2003 AMC 12B Problems/Problem 22

Problem

Let $ABCD$ be a rhombus with $AC = 16$ and $BD = 30$ . 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. Which of the following is closest to the minimum possible value of $PQ$ ?

Let $\overline{AC}$ and $\overline{BD}$ intersect at $O$ . Since $ABCD$ is a rhombus, then $\overline{AC}$ and $\overline{BD}$ are perpendicular bisectors. Thus $\angle POQ = 90^{\circ}$ , so $OPNQ$ is a rectangle. Since the diagonals of a rectangle are of equal length, $PQ = ON$ , so we want to minimize $ON$ . It follows that we want $ON \perp AB$ .