Vijetnam 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, $CF\parallel AB$. $$\begin{gathered} \text{Hence }\overline{CP}=\overline{FP} \\ \overline{CA}=\overline{FB}. \end{gathered}$$ 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 $PA=\sqrt{P{B}^2+A{B}^2}=\sqrt{x^2+4R^2}$. Consequently $PC=\frac{P{B}^2}{PA}=\frac{x^2}{\sqrt{x^2+4R^2}}$ 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 $$BM=\frac{PA\cdot BC}{PC+PA}=\frac{PB\cdot AB}{PC+PA}=\frac{Rx\sqrt{x^2+4R^2}}{x^2+2R^2}.$$ Thus $S_{AMB}=\frac{1}{2}AB\cdot BM\cdot \sin \widehat{ABM}=\frac{1}{2}\cdot 2R\cdot \frac{Rx\sqrt{x^2+4R^2}}{x^2+2R^2}\cdot \frac{AC}{2R}=\frac{2R^{3}x}{x^2+2R^2}.$ It follows that $S_{AMB}\le \frac{2R^{3}x}{2\sqrt{2}xR}=\frac{R^2}{\sqrt{2}}$ and $S_{AMB}=\frac{R^2}{\sqrt{2}}\iff x^2=2R^2\iff x=\sqrt{2}R$. 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 $S_{AMB}=\frac{R^2}{\sqrt{2}}$.