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>. | ||
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| + | 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 07:05, 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?
![[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); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); [/asy]](http://latex.artofproblemsolving.com/1/2/5/12524bb96209b0c956b2060192deb8429c73e3f3.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
