Not as They Seem to Be

Let $\angle BAM = \theta$ and $BM = a$ . Therefore, $\angle BDM = 3\theta$

$\Rightarrow \tan {\theta} = \dfrac {a}{11}\quad$ and $\quad \tan{3\theta} = a \quad \Rightarrow \tan {3\theta} = 11 \tan{\theta}$

Let $t = \tan{\theta}$ then we have:

$\tan {3\theta} = 11 \tan{\theta} \quad \Rightarrow \dfrac {t+\frac{2t}{1-t^2}}{1-t\left( \frac{2t}{1-t^2} \right)} = 11 t \quad \Rightarrow \dfrac {1+\frac{2}{1-t^2}}{1-t\left( \frac{2t}{1-t^2} \right)} = 11$

$\Rightarrow \dfrac {1-t^2+2}{1-t^2 - 2t^2} = 11 \quad \Rightarrow \dfrac {3-t^2}{1-3t^2} = 11 \quad \Rightarrow 3-t^2 = 11-33t^2$

$\Rightarrow 32t^2 = 8\quad \Rightarrow t = \frac{1}{2}$

$\Rightarrow a = \frac {11}{2} \quad \Rightarrow AB = \frac {11}{2}\sqrt{5}$

Therefore, the perimeter is $11+11\sqrt{5} = 11 + \sqrt{605}$

$\quad \Rightarrow a + b = 11 + 605 = \boxed{616}$

$\angle BAM=\alpha$ , $\angle BDM=3\alpha$ and $BM=x$

$\tan \alpha=\frac{x}{11}$ and $\tan 3\alpha=x$ so $\tan 3\alpha=11\tan \alpha$ .

Now

$\tan 3\alpha=\frac{\sin \alpha \cos 2\alpha+\sin 2\alpha \cos \alpha}{\cos \alpha \cos 2\alpha-\sin \alpha \sin 2\alpha}$

$\tan 3\alpha=\frac{\sin \alpha( \cos 2\alpha+2\cos^2 \alpha)}{\cos \alpha( \cos 2\alpha-2\sin^2 \alpha)}$

We get that

$\frac{\sin \alpha( \cos 2\alpha+2 \cos^2 \alpha)}{\cos \alpha( \cos 2\alpha-2\sin ^2 \alpha)}=11 \tan \alpha$

$\frac{\cos 2\alpha+2\cos^2 \alpha}{\cos 2\alpha-2\sin^2 \alpha}=11$

Smplifying

$\cos^2 \alpha=\frac{32}{40}=0.8$ but $\cos^2 \alpha=(\frac{11}{\sqrt{x^2+121}})^2=\frac{121}{x^2+121}$

$\frac{121}{x^2+121}=0.8 \Rightarrow \boxed{x=5.5}$

$\triangle ABC$ perimeter is $2x+2\sqrt{x^2+121}$

Perimeter is $11+2\sqrt{(5.5)^2+121}=11+2\sqrt{\frac{605}{4}}=\boxed{11+\sqrt{605}}$

Finally, $\boxed{a+b=11+605=616}$

My approach was similar but using the Archimedes method for trisection.

calling BD=R we get Rcos(3x)= 1 and 2Rcos(x)=10

Then cos(x)=5cos(3x).

From that point the nitty gritty and hateful trigonometry manipulation. As said for somebody it had been kinder express the result as 11(1+sqrt(5))