Difference between revisions of "1975 AHSME Problems/Problem 28"
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+ | ==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). | ||
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+ | ~afroromanian | ||
==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 01:17, 8 October 2022
Contents
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
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.