Written by Colin+ in ask uncle colin.

Dear Uncle Colin

I wonder: at what height is the volume of a cone above that height equal to the volume below? What about the surface area? Are there any cones where it’s the same height?

Finally Researching Uniformly Splitting Things Up, Mate

Hi, FRUSTUM, and thanks for your message!

Let’s suppose our cone has a height of 1 and radius $r$. If we cut the cone parallel to its base a distance $h$ below the point, the upper cone is similar to the original cone, but with a distance scale factor of $h$. That means its volume scale factor is $h^3$, and half of the volume is below a point $2^{-1/3} \approx 0.8$ from the tip.

*whoosh* The cube root of 2 is very close to 1.26 - obviously…

“Obviously.”

"**Obviously**, $\left( \frac{5}{4} \right)^3 = \frac{125}{64}$, and $\left(k^3 + 3\epsilon \right)^{\frac{1}{3}} \approx k + \frac{\epsilon}{k^2}$.

“… with you?”

"Here, $k = \frac{5}{4}$ and $\epsilon = \frac{1}{64}$, so we have $2^{\frac{1}{3}} \approx \frac{5}{4} + \frac{\frac{1}{64}}{\frac{25}{16}}$

“Which is easy to do in one’s head.”

It’s a hundredth.

“Oh. But how about the reciprocal?”

It’s $\frac{50}{63}$.

“Uh-huh.”

A sixty-third is half of $\left( \frac{1}{7}-\frac{1}{9}\right)$. That bracket is 0.031746 recurring…

“Nice!”

And we want 25 of those…

“So we can multiply by 100 and divide by 4 to get 0.7937 or so.”

Or so being good to one part in a million and a half.

"Good enough for government work.

**Whoosh**

The surface area of a cone of unit height is $\pi r (l + r)$, including the circle on the bottom. If we slice parallel to the base at a distance $h$ from the tip, we have a cone and a frustum. The cone is similar to the original cone, so its surface area is $\pi h^2 r (l + r)$. The frustum has three surfaces: the top circle (with area $\pi h^2 r^2$), the bottom circle (with area $\pi r^2$) and the slanted side (with area $\pi (1-h^2) rl$).

So, with a bit of cancelling, we need $h^2 (l+r) = h^2 r + r + (1-h^2)l$

We can rearrange that: $2h^2 l = r + l$, and we know from trigonometry that $l^2 = 1 + r^2$.

So $h^2 = \frac{1}{2}\left( \frac{r}{\sqrt{1+r^2}} + 1 \right)$.

Well, we know what $h$ is! It’s the cube root of a half. And if we square it and double the equation, we get $2^{\frac{1}{3}} = \frac{r}{\sqrt{1 + r^2}} + 1$. We’re looking for the radius that makes this work.

Letting $k = 2^\frac{1}{3} - 1$, we get $k^2 = \frac{r^2}{1+r^2}$ by squaring everything. This is a rabbit-hole of rearrangement: If we invert everything, $k^{-2} = 1 + r^{-2}$, so $r^2 = \frac{1}{k^{-2} - 1}$.

Alternatively, that’s $\frac{k^2}{1 - k^2}$, and $k^2 = 2^{\frac{2}{3}} - 2^{\frac{4}{3}} + 1$.

We end up with $\frac{1 + 2^{\frac{2}{3}} - 2^{\frac{4}{3}}}{2^{\frac{4}{3}} - 2^{\frac{2}{3}}}$.

Does that simplify? Let’s take $t = 2^{\frac{2}{3}}$ to make life easier: we have $r^2 = \frac{1 + t - t^2}{t^2 - t}$. None of that factorises, but it does simplify a little, to $\frac{1}{t^2 - t} - 1$.

Eyeballing it (sorry, sensei, I’ve got a school run to do), $t^2$ is $2^{\frac{4}{3}}$, which is 2.52, and $2^{\frac{2}{3}}$ is about 1.59, so the bottom is about $1 - 0.07$. The top, then, is about 1.07, so the radius squared is 0.07 and the radius? Somewhere in the region of 0.27.

Hope that helps!

- Uncle Colin