1966 AMC 12 Problem 38

Attempt Problem 38 of the 1966 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1966 AMC 12 solutions, or check the answer key.

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38.

In triangle ABCABC the medians AMAM and CNCN to sides BCBC and AB,AB, respectively, intersect in point O.O. PP is the midpoint of side AC,AC, and MPMP intersects CNCN in Q.Q. If the area of triangle OMQOMQ is n,n, then the area of triangle ABCABC is:
16n16n
18n18n
21n21n
24n24n
27n27n
Answer: D
Concepts:area ratiocentroidcoordinate geometrymedian (geometry)
Small hint:
Use an affine model A=(0,0),A=(0,0), B=(2,0),B=(2,0), and C=(0,2)C=(0,2)
Big hint:
Find M,O,P,QM,O,P,Q and compare the two areas
Solution:
Area ratios are affine-invariant, so take A=(0,0),A=(0,0), B=(2,0),B=(2,0), and C=(0,2).C=(0,2). Then M=(1,1),O=(23,23),P=(0,1). \begin{gathered} M=(1,1),\\ O=(\frac{2}{3},\frac{2}{3}),\\ P=(0,1). \end{gathered} Line MPMP is y=1,y=1, and it meets median CNCN at Q=(12,1).Q=(\frac{1}{2},1). Thus [OMQ]=112,[OMQ]=\frac{1}{12}, while [ABC]=2.[ABC]=2. Their ratio is 24,24, so [ABC]=24n.[ABC]=24n.
Therefore, the correct answer is D.
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