2025/26 AHSMC Part 1 Problem 9

We have $AM = MB = 1/2$, $NC = 1/4$ and $DN = 3/4$. Since $AB \parallel DC$, $\triangle AMP \sim \triangle NDP$ with height ratio $\dfrac{AM}{DN} = \dfrac{2}{3}$. The heights from $P$ to $AM$ and $DN$ sum to $1$, so they are $2/5$ and $3/5$. Thus $$\mathrm{Area}(\triangle DNP) = \frac{1}{2}\cdot\frac{3}{5}\cdot\frac{3}{4} = \frac{9}{40}.$$ Likewise $\triangle BMQ \sim \triangle NCQ$ with base ratio $2$, so the height of $\triangle CQN$ from $Q$ is $1/3$ and $$\mathrm{Area}(\triangle CQN) = \frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4} = \frac{1}{24}.$$ Therefore $$\mathrm{Area}(MPNQ) = \mathrm{Area}(\triangle DMC) - \mathrm{Area}(\triangle DPN) - \mathrm{Area}(\triangle CQN) = \frac{1}{2} - \frac{9}{40} - \frac{1}{24} = \frac{7}{30}.$$ The answer is (B).