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

(sol, img needed)
 
(asy)
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== Problem ==
 
== Problem ==
 
Let <math>ABCD</math> be a [[rhombus]] with <math>AC = 16</math> and <math>BD = 30</math>. Let <math>N</math> be a point on <math>\overline{AB}</math>, and let <math>P</math> and <math>Q</math> be the feet of the perpendiculars from <math>N</math> to <math>\overline{AC}</math> and <math>\overline{BD}</math>, respectively. Which of the following is closest to the minimum possible value of <math>PQ</math>?
 
Let <math>ABCD</math> be a [[rhombus]] with <math>AC = 16</math> and <math>BD = 30</math>. Let <math>N</math> be a point on <math>\overline{AB}</math>, and let <math>P</math> and <math>Q</math> be the feet of the perpendiculars from <math>N</math> to <math>\overline{AC}</math> and <math>\overline{BD}</math>, respectively. Which of the following is closest to the minimum possible value of <math>PQ</math>?
 +
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<center><asy>
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size(200);
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defaultpen(0.6);
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pair O = (15*15/17,8*15/17), C = (17,0), D = (0,0), P = (25.6,19.2), Q = (25.6, 18.5);
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pair A = 2*O-C, B = 2*O-D;
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pair P = (A+O)/2, Q=(B+O)/2, N=(A+B)/2;
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draw(A--B--C--D--cycle);
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draw(A--O--B--O--C--O--D);
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draw(P--N--Q);
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label("\(A\)",A,WNW);
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label("\(B\)",B,ESE);
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label("\(C\)",C,ESE);
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label("\(D\)",D,SW);
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label("\(P\)",P,SSW);
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label("\(Q\)",Q,SSE);
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label("\(N\)",N,NNE);
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</asy></center>
  
 
<math>\mathrm{(A)}\ 6.5
 
<math>\mathrm{(A)}\ 6.5
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\qquad\mathrm{(D)}\ 7.25
 
\qquad\mathrm{(D)}\ 7.25
 
\qquad\mathrm{(E)}\ 7.5</math>
 
\qquad\mathrm{(E)}\ 7.5</math>
−
 
−
{{image}}
 
 
== Solution ==
 
== Solution ==
 
Let <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>O</math>. Since <math>ABCD</math> is a rhombus, then <math>\overline{AC}</math> and <math>\overline{BD}</math> are [[perpendicular]] [[bisector]]s. Thus <math>\angle POQ = 90^{\circ}</math>, so <math>OPNQ</math> is a [[rectangle]]. Since the diagonals of a rectangle are of equal length, <math>PQ = ON</math>, so we want to minimize <math>ON</math>. It follows that we want <math>ON \perp AB</math>.
 
Let <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect at <math>O</math>. Since <math>ABCD</math> is a rhombus, then <math>\overline{AC}</math> and <math>\overline{BD}</math> are [[perpendicular]] [[bisector]]s. Thus <math>\angle POQ = 90^{\circ}</math>, so <math>OPNQ</math> is a [[rectangle]]. Since the diagonals of a rectangle are of equal length, <math>PQ = ON</math>, so we want to minimize <math>ON</math>. It follows that we want <math>ON \perp AB</math>.

Revision as of 13:51, 3 February 2008

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

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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