Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 16"
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== Solution == | == Solution == | ||
| + | <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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Latest revision as of 11:45, 17 August 2006
Problem
In the triangle below, 
and 
are the midpoints of 
and 
respectively. 
and 
intersect at 
. If the length of 
is 4, then what is the length of 
?
Solution
and are the medians of 
, so their intersection point is the centroid of the triangle. Also, 
so 
is parallel to 
and thus 
and 
. Then 
. Since the centroid trisects the medians, 
and 
which is answer choice 
.
