Difference between revisions of "1975 AHSME Problems/Problem 28"
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+ | ==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 00:46, 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:
\begin{align*}
\angle MAC=\alpha, \; \angle MAB=\beta, \; \angle AGE=\gamma,\; \angle AMC=\delta
\end{align*}
Then, in the
and
:
\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
and
:
\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
.
-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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