Difference between revisions of "2014 AMC 10A Problems/Problem 16"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | Note that the region is a kite; hence its diagonals are perpendicular and it has area <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>. Since <math>HF=1</math> as both <math>H</math> and <math>F</math> are midpoints of parallel sides of rectangle <math>GECD</math> and <math>CE=1</math>, we let <math>b=HF=1</math>. Now all we need to do is to find <math>a</math>. | ||
+ | |||
+ | Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. This gives us <math>a=IJ</math>. Now let <math>D=(0,0)</math>. We thus find that the equation of <math>\overleftrightarrow{AF}</math> is <math>4x+y=2</math> and that of <math>\overleftrightarrow{DH}</math> is <math>2x-y=0</math>. Solving this system gives us <math>x=\dfrac{1}{3}</math>, so the <math>x</math>-coordinate of <math>I</math> is <math>\dfrac{1}{3}</math>; in other words, <math>I</math> is <math>\dfrac{1}{3}</math> from <math>\overline{AD}</math>. By symmetry, <math>J</math> is also the same distance from <math>\overline{BC}</math>, so as <math>CD=1</math> we have <math>a=IJ=1-\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{1}{3}</math>. Hence the area of the kite is <math>\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(C)}\ \dfrac{1}{6}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 08:05, 7 February 2014
Contents
Problem
In rectangle ,
,
, and points
,
, and
are midpoints of
,
, and
, respectively. Point
is the midpoint of
. What is the area of the shaded region?
Solution
Solution 1
Note that the region is a kite; hence its diagonals are perpendicular and it has area for diagonals of length
and
. Since
as both
and
are midpoints of parallel sides of rectangle
and
, we let
. Now all we need to do is to find
.
Let the other two vertices of the kite be and
with
closer to
than
. This gives us
. Now let
. We thus find that the equation of
is
and that of
is
. Solving this system gives us
, so the
-coordinate of
is
; in other words,
is
from
. By symmetry,
is also the same distance from
, so as
we have
. Hence the area of the kite is
.
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
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