Difference between revisions of "2014 AMC 10A Problems/Problem 16"
(→Solution 2: boxed answer) |
(Cleaned up the previous solution, fixed wording in some places. also added diagram.) |
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==Solution 1== | ==Solution 1== | ||
− | + | 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; | ||
+ | size(9cm); | ||
+ | pen dps = fontsize(10); defaultpen(dps); | ||
+ | pair D = (0,0); | ||
+ | pair F = (1/2,0); | ||
+ | pair C = (1,0); | ||
+ | pair G = (0,1); | ||
+ | pair E = (1,1); | ||
+ | pair A = (0,2); | ||
+ | pair B = (1,2); | ||
+ | pair H = (1/2,1); | ||
+ | |||
+ | // do not look | ||
+ | pair X = (1/3,2/3); | ||
+ | pair Y = (2/3,2/3); | ||
+ | |||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(G--E); | ||
+ | draw(A--F--B); | ||
+ | draw(D--H--C); | ||
+ | filldraw(H--X--F--Y--cycle,grey); | ||
+ | draw(X--Y,dashed); | ||
+ | |||
+ | |||
+ | |||
+ | label("<math>A\: (0,2)</math>",A,NW); | ||
+ | label("<math>B</math>",B,NE); | ||
+ | label("<math>C</math>",C,SE); | ||
+ | label("<math>D \: (0,0)</math>",D,SW); | ||
+ | label("<math>E</math>",E,E); | ||
+ | label("<math>F\: (\frac12,0)</math>",F,S); | ||
+ | label("<math>G</math>",G,W); | ||
+ | label("<math>H \: (\frac12,1)</math>",H,N); | ||
+ | label("<math>Y</math>",Y,E); | ||
+ | label("<math>X</math>",X,W); | ||
+ | |||
+ | |||
+ | label("<math>\displaystyle\frac12</math>",(0.25,0),S); | ||
+ | label("<math>\displaystyle\frac12</math>",(0.75,0),S); | ||
+ | label("<math>1</math>",(1,0.5),E); | ||
+ | label("<math>1</math>",(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
.
[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1);
// do not look pair X = (1/3,2/3); pair Y = (2/3,2/3);
draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed);
label("",A,NW);
label("
",B,NE);
label("
",C,SE);
label("
",D,SW);
label("
",E,E);
label("
",F,S);
label("
",G,W);
label("
",H,N);
label("
",Y,E);
label("
",X,W);
label("",(0.25,0),S);
label("
",(0.75,0),S);
label("
",(1,0.5),E);
label("
",(1,1.5),E);
[/asy]
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.