Difference between revisions of "2003 AMC 12B Problems/Problem 22"

Latest revision as of 13:17, 23 June 2022

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.

@@ Line 35: / Line 35: @@
 -MP8148
+== Solution 3 ==
+Let the intersection <math>\overline{AC}</math> and <math>\overline{BD}</math> be <math>E.</math> Let <math>\overline{PE}=y \implies \overline{AP}=8-y</math> and <math>\overline{EQ}=x \implies \overline {QB}=15-x.</math> Since <math>\triangle NQB \sim \triangle APN,</math> we have <math>\frac{8-y}{x}=\frac{y}{15-x} \implies 120=15y+8x.</math> We want to minimize <math>\overline{PQ}=\sqrt{x^2+y^2}.</math> By Cauchy, <math>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.</math> So, choice <math>\boxed{C}</math> is closest to the minimum.
 == See also ==

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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