Difference between revisions of "2014 AMC 10B Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | In rectangle <math>ABCD</math>, <math>DC = | + | In rectangle <math>ABCD</math>, <math>DC = 2 \cdot CB</math> and points <math>E</math> and <math>F</math> lie on <math>\overline{AB}</math> so that <math>\overline{ED}</math> and <math>\overline{FD}</math> trisect <math>\angle ADC</math> as shown. What is the ratio of the area of <math>\triangle DEF</math> to the area of rectangle <math>ABCD</math>? |
<asy> | <asy> |
Revision as of 16:18, 12 July 2020
Contents
Problem
In rectangle ,
and points
and
lie on
so that
and
trisect
as shown. What is the ratio of the area of
to the area of rectangle
?
Solution 1
Let the length of be
, so that the length of
is
and
.
Because is a rectangle,
, and so
. Thus
is a
right triangle; this implies that
, so
. Now drop the altitude from
of
, forming two
triangles.
Because the length of is
, from the properties of a
triangle the length of
is
and the length of
is thus
. Thus the altitude of
is
, and its base is
, so its area is
.
To finish, .
Solution 2
WLOG, let and
. Furthermore, drop an an altitude from
to
, which meets
at
. Since
is right and has been trisected, it follows that
and
are both
triangles. Therefore,
, and
. Hence, it follows that
. Since
is right, the height and base of
are
and
, respectively. Thus, the area of
is
, and the area of rectengle
is
, so the ratio beween the area of
and
is
. Note that we are able to assume that
and
because we were asked to find the ratio between two areas.
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.