Problem 1182

Let $Q$ be the point on $AB$ so that $AQ=2BQ$, and let $X$ be the intersection of $BP$ and $CQ$. The key observation that, as we will show, $BX$ and $CX$ are fixed lengths, and the ratio of areas $[ABC]/[BCX]$ is constant. So, to maximize $[ABC]$, it is equivalent to maximize $[BCX]$. Using Menelaus' theorem on $ABP$, we have $$\frac{BX \cdot PC \cdot AQ}{XP \cdot CA \cdot QB}=1$$ Since $PC/CA=1/3$ and $AQ/QB=2$, we get $BX/XP=3/2$. It follows that $BX=3/5$. By symmetry, $CX=3/5$. Also, we have $$[ABC]=3[BCP]=3 \cdot \frac{5}{3}[BXC]=5[BXC]$$ Note that $[BXC]$ is maximized when $\angle BXC=90^{\circ}$ (one can check that this configuration is indeed possible). Thus, the maximum value of $[BXC]$ is $\frac{1}{2}BX \cdot CX=\frac{1}{2}\left(\frac{3}{5}\right)^{2}=\frac{9}{50}$. It follows that the maximum value of $[ABC]$ is $\frac{9}{10}$.