, 2011

1/ Let $F$ be the intersection of lines $AE$ and $BP$.
We have $\overline{ACE} = 90^\circ + \overline{BCE} = 90^\circ + \overline{FAB} = \overline{EFP}$. Consequently $\overline{EFP} + \overline{ECP} = 180^\circ$.

Hence $CEFP$ is a cyclic quadrilateral. Consequently $\overline{CFP} = \overline{CEP} = 90^\circ$. Thus, $CF \parallel AB$.
$$\text{Hence } \overline{CP} = \overline{FP} \\ \overline{CA} = \overline{FB}.$$
Whence, considering the triangle $ABP$, we have
$$\frac{\overline{CP}}{\overline{CA}} = \frac{\overline{OA}}{\overline{OB}} = \frac{\overline{FB}}{\overline{FP}} = \frac{\overline{OA}}{\overline{OB}} = -1.$$
Hence, according to Ceva theorem, the lines $PO$, $AE$ and $BC$ are concurrent.

2/ Let $BP = x$ and denote $R$ the radius of $(O)$.
Consider the right $ABP$, we have $PA = \sqrt{PB^2 + AB^2} = \sqrt{x^2 + 4R^2}$.
Consequently $PC = \frac{PB^2}{PA} = \frac{x^2}{\sqrt{x^2 + 4R^2}}$ and $AC = PA - PC = \frac{4R^2}{\sqrt{x^2 + 4R^2}}$.
Since $CF \parallel AB$ (see proof above), one has $\frac{MC}{MB} = \frac{CF}{AB} = \frac{PC}{PA}$.
Consequently $\frac{BC}{MB} = \frac{PC}{PA} + 1 = \frac{PC + PA}{PA}$. Hence
$$BM = \frac{PA \cdot BC}{PC + PA} = \frac{PB \cdot AB}{PC + PA} = \frac{R x \sqrt{x^2 + 4R^2}}{x^2 + 2R^2}.$$
Thus $S_{AMB} = \frac{1}{2} AB \cdot BM \cdot \sin \widehat{ABM} = \frac{1}{2} \cdot 2R \cdot \frac{R x \sqrt{x^2 + 4R^2}}{x^2 + 2R^2} \cdot \frac{AC}{2R} = \frac{2R^3 x}{x^2 + 2R^2}.$
It follows that $S_{AMB} \le \frac{2R^3 x}{2\sqrt{2}xR} = \frac{R^2}{\sqrt{2}}$ and $S_{AMB} = \frac{R^2}{\sqrt{2}} \Leftrightarrow x^2 = 2R^2 \Leftrightarrow x = \sqrt{2}R$.
Thus, the area of triangle $AMB$ attains its maximum if and only if the distance between $P$ and $B$ is equal to $\sqrt{2}R$ (there are two such places); in those cases $S_{AMB} = \frac{R^2}{\sqrt{2}}$.