An interesting one:

A triangle is right-angled at O with its legs parallel to the axes. Its hypotenuse passes through the point with coordinates $(a,b)$; what is the shortest possible hypotenuse?

It’s interesting (to me) because there are several different approaches; all of them sort of threaten to turn out nicely, but I don’t have a really elegant solution.

I’ll share one solution below the line.

Suppose the angle between the hypotenuse and the $x$-axis is $\theta$.

Then (after a bit of sketching), the length of the hypotenuse is $h = a \sec(\theta) + b \cosec(\theta)$. All we need to do is differentiate that and set it to zero.

$\diff{h}{\theta} = a \sec(\theta)\tan(\theta) - b \cosec(\theta) \cot(\theta) = 0$

Multiply everything by $\sin(\theta)\tan(\theta)$ and rearrange((I actually mutiplied by $\cos^2(\theta)\sin^2(\theta)$ and then divided by the $\cos^3(\theta)$ that turned up.)):

$\tan^3(\theta) = \frac{b}{a}$, so $\tan(\theta) = \sqrt[3]{\frac{b}{a}}$.

That’s a bit ugly. I’m going to let $B = \sqrt[3]{b}$ and $A = \sqrt[3]{a}$, and let $C = \sqrt{A^2 + B^2}$, mainly to save on LaTeX.

So, $\tan(\theta) = \frac{B}{A}$, which means $\sec(\theta) = \frac{C}{A}$ and $\cosec(\theta) = \frac{C}{B}$.

The hypotenuse is $h = \frac{aC}{A} + \frac{bC}{B}$. We know that $\frac{a}{A} = a^{2/3}$ and similarly for $B$; we can also write $C$ as $\sqrt{a^{2/3} + b^{2/3}}$.

So we now have $h = \sqrt{a^{2/3} + b^{2/3}}\left(a^{2/3} + b^{2/3}\right)$, or $\left(a^{2/3} + b^{2/3}\right)^{3/2}$.

I think it’s neater to write it as $h^{2/3} = a^{2/3}+b^{2/3}$, which looks sort of Pythagoras-shaped, but I can’t see an immediate link to a triangle with cube-root-flavoured sides.

Any other methods? I’d love to hear about them.