University of South Carolina High School Math Contest/1993 Exam/Problem 16

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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