1975 AHSME Problems/Problem 28

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 $\frac{\text{m}F}{\text{m}E} = \frac {\frac{24}{x}}{\frac{16}{x}} = \frac{3}{2}$ $\boxed{A}$

~JustinLee2017

Page

Toolbox

Search

1975 AHSME Problems/Problem 28

Solution