# 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 1

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

## Solution 2 (semi-bash)

Let the intersection of $\overline{AC}$ and $\overline{BD}$ be $E$. Since $ABCD$ is a rhombus, we have $\overline{AC} \perp \overline{BD}$ and $AE = CE = \dfrac{AC}{2} = 8$. Since $\overline{NQ} \perp \overline{BD}$, we have $\overline{NQ} \parallel \overline{AC}$, so $\triangle{BNQ} \sim \triangle{BAE} \sim \triangle{NAP}$. Therefore, $$\dfrac{NP}{AP} = \dfrac{NP}{8 - NQ} = \dfrac{BE}{AE} = \dfrac{15}{8} \Rightarrow NP = 15 - \dfrac{15}{8} NQ.$$ By Pythagorean Theorem, $$PQ^2 = NQ^2 + NP^2 = NQ^2 + \left(15 - \dfrac{15}{8} NQ \right)^2 = \dfrac{289}{64} NQ^2 - \dfrac{225}{4} NQ + 225.$$ The minimum value of $PQ^2$ would give the minimum value of $PQ$, so we take the derivative (or use vertex form) to find that the minimum occurs when $NQ = \dfrac{225 \cdot 8}{289}$ which gives $PQ^2 = \dfrac{225 \cdot 64}{289}$. Hence, the minimum value of $PQ$ is $\sqrt{\dfrac{225 \cdot 64}{289}} = \dfrac{120}{17}$, which is closest to $7 \Rightarrow \boxed{\textbf{C}}$.

-MP8148

## Solution 3

Let the intersection $\overline{AC}$ and $\overline{BD}$ be $E.$ Let $\overline{PE}=y \implies \overline{AP}=8-y$ and $\overline{EQ}=x \implies \overline {QB}=15-x.$ Since $\triangle NQB \sim \triangle APN,$ we have $\frac{8-y}{x}=\frac{y}{15-x} \implies 120=15y+8x.$ We want to minimize $\overline{PQ}=\sqrt{x^2+y^2}.$ By Cauchy, $17^2(x^2+y^2)=(x^2+y^2)(8^2+15^2) \ge (8x+15y)^2=120^2 \implies \sqrt{x^2+y^2} \ge \frac{120}{17} \approx 7.$ So, choice $\boxed{C}$ is closest to the minimum.