In a right triangle $ABC$ (right-angled at $B$), $D$ and $E$ are points of $\overline{AB}$ and $\overline{BC}$ respectively such that $\overline{CD}$ and $\overline{AE}$ are the angle bisectors of the acute angles of the triangle. Given that $AE=9$ and $CD=8\sqrt{2}$, find the length of the hypotenuse $\overline{AC}$.
Let $\alpha = \angle BAE$. Let $\beta = \angle BCD$. What we want is $AC = \sqrt{AB^2 + BC^2}$. In other words, we want to solve for $\alpha, \beta, AB, BC$ such that
$$AB = 9 \cos\alpha \quad\quad BC = 8\sqrt{2} \cos\beta \quad\quad \alpha + \beta = \frac{\pi}{4} \quad\quad \frac{BC}{AB} = \tan 2\alpha$$
From the last equation we have
$$\begin{align*} \frac{8\sqrt{2}\cos\beta}{9\cos\alpha} &= \tan 2\alpha \\ \frac{8\sqrt{2}}{9} \cdot \frac{\cos(\pi/4 - \alpha)}{\cos\alpha} &= \tan 2\alpha\\ \frac{8\sqrt{2}}{9} \cdot \frac{(\cos\alpha + \sin\alpha)/\sqrt{2}}{\cos\alpha} &= \tan2\alpha \\ \frac{8}{9}(1+\tan\alpha) &= \frac{2\tan\alpha}{1 - \tan^2 \alpha} \\ 4(1 -\tan^2\alpha + \tan\alpha - \tan^3 \alpha) &= 9\tan\alpha \\ 4\tan^3 \alpha + 4\tan^2 \alpha + 5\tan\alpha - 4 &= 0 \end{align*}$$
At this point we have a cubic equation to solve. But worry not: using the rational root theorem we can search for rational roots relatively easily. It turns out that $\tan\alpha = 1/2$ is a root. Is this the only solution we want? Let's factorise the equation:
$$(2\tan\alpha - 1)(2\tan^2 \alpha + 3\tan\alpha + 4) = 0$$
The discriminant of the quadratic factor is $3^2 - 4\cdot 2 \cdot 4 < 0$. Indeed, $\tan\alpha = 1/2$ is the only real root.
We find $AB = 18/\sqrt{5}$, $BC = 24/\sqrt{5}$. So $AC = 6\sqrt{5}$.
