Problem (Geometry):
In parallelogram $A R M L$, points $P$ and $Q$ are the midpoints of sides $\overline{R M}$ and $\overline{A L}$, respectively. Point $X$ lies on segment $\overline{P Q}$, and $P X=3, R X=4$, and $P R=5$. Point $I$ lies on segment $\overline{R X}$ such that $I A=I L$. Compute the maximum possible value of $\frac{[P Q R]}{[L I P]}$.

Solution:
Because $A I=L I$ and $A Q=L Q$, line $I Q$ is the perpendicular bisector of $\overline{A L}$. Because $A R M L$ is a parallelogram, $\overline{Q I} \perp \overline{R P}$. Note also that $\mathrm{m} \angle R X P=90^{\circ}$. Thus $I$ is the orthocenter of triangle $P Q R$, from



which it follows that $\overleftrightarrow{P I} \perp \overline{R Q}$ and $\overline{P I} \perp \overline{P L}$ (because $P R Q L$ is a parallelogram). Extend $\overline{P I}$ through $I$ to meet $\overline{R Q}$ at $D$. Then $2[P Q R]=R Q \cdot P D$ and $2[L I P]=P I \cdot P L=P I \cdot R Q$. Hence the problem is equivalent to determining the maximum value of $P D / P I$.

Set $\mathrm{m} \angle R P D=\mathrm{m} \angle R P I=\alpha$ and $\mathrm{m} \angle I P X=\beta$, and note that $P D=P R \cos \alpha=5 \cos \alpha$ and $P I=P X / \cos \beta=$ $3 / \cos \beta$. It follows that

$$
\frac{P D}{P I}=\frac{5 \cos \alpha \cos \beta}{3}=\frac{5(\cos (\alpha+\beta)+\cos (\alpha-\beta))}{6} \leq \frac{5(3 / 5+1)}{6}=\frac{\mathbf{4}}{\mathbf{3}}
$$

with equality when $\alpha=\beta$.



Final answer: $\frac{4}{3}$