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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:

$\angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta$

Then, in the $\triangle AMC$ and $\triangle AMB$:

$\frac{MC}{sin\alpha}=\frac{AC}{sin\delta};$ $\frac{MB}{sin\beta}=\frac{AB}{sin\delta}=\frac{MC}{sin\alpha}\cdot\frac{3}{4};$ $\frac{sin\alpha}{sin\beta}=\frac{3}{4}$

In the $\triangle AGE$ and $\triangle AGF$:

$\frac{FG}{sin\beta}=\frac{AF}{sin\gamma};$ $\frac{EG}{sin\alpha}=\frac{AE}{sin\gamma};$ $\frac{2FG}{sin\beta}=\frac{EG}{sin\alpha};$ $\frac{EG}{FG}=\frac{2sin\alpha}{sin\beta}=\frac{3}{2}$

Hence, our answer is A.

-VSN

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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