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

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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?

Image: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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