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"

(Created page with "==Solution== Here, we use Mass Points. Let <math>AF = x</math>. We then have <math>AE = 2x</math>, <math>EC = 16-2x</math>, and <math>FB = 12 - x</math> Let <math>B</math> hav...")
 
(→‎Solution 3)
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 +
== Problem 28 ==
 +
 +
In <math>\triangle ABC</math> shown in the adjoining figure, <math>M</math> is the midpoint of side <math>BC, AB=12</math> and <math>AC=16</math>. Points <math>E</math> and <math>F</math> are taken on <math>AC</math>
 +
and <math>AB</math>, respectively, and lines <math>EF</math> and <math>AM</math> intersect at <math>G</math>. If <math>AE=2AF</math> then <math>\frac{EG}{GF}</math> 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>
 +
 +
<math>\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} </math>
 +
 
==Solution==
 
==Solution==
 
Here, we use Mass Points.
 
Here, we use Mass Points.
Line 22: Line 46:
  
 
~JustinLee2017
 
~JustinLee2017
 +
 +
==Solution 2==
 +
Since we only care about a ratio <math>EG/GF</math>, and since we are given <math>M</math> being the midpoint of <math>\overline{BC}</math>, we realize we can conveniently also choose <math>E</math> to be the midpoint of <math>\overline{AC}</math>. (we're free to choose any point <math>E</math> on <math>\overline{AC}</math> as long as <math>AE</math> is twice <math>AF</math>, the constraint given in the problem). This means <math>AE = 16/2 = 8</math>, and <math>AF = 4</math>. We then connect <math>ME</math> which creates similar triangles <math>\triangle EMC</math> and <math>\triangle ABC</math> by SAS, and thus generates parallel lines <math>\overline{EM}</math> and <math>\overline{AB}</math>. This also immediately gives us similar triangles <math>\triangle AFG \sim \triangle MEG, => EG/FG = ME/AF = 6/4 = \boxed{3/2}</math> (note that <math>ME = 6</math> because <math>ME/AB</math> is in <math>1:2</math> ratio).
 +
 +
~afroromanian
 +
 +
 +
==Solution 3==
 +
In order to find <math>\frac{EG}{FG}</math>, we can apply the law of sine to this model. Let:
 +
 +
<math>\angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta</math>
 +
 +
Then, in the <math>\triangle AMC</math> and <math>\triangle AMB</math>:
 +
 +
<math>\frac{MC}{sin\alpha}=\frac{AC}{sin\delta};</math>
 +
<math>\frac{MB}{sin\beta}=\frac{AB}{sin\delta}=\frac{MC}{sin\alpha}\cdot\frac{3}{4};</math>
 +
<math>\frac{sin\alpha}{sin\beta}=\frac{3}{4}</math>
 +
 +
In the <math>\triangle AGE</math> and <math>\triangle AGF</math>:
 +
 +
<math>\frac{FG}{sin\beta}=\frac{AF}{sin\gamma};</math>
 +
<math>\frac{EG}{sin\alpha}=\frac{AE}{sin\gamma};</math>
 +
<math>\frac{2FG}{sin\beta}=\frac{EG}{sin\alpha};</math>
 +
<math>\frac{EG}{FG}=\frac{2sin\alpha}{sin\beta}=\frac{3}{2}</math>
 +
 +
Hence, our answer is A.
 +
 +
-VSN
 +
 +
==See Also==
 +
{{AHSME box|year=1975|num-b=27|num-a=29}}
 +
{{MAA Notice}}

Latest revision as of 23:49, 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:

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

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

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=193028"