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...") |
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+ | == 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 00:49, 12 May 2023
Problem 28
In shown in the adjoining figure,
is the midpoint of side
and
. Points
and
are taken on
and
, respectively, and lines
and
intersect at
. If
then
equals
Solution
Here, we use Mass Points.
Let . We then have
,
, and
Let
have a mass of
. Since
is the midpoint,
also has a mass of
.
Looking at segment
, we have
So
Looking at segment
,we have
So
From this, we get
and
We want the value of
. This can be written as
Thus
~JustinLee2017
Solution 2
Since we only care about a ratio , and since we are given
being the midpoint of
, we realize we can conveniently also choose
to be the midpoint of
. (we're free to choose any point
on
as long as
is twice
, the constraint given in the problem). This means
, and
. We then connect
which creates similar triangles
and
by SAS, and thus generates parallel lines
and
. This also immediately gives us similar triangles
(note that
because
is in
ratio).
~afroromanian
Solution 3
In order to find , we can apply the law of sine to this model. Let:
Then, in the and
:
In the and
:
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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