Trig-gono-metry-1

Let $AP =x$ and $\angle CPB = \theta$ . Therefore, $\angle APD = 2 \angle CPB = 2 \theta$ . Then, we have:

$\begin{cases} \tan 2\theta = \dfrac 3x \\ \tan \theta = \dfrac 3{6-x} \end{cases}$

$\begin{aligned} \tan 2 \theta & = \frac {2 \tan \theta}{1-\tan^2 \theta} \\ \implies \frac 3x & = \frac {\frac 6{6-x}}{1- \left(\frac 3{6-x}\right)^2} \\ & = \frac {6(6-x)}{(6-x)^2-9} \\ \implies \frac 3x & = \frac {36-6x}{x^2-12x+27} \\ \frac 1x & = \frac {12-2x}{x^2-12x+27} \\ x^2-12x+27 & = 12x - 2x^2 \\ 3x^2 - 24x + 27 & = 0 \\ x^2 - 8x + 9 & = 0 \\ \implies x & = \frac {8 \pm \sqrt{28}}2 \\ & = 4 - \sqrt{7} & \small \color{#3D99F6}{ 4+\sqrt 7 > 6 \text{ rejected.}} \end{aligned}$

$\implies a+b+c = 4+1+7 = \boxed{12}$