1975 AHSME Problems/Problem 28

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Solution

Here, we use Mass Points. Let $AF = x$. We then have $AE = 2x$, $EC = 16-2x$, and $FB = 12 - x$ Let $B$ have a mass of $2$. Since $M$ is the midpoint, $C$ also has a mass of $2$. Looking at segment $AB$, we have \[2 \cdot (12-x) = \text{m}A_{AB} \cdot x\] So \[\text{m}A_{AB} = \frac{24-2x}{x}\] Looking at segment $AC$,we have \[2 \cdot (16-2x) = \text{m}A_{AC} \cdot 2x\] So \[\text{m}A_{AC} = \frac{16-2x}{x}\] From this, we get \[\text{m}E = \frac{16-2x}{x} + 2 \Rrightarrow \text{m}E = \frac{16}{x}\] and \[\text{m}F = \frac{24-2x}{x} + 2 \Rrightarrow \text{m}F = \frac{24}{x}\] We want the value of $\frac{EG}{GF}$. This can be written as \[\frac{EG}{GF} = \frac{\text{m}F}{\text{m}E}\] Thus \[\frac{\text{m}F}{\text{m}E} = \frac {\frac{24}{x}}{\frac{16}{x}} = \frac{3}{2}\] $\boxed{A}$

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