1975 AHSME Problems/Problem 28

Problem 28

In $\triangle ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC, AB=12$ and $AC=16$ . Points $E$ and $F$ are taken on $AC$ and $AB$ , respectively, and lines $EF$ and $AM$ intersect at $G$ . If $AE=2AF$ then $\frac{EG}{GF}$ equals

$[asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label("A", (0,0), S); label("B", (12,0), S); label("C", (14,7.75), E); label("E", (8.75,4.84), N); label("F", (5,0), S); label("M", (13,3.875), E); label("G", (7,1)); [/asy]$

$\textbf{(A)}\ \frac{3}{2} \qquad \textbf{(B)}\ \frac{4}{3} \qquad \textbf{(C)}\ \frac{5}{4} \qquad \textbf{(D)}\ \frac{6}{5}\\ \qquad \textbf{(E)}\ \text{not enough information to solve the problem}$

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

Solution 2

Since we only care about a ratio $EG/GF$ , and since we are given $M$ being the midpoint of $\overline{BC}$ , we realize we can conveniently also choose $E$ to be the midpoint of $\overline{AC}$ . (we're free to choose any point $E$ on $\overline{AC}$ as long as $AE$ is twice $AF$ , the constraint given in the problem). This means $AE = 16/2 = 8$ , and $AF = 4$ . We then connect $ME$ which creates similar triangles $\triangle EMC$ and $\triangle ABC$ by SAS, and thus generates parallel lines $\overline{EM}$ and $\overline{AB}$ . This also immediately gives us similar triangles $\triangle AFG \sim \triangle MEG, => EG/FG = ME/AF = 6/4 = \boxed{3/2}$ (note that $ME = 6$ because $ME/AB$ is in $1:2$ ratio).

~afroromanian

Solution 3

In order to find $\frac{EG}{FG}$ , we can apply the law of sine to this model. Let: \begin{align*} \angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta \end{align*} Then, in the $\triangle AMC$ and $\triangle AMB$ : \begin{align*} \frac{MC}{sin\alpha}&=\frac{AC}{sin\delta}\\ \frac{MB}{sin\beta}&=\frac{AB}{sin\delta}=\frac{MC}{sin\alpha}\cdot\frac{3}{4}\\ \frac{sin\alpha}{sin\beta}&=\frac{3}{4} \end{align*} In the $\triangle AGE$ and $\triangle AGF$ : \begin{align*} \frac{FG}{sin\beta}&=\frac{AF}{sin\gamma}\\ \frac{EG}{sin\alpha}&=\frac{AE}{sin\gamma}\\ \frac{2FG}{sin\beta}&=\frac{EG}{sin\alpha}\\ \frac{EG}{FG}=\frac{2sin\alpha}{sin\beta}=\frac{3}{2} \end{align*} Hence, our answer is $A$ .

-VSN

1975 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1975 AHSME Problems/Problem 28

Contents

Problem 28

Solution

Solution 2

Solution 3

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