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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 16"

 
 
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== Problem ==
 
== Problem ==
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In the triangle below, <math>\displaystyle M, N, </math> and <math>P</math> are the midpoints of <math>BC, AB,</math> and <math>AC</math> respectively.  <math>CN</math> and <math>AM</math> intersect at <math>O</math>.  If the length of <math>CQ</math> is 4, then what is the length of <math>OQ</math>?
  
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<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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<center>[[Image:Usc93.16.PNG]]</center>
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<center><math> \mathrm{(A) \ }1 \qquad \mathrm{(B) \ }4/3 \qquad \mathrm{(C) \ }\sqrt{2} \qquad \mathrm{(D) \ }3/2 \qquad \mathrm{(E) \ }2 </math></center>
  
 
== Solution ==
 
== Solution ==
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<math>AM</math> and <math>CN</math> are the [[median of a triangle|medians]] of <math>\triangle ABC</math>, so their intersection point <math>O</math> is the [[centroid]] of the [[triangle]].  Also, <math>\frac{CM}{MB} = \frac{CP}{PA} = 1</math> so <math>MP</math> is [[parallel]] to <math>AB</math> and thus <math>\frac{CQ}{QN} = 1</math> and <math>CQ = QN = 4</math>.  Then <math>CN = CQ + QN = 8</math>.  Since the centroid trisects the medians, <math>CO = \frac23 CN = \frac{16}3</math> and <math>OQ = CO - CQ = \frac{16}3 - 4 = \frac43</math> which is answer choice <math>\mathrm{(B)}</math>.
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 15|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 17|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
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== See also ==
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[[Category:Intermediate Geometry Problems]]
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* [[University of South Carolina High School Math Contest/1993 Exam]]
 

Latest revision as of 11:45, 17 August 2006

Problem

In the triangle below, $\displaystyle M, N,$ and $P$ are the midpoints of $BC, AB,$ and $AC$ respectively. $CN$ and $AM$ intersect at $O$. If the length of $CQ$ is 4, then what is the length of $OQ$?

Usc93.16.PNG
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }4/3 \qquad \mathrm{(C) \ }\sqrt{2} \qquad \mathrm{(D) \ }3/2 \qquad \mathrm{(E) \ }2$

Solution

$AM$ and $CN$ are the medians of $\triangle ABC$, so their intersection point $O$ is the centroid of the triangle. Also, $\frac{CM}{MB} = \frac{CP}{PA} = 1$ so $MP$ is parallel to $AB$ and thus $\frac{CQ}{QN} = 1$ and $CQ = QN = 4$. Then $CN = CQ + QN = 8$. Since the centroid trisects the medians, $CO = \frac23 CN = \frac{16}3$ and $OQ = CO - CQ = \frac{16}3 - 4 = \frac43$ which is answer choice $\mathrm{(B)}$.


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