Stuck on step of the solution Circle theorems/Triangles

Question

In $ABC$ right triangle, $AC=2+\sqrt3$ and $BC=3+2\sqrt3$. Circle goes on point $C$ and its center is on $AC$ cathetus, $C$ Cuts the circle in point $M$ and it touches $AB$ hypotenuse at point $D$. Find the area of shaded part.

Now my solutions and where I'm stuck at:

I found that $\tan \angle A=\sqrt3$ which means $\angle CAB=60^\circ$ then i used formula $\angle CAB=\frac{CD-MD}2$ got to $\stackrel{\frown}{CD}=150^\circ$, $\stackrel{\frown}{MD}=30^\circ$. $\angle MOD=30^\circ$ $\angle OAD=60^\circ$, so $OAD$ is a right triangle.

Then bring in ratios: $AD=x, AO=2x, OD=x\sqrt3$

Now I'm stuck here $AD=2x-x\sqrt3$ , why? I might be missing some rule here, but the next one also confused me. We use another formula $$AD^2=AM\times AC,\tag{1}$$ I understand why we need to use this, but the calculation confuses me. $x^2=(2x-x\sqrt3)(2+\sqrt3)$ I don't understand how we come to this, I'd guess it should be like this instead $(2x-x\sqrt3)^2=x(2+\sqrt3)$ cause it fits the formula (1)

Answer 1

There is a typo. It should be $OD=x\sqrt{3}$ and $AM=AO-OD=2x-x\sqrt{3}$, since $OD=OM$ is the radius of the circle.

Answer 2

As you mentioned, the AOD triangle is key to get to the right answer.

If we take OD=r for radius, then $AD=\frac{r}{\sqrt{3}}$ and $OA=\frac{2r}{\sqrt{3}}$.

Then, the shaded area results from the subtraction of the area of the triangle AOD with the area of the circular sector ODM, so $S_{AMD}=S_{\triangle AOD}-S_{OMD}$

The area of the triangle AOD is $S_{\triangle AOD}=\frac{r^2}{2\sqrt{3}}$

and the area of the circular sector is $S_{OMD}=\pi r^2\frac{30}{360}=\frac{\pi r^2}{12}$.

Then, the area of the shaded area $S_{AMD}=\frac{r^2}{2\sqrt{3}}-\frac{\pi r^2}{12}$.

On the other hand, to get the value of r, we focus on AC: $AC=AO+OC=2+\sqrt{3}$. Then, $2+\sqrt{3}=\frac{2r}{\sqrt{3}}+r$ and $r=\sqrt{3}$.

Using this on the last equation we have, $S_{AMD}=\frac{\sqrt{3}}{2}-\frac{\pi}{4}$