Difference between revisions of "2014 AMC 10B Problems/Problem 15"
m (→Solution 2) |
|||
Line 28: | Line 28: | ||
===Solution 2=== | ===Solution 2=== | ||
− | WLOG, let <math>AD = 1</math> and <math>BC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. | + | WLOG, let <math>AD = 1</math> and <math>BC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. Note that we are able to assume that <math>AD=1</math> and <math>BC = 2</math> because we were asked to find the ratio between two areas. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:06, 1 January 2018
Contents
[hide]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
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.