Difference between revisions of "2004 AMC 10B Problems/Problem 18"
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− | ==Solution== | + | ==Solution 1== |
+ | |||
+ | Let <math>x = [DBF]</math>. Because <math>\triangleACE</math> is divided into four triangles, <math>[ACE] = [BCD] + [ABF] + [DEF] + x</math>. | ||
+ | Because the area of <math>\triangleXYZ = \frac12 XY \cdot XZ \cdot sin(\angle X), \frac12 12 \cdot 16 = \frac12 9 \cdot 4 + \frac12 3 \cdot 15 \cdot sin(\angle A) + \frac12 5 \cdot 12 \cdot sin(\angle E) + x</math>. | ||
+ | <math>sin(\angle A) = 16 / 20</math> and <math>sin(\angle E) = 12 / 20</math>, so <math>96 = 18 + 18 + 18 + x</math>. | ||
+ | <math>x = 42</math>, so <math>[DBF] / [ACE] = 42 / 96 = 7 / 16 \Rightarrow \boxed{\frac 7{16}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
First of all, note that <math>\frac{AB}{AC} = \frac{CD}{CE} = \frac{EF}{EA} = \frac 14</math>, and therefore | First of all, note that <math>\frac{AB}{AC} = \frac{CD}{CE} = \frac{EF}{EA} = \frac 14</math>, and therefore |
Revision as of 21:29, 24 January 2015
Contents
Problem
In the right triangle , we have
,
, and
. Points
,
, and
are located on
,
, and
, respectively, so that
,
, and
. What is the ratio of the area of
to that of
?
Solution 1
Let . Because $\triangleACE$ (Error compiling LaTeX. Unknown error_msg) is divided into four triangles,
.
Because the area of $\triangleXYZ = \frac12 XY \cdot XZ \cdot sin(\angle X), \frac12 12 \cdot 16 = \frac12 9 \cdot 4 + \frac12 3 \cdot 15 \cdot sin(\angle A) + \frac12 5 \cdot 12 \cdot sin(\angle E) + x$ (Error compiling LaTeX. Unknown error_msg).
and
, so
.
, so
.
Solution 2
First of all, note that , and therefore
.
Draw the height from onto
as in the picture below:
Now consider the area of . Clearly the triangles
and
are similar, as they have all angles equal. Their ratio is
, hence
.
Now the area
of
can be computed as
=
.
Similarly we can find that as well.
Hence , and the answer is
.
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.