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)
Best Answer
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.