Somewhere deep in the recesses of my email folder lurks a puzzle that looks simple enough, but that several of my so-inclined friends haven’t found easy:

A circle of radius $r$, has centre $C\ (0,r)$. A tangent to the circle touches the axes at $A\ (9,0)$ and $B\ (0, 2r+3)$. Find $r$.

Now, I have a solution to this, but it involves solving a cubic. This is (of course) possible, but not exactly the done thing in polite society, so I’m especially interested in other solutions.

My approach was this:

Consider the angle, $\theta$ between the $x$-axis and the tangent. $\tan(\theta) = \frac{r}{9}$, while $\tan(2\theta) = \frac{2r+3}{9}$.

Now, $\tan(2\theta) \equiv \frac{2 \tan(\theta)}{1 - \tan^2(\theta)} = \frac{\frac{2r}{9}}{1 - \frac{r^2}{81}}$.

That’s an unholy mess. Let’s tidy it up: $\frac{18r}{81 - r^2} = \frac{2r+3}{9}$

… which means $162r = (2r+3)(81-r^2)$ …

… so $162r = -2r^3 - 3r^2 + 162r + 243$ …

… or $2r^3 + 3r^2 - 243 = 0$

Assuming this has a nice factor, it needs to be something that goes cleanly into $\frac{243}{2}$, which means it’s either a power of 3 or half of a power of 3. It also needs to be between 0 and 9 for the tangent to go somewhere sensible, which gives us a few options: $\frac {1}{2},\ 1,\ \frac{3}{2},\ 3,$ or $\frac{9}{2}$.

Turns out it’s the last one, and the quadratic factor remaining, $r^2 + 6r + 27$, has no real solutions.

After all that, it turns out that this succumbs to Beveridge’s Second Law of MathsJam: if it looks like it might be a 3-4-5 triangle, it’s probably worth checking whether a 3-4-5 triangle works.

* Edited 2016-04-11. Thanks to Mark Ritchings for pointing out a transposition of powers.