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(C) Let $AM = x AB$ and $AN = y AC$. Since $G$ is the centroid,the position vector of $G$ is $\frac{A+B+C}{3}$.Since $M, G, N$ are collinear,there exi

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$4/9 \leq r < 1/2$

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(C) Let $AM = x AB$ and $AN = y AC$. Since $G$ is the centroid,the position vector of $G$ is $\frac{A+B+C}{3}$.Since $M, G, N$ are collinear,there exists a scalar $k$ such that $\vec{G} = (1-k)\vec{M} + k\vec{N}$.Using the property of the centroid and the given conditions,it can be shown that $\frac{1}{3x} + \frac{1}{3y} = 1$,or $\frac{1}{x} + \frac{1}{y} = 3$.The ratio of the areas is $r = \frac{	ext{Area}(	riangle AMN)}{	ext{Area}(	riangle ABC)} = xy$.From $\frac{1}{x} + \frac{1}{y} = 3$,we have $y = \frac{x}{3x-1}$.Thus $r = \frac{x^2}{3x-1}$.Since $M$ and $N$ are in the interior of the segments,$0 For $y  1/2$.Also,the minimum value of $r$ occurs when $x=y=2/3$,giving $r = (2/3)(2/3) = 4/9$.As $x 	o 1/2$,$y 	o 1$,so $r 	o 1/2$.Thus,$4/9 \leq r < 1/2$.

In a $\triangle ABC$,let $G$ denote its centroid and let $M, N$ be points in the interiors of the segments $AB, AC$,respectively,such that $M, G, N$ are collinear. If $r$ denotes the ratio of the area of $\triangle AMN$ to the area of $\triangle ABC$,then

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