Trisected Triangle Perimeter ICTM State 2017 Division AA Precalculus Individual Question 17 reads:
$C$ and $D$ lie on $BE$ such that $AC$ and $AD$ trisect $\angle$BAE in $△ABE$. $BC = 2$, $CD = 3$, and $DE = 6$. Then the perimeter of $△ABC$ may be expressed as $P = f + k \sqrt{w} + p \sqrt{q}$ in simplified and reduced radical form. Determine the sum $(f+k+w+p+q)$. See figure below

What I tried.
I tried to create a system of equations equating the angles of the three triangles to 180°. I also equated the supplementary angles.
There were not enough equations to solve for every variable.

Let $AB=x$.
Thus, since $$\frac{AD}{AB}=\frac{CD}{BC},$$ we obtain
$$\frac{AD}{x}=\frac{3}{2},$$ which gives
$$AD=\frac{3}{2}x.$$
Also, let $AC=y$.
Thus, since $$\frac{AE}{AC}=\frac{DE}{DC},$$ we obtain $AE=2y$.
But, $$AC^2=AB\cdot AD-BC\cdot CD$$ and
$$AD^2=AC\cdot AE-CD\cdot DE,$$ which gives
$$y^2=\frac{3}{2}x^2-6$$ and
$$\frac{9}{4}x^2=2y^2-18.$$
After solving of this system we obtain: $x=2\sqrt{10},$ $y=3\sqrt{6}$, $P=11+2\sqrt{10}+6\sqrt{6},$ 
which gives the answer: $35$.
