24. [10] Suppose that $A B C$ is an isosceles triangle with $A B=A C$. Let $P$ be the point on side $A C$ so that $A P=2 C P$. Given that $B P=1$, determine the maximum possible area of $A B C$.
and $C Q$. The key observation that, as we will show, $B X$ and $C X$ are fixed lengths, and the ratio of areas $[A B C] /[B C X]$ is constant. So, to maximize $[A B C]$, it is equivalent to maximize $[B C X]$.
Using Menelaus' theorem on $A B P$, we have
$$
\frac{B X \cdot P C \cdot A Q}{X P \cdot C A \cdot Q B}=1 \text {. }
$$

Since $P C / C A=1 / 3$ and $A Q / Q B=2$, we get $B X / X P=3 / 2$. It follows that $B X=3 / 5$. By symmetry, $C X=3 / 5$.
Also, we have
$$
[A B C]=3[B P C]=3 \cdot \frac{5}{3}[B X C]=5[B X C] .
$$

Note that $\left[B X C\right.$ ] is maximized when $\angle B X C=90^{\circ}$ (one can check that this configuration is indeed possible). Thus, the maximum value of $[B X C]$ is $\frac{1}{2} B X \cdot C X=\frac{1}{2}\left(\frac{3}{5}\right)^{2}=\frac{9}{50}$. It follows that the maximum value of $[A B C]$ is $\frac{9}{10}$.