we can now use the rule that corresponding sides in similar triangles are in the same ratio: $\begin{aligned}\frac{BD}{AD}&=\frac{DC}{AC}\\ \frac{r}{h}&=\frac{s}{\sqrt{s^2+h^2}}\\ r&=\frac{sh}{\sqrt{s^2+h^2}}\\ r&=\frac{s\frac{3V}{2\pi s^2}}{\sqrt{s^2+\left(\frac{3V}{2\pi s^2}\right)^2}}\\ r&=\frac{3Vs}{\sqrt{4\pi^2s^6+9V^2}}\end{aligned}$ Intuitively, the maximum possible radius of the sphere would give the maximum possible volume as well, so we can now do calculus to our $r$ variable. and substitute it into the formula for the volume of a sphere later. To do maxima and minima, we have to take the derivative, using the quotient and chain rules: $\begin{aligned}\frac{dr}{ds}&=3V\left(\frac{\sqrt{4\pi^2s^6+9V^2}-s\frac{1}{2}\frac{1}{\sqrt{4\pi^2s^6+9V^2}}\left(24\pi^2s^5\right)}{4\pi^2s^6+9V^2}\right)\\ &=3V\left(\frac{\left(4\pi^2s^6+9V^2\right)-s\frac{1}{2}\left(24\pi^2s^5\right)}{\left(4\pi^2s^6+9V^2\right)\sqrt{4\pi^2s^6+9V^2}}\right)\\ &=3V\left(\frac{4\pi^2s^6+9V^2-12\pi^2s^6}{\sqrt{\left(4\pi^2s^6+9V^2\right)^3}}\right)\\ &=3V\left(\frac{9V^2-8\pi^2s^6}{\sqrt{\left(4\pi^2s^6+9V^2\right)^3}}\right)\end{aligned}$ Now we let the derivative equal zero to find a maximum or minimum turning point. The bottom of the fraction is irrelevant for this: $\begin{aligned}3V\left(\frac{9V^2-8\pi^2s^6}{\sqrt{\left(4\pi^2s^6+9V^2\right)^3}}\right)&=0\\ 9V^2-8\pi^2s^6&=0\\ s^6&=\frac{9V^2}{8\pi^2}\\ s^3&=\frac{3V}{2\sqrt2\pi}\\ s&=\sqrt[3]{\frac{3V}{2\sqrt2\pi}}\end{aligned}$ If we let $V$ equal a manageable number like $1$ , We can now check whether this is a maximum or minimum by doing a first derivative test:

Let $V=1$ , $∴s=\sqrt[3]{\frac{3}{2\sqrt2\pi}}=0.6963\dots$

Since the function is increasing when $s<\sqrt[3]{\frac{3V}{2\sqrt2\pi}}$ and decreasing when $s>\sqrt[3]{\frac{3V}{2\sqrt2\pi}}$ , we can conclude that $s=\sqrt[3]{\frac{3V}{2\sqrt2\pi}}$ is a maximum turning point.

Now that we know that $s=\sqrt[3]{\frac{3V}{2\sqrt2\pi}}$ gives us the maximum radius of the sphere, we can now substitute this into our radius formula: $\begin{aligned}r&=\frac{3Vs}{\sqrt{4\pi^2s^6+9V^2}}\\ &=\frac{3V\sqrt[3]{\frac{3V}{2\sqrt2\pi}}}{\sqrt{4\pi^2\left(\sqrt[3]{\frac{3V}{2\sqrt2\pi}}\right)^6+9V^2}}\\ &=\frac{3V\sqrt[3]{\frac{3V}{2\sqrt2\pi}}}{\sqrt{4\pi^2\left(\frac{9V^2}{8\pi^2}\right)+9V^2}}\\ &=\frac{3V\sqrt[3]{\frac{3V}{2\sqrt2\pi}}}{3V\sqrt{\frac{1}{2}+1}}\\ &=\frac{\sqrt[3]{\frac{3V}{2\sqrt2\pi}}}{\sqrt{\frac{3}{2}}}\\ &=\sqrt[3]{\frac{3V}{2\sqrt2\pi}}\times\sqrt[3]{\frac{2\sqrt2}{3\sqrt3}}\\ r&=\sqrt[3]{\frac{V}{\pi\sqrt3}}\end{aligned}$ Finally, we can use the formula for the volume of a sphere: $\begin{aligned}v&=\frac{4}{3}\pi r^3\\ &=\frac{4}{3}\pi\left(\sqrt[3]{\frac{V}{\pi\sqrt3}}\right)^3\\ &=\frac{4}{3}\pi\frac{V}{\pi\sqrt3}\\ &=\frac{4}{3\sqrt3}V\\ &=\frac{4\sqrt3}{9}V\end{aligned}$ So $a=4$ , $b=3$ and $c=9$ , so the answer is $4+3+9=\boxed{16}$