Difference between revisions of "2014 AMC 10A Problems/Problem 16"
(Cleaned up the previous solution, fixed wording in some places. also added diagram.) |
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Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. | Denote <math>D=(0,0)</math>. Then <math>A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)</math>. Let the intersection of <math>AF</math> and <math>DH</math> be <math>X</math>, and the intersection of <math>BF</math> and <math>CH</math> be <math>Y</math>. Then we want to find the coordinates of <math>X</math> so we can find <math>XY</math>. From our points, the slope of <math>AF</math> is <math>\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4</math>, and its <math>y</math>-intercept is just <math>2</math>. Thus the equation for <math>AF</math> is <math>y = -4x + 2</math>. We can also quickly find that the equation of <math>DH</math> is <math>y = 2x</math>. Setting the equations equal, we have <math>2x = -4x +2 \implies x = \frac13</math>. Because of symmetry, we can see that the distance from <math>Y</math> to <math>BC</math> is also <math>\frac13</math>, so <math>XY = 1 - 2 \cdot \frac13 = \frac13</math>. Now the area of the kite is simply the product of the two diagonals over <math>2</math>. Since the length <math>HF = 1</math>, our answer is <math>\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}</math>. | ||
− | + | <asy> | |
import graph; | import graph; | ||
size(9cm); | size(9cm); | ||
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− | label(" | + | label("$A\: (0,2)$",A,NW); |
− | label(" | + | label("$B$",B,NE); |
− | label(" | + | label("$C$",C,SE); |
− | label(" | + | label("$D \: (0,0)$",D,SW); |
− | label(" | + | label("$E$",E,E); |
− | label(" | + | label("$F\: (\frac12,0)$",F,S); |
− | label(" | + | label("$G$",G,W); |
− | label(" | + | label("$H \: (\frac12,1)$",H,N); |
− | label(" | + | label("$Y$",Y,E); |
− | label(" | + | label("$X$",X,W); |
− | label(" | + | label("$\frac12$",(0.25,0),S); |
− | label(" | + | label("$\frac12$",(0.75,0),S); |
− | label(" | + | label("$1$",(1,0.5),E); |
− | label(" | + | label("$1$",(1,1.5),E); |
− | + | </asy> | |
==Solution 2== | ==Solution 2== |
Revision as of 02:13, 11 February 2014
Contents
[hide]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 1
Denote . Then
. Let the intersection of
and
be
, and the intersection of
and
be
. Then we want to find the coordinates of
so we can find
. From our points, the slope of
is
, and its
-intercept is just
. Thus the equation for
is
. We can also quickly find that the equation of
is
. Setting the equations equal, we have
. Because of symmetry, we can see that the distance from
to
is also
, so
. Now the area of the kite is simply the product of the two diagonals over
. Since the length
, our answer is
.
Solution 2
Let the area of the shaded region be . Let the other two vertices of the kite be
and
with
closer to
than
. Note that
. The area of
is
and the area of
is
. We will solve for the areas of
and
in terms of x by noting that the area of each triangle is the length of the perpendicular from
to
and
to
respectively. Because the area of
=
based on the area of a kite formula,
for diagonals of length
and
,
. So each perpendicular is length
. So taking our numbers and plugging them into
gives us
Solving this equation for
gives us
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.