Consider an empty spherical bowl of radius $r$. I was trying to find the height to which I would need to fill the bowl with water so that it would be one quarter full (in terms of volume).

The total volume is $\frac{4}{3}\pi r^3$ and the volume filled be water up to a height $h$ is $\pi r h^2 - \frac{1}{3}\pi h^3$. For the bowl to be one quarter full (in terms of volume), $h$ needs to satisfy $h^3 - 3rh^2 + r^3 = 0$.

I couldn't solve this by hand and so used Maple, expecting only a numerical solution. Astonishingly, Maple was able to give two beautiful solutions. The positive one is such that $$\color{blue}{\frac{h}{r} = \sqrt{3}\sin\frac{\pi}{9}-\cos\frac{\pi}{9}+1}$$

I've tried making the substitution $h = \rho\mathrm{e}^{\mathrm{i}\theta}$ into $h^3 - 3rh^2 + r^3 = 0$ and then equating real and imaginary parts. Taking the imaginary parts gives $4\rho\cos^2\theta - 6r\cos\theta - \rho = 0$. This seemed promising, but solving doesn't give anything useful.

I can see that the solution could have been found by solving a general cubic and then simplifying the ghastly expressions by applying de Moivre's formula.

• Is there an elegant and simple way of arriving at the result by hand?
• Under what conditions are such nice solutions possible?
• On the wikipedia page for cubic equations there is a section titled " Trigonometric (and hyperbolic) method" that will probably be useful. I know needham also has a section on this in his visual complex analysis book. – Kai Sikorski Apr 5 '14 at 21:01
• @Sabyasachi Thanks for your kind words. Two people had commented on it and neither had up-voted so I assumed it was a dumb question. (It turned out that Archimedes had worked it out, without calculus, 2300 years ago.) – Fly by Night Apr 6 '14 at 18:19
• @Sabyasachi I've undeleted it. Thank you very much for your comments. – Fly by Night Apr 6 '14 at 18:24
• @FlybyNight okay. I will delete my comments here to clean up. – Guy Apr 6 '14 at 18:25

After some reading around, I think that I have found the solution.

Starting with $h^3-3rh^2+r^3=0$, we make the substitution $h=x+r$ to get the reduced cubic $x^3 -3r^2x-r^3$, or equivalently $x^3 = 3r^2x+r^3$.

Next, we make the substitution $x=2r\cos\theta$ which yields $8r^3\cos^3\theta = 6r^3\cos\theta + r^3$. Since $r>0$ in the context of the original problem, we have $8\cos^3\theta = 6\cos\theta + 1$, or equivalently $$4\cos^3\theta - 3\cos\theta = \frac{1}{2}$$

At this point we apply the identity $\cos3\theta \equiv 4\cos^3\theta - 3\cos\theta$ to give $\displaystyle{\cos3\theta = \frac{1}{2}}$. This gives $$3\theta = \pm\frac{\pi}{3},\pm\frac{5\pi}{3},\pm\frac{7\pi}{3},\pm\frac{11\pi}{3},\cdots$$ $$\theta = \pm\frac{\pi}{9},\pm\frac{5\pi}{9},\pm\frac{7\pi}{9},\pm\frac{11\pi}{9},\cdots$$ Since $x=2r\cos\theta$ and $h=x+r$, the solutions are $$\color{blue}{h = 2r\cos\frac{\pi}{9}+r, \ 2r\cos\frac{5\pi}{9}+r, \ 2r\cos\frac{7\pi}{9}+r}$$ The negative solutions for $\theta$ are not used because $\cos\theta \equiv \cos(-\theta)$. The later solutions for $\theta$ are not used because, due to cosine's periodicity, they start to repeat the same answers.

Note that these three solutions take the same values as the ones in the OP.

Well I guess it depends if the solutions to depressed cubics can be counted as "getting there by hand" (by which I mean using the formula.. I'm not sure how hairy it is to be honest), but if they can, this is how I got to a depressed cubic:

$$h^3-3rh^2+r^3=0$$

$$h^2r(\frac{h}{r}-3)+r^3=0$$

$$\frac{h^2}{r^2}(\frac{h}{r}-3)+1=0$$

Set $x=\frac{h}{r}$

$$x^2(x-3)+1=0$$

$$x^3-3x^2+1=0$$

Since $(x-1)^3 = x^3-3x^2+3x-1$, we can write the above as

$$(x-1)^3-3x+2=0$$

Set $z=x-1$

$$z^3-3z+5=0$$

Then I guess a formula would do from here.

• Thank you for your reply. However, it does not address either of my questions. How does one get from your $z^3-3z+5=0$ to an answer involving sine and cosine? – Fly by Night Apr 5 '14 at 21:42