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 $PQ$ ?

$[asy] size(200); defaultpen(0.6); pair O = (15*15/17,8*15/17), C = (17,0), D = (0,0), P = (25.6,19.2), Q = (25.6, 18.5); pair A = 2*O-C, B = 2*O-D; pair P = (A+O)/2, Q=(B+O)/2, N=(A+B)/2; draw(A--B--C--D--cycle); draw(A--O--B--O--C--O--D); draw(P--N--Q); label("$A$",A,WNW); label("$B$",B,ESE); label("$C$",C,ESE); label("$D$",D,SW); label("$P$",P,SSW); label("$Q$",Q,SSE); label("$N$",N,NNE); [/asy]$

$\mathrm{(A)}\ 6.5 \qquad\mathrm{(B)}\ 6.75 \qquad\mathrm{(C)}\ 7 \qquad\mathrm{(D)}\ 7.25 \qquad\mathrm{(E)}\ 7.5$

Solution

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 $ON \perp AB$ .

Finding the area in two different ways, $\frac{1}{2} AO \cdot BO = 60 = \frac{1}{2} ON \cdot AB = \frac{\sqrt{8^2 + 15^2}}{2} \cdot ON \Longrightarrow ON = \frac{120}{17} \approx 7.06 \Rightarrow \mathrm{(C)}$

2003 AMC 12B (Problems • Answer Key • Resources)
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2003 AMC 12B Problems/Problem 22

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