Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1975 AHSME Problems/Problem 28"
(→‎Solution 2)
Line 51: Line 51:
  
 
~afroromanian
 
~afroromanian
 +
 +
 +
==Solution 3==
 +
In order to find <math>\frac{EG}{FG}</math>, 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 <math>\triangle AMC</math> and <math>\triangle AMB</math>:
 +
\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 <math>\triangle AGE</math> and <math>\triangle AGF</math>:
 +
\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 <math>A</math>.
 +
 +
-VSN
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1975|num-b=27|num-a=29}}
 
{{AHSME box|year=1975|num-b=27|num-a=29}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:46, 11 May 2023

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_AHSME_Problems/Problem_28&oldid=193026"