Trisect Me

Let $\theta = \angle ACG = \angle GCB = \angle BCD$ .

$AE$ is the height to the hypotenuse of right $\triangle ABC$ , thus, $\begin{aligned} & A{{C}^{2}}=CE\cdot CB\Rightarrow 36=CE\cdot \left( CE+1 \right) \\ & \Rightarrow C{{E}^{2}}+CE-36=0 \\ & \Rightarrow CE=\dfrac{-1+\sqrt{145}}{2} \\ \end{aligned}$ By Pythagorean theorem, $\begin{aligned} A{{B}^{2}} & =C{{B}^{2}}-A{{C}^{2}} \\ & ={{\left( CE+EB \right)}^{2}}-A{{C}^{2}} \\ & ={{\left( \dfrac{-1+\sqrt{145}}{2}+1 \right)}^{2}}-{{6}^{2}} \\ & =\dfrac{1+\sqrt{145}}{2} \\ \end{aligned}$ $AB=\sqrt{\dfrac{1+\sqrt{145}}{2}}$ By angle bisector theorem on $ABC$ , $\begin{aligned} & \dfrac{AG}{GB}=\dfrac{AC}{CB}\Rightarrow \dfrac{AG}{GB+AG}=\dfrac{AC}{CB+AC} \\ & \Rightarrow \dfrac{AG}{AB}=\dfrac{\sqrt{\dfrac{1+\sqrt{145}}{2}}}{\dfrac{1+\sqrt{145}}{2}+6} \\ & \Rightarrow \tan \theta =\dfrac{1}{\sqrt{12+\sqrt{145}}} \\ \end{aligned}$ Finally, for the area $A$ of $\triangle ACD$ we have $\begin{aligned} A & =\dfrac{1}{2}AC\cdot AD \\ & =\dfrac{1}{2}A{{C}^{2}}\tan 3\theta \\ & =18\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \\ \end{aligned}$ Substituting, we find $A=\dfrac{9}{2}\sqrt{\dfrac{5}{2}\left( \sqrt{145}-9 \right)}$ Consequently, $\begin{aligned} {{A}^{2}} & =\dfrac{405}{8}\left( \sqrt{145}-9 \right) \\ & \Leftrightarrow 8{{A}^{2}}+3645=405\sqrt{145} \\ & \Leftrightarrow {{\left( 8{{A}^{2}}+3645 \right)}^{2}}=\text{23783625} \\ & \Leftrightarrow 64{{A}^{4}}+58320{{A}^{2}}-10497600=0 \\ & \Leftrightarrow 4{{A}^{4}}+3645{{A}^{2}}-656100=0 \\ \end{aligned}$ i.e. $A$ is a root of the polynomial $f\left( x \right)=4{{x}^{4}}+3645{{x}^{2}}-656100$ .

For the answer, $f\left( 13 \right)=4\times {{13}^{4}}+3645\times {{13}^{2}}-656100=\boxed{74149}$ .