Let $ABCD$ be a rhombus with $AC = 30$ and $BD = 40$. Let $N$ be a point on $\overline{AB}$, and let $P$ and $Q$ be the feet of the perpendiculars from $N$ to $\overline{AC}$ and $\overline{BD}$, respectively. What is the minimum possible value of $PQ$?

1. Let $O$ be the center. As in the problem, $NPOQ$ is a rectangle, so $PQ=ON$, which is smallest when $ON\perp AB$. In the right triangle $AOB$ with legs $15,20$ and hypotenuse $25$, this altitude is $\frac{15\cdot 20}{25}=12$.
2. The minimum is $12$.