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Difference between revisions of "2003 AMC 12A Problems/Problem 14"
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==Problem==
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Points <math>K, L, M,</math> and <math>N</math> lie in the plane of the square <math>ABCD</math> such that <math>AKB</math>, <math>BLC</math>, <math>CMD</math>, and <math>DNA</math> are equilateral triangles. If <math>ABCD</math> has an area of 16, find the area of <math>KLMN</math>.
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<asy>
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unitsize(2cm);
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defaultpen(fontsize(8)+linewidth(0.8));
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pair A=(-0.5,0.5), B=(0.5,0.5), C=(0.5,-0.5), D=(-0.5,-0.5);
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pair K=(0,1.366), L=(1.366,0), M=(0,-1.366), N=(-1.366,0);
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draw(A--N--K--A--B--K--L--B--C--L--M--C--D--M--N--D--A);
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label("$A$",A,SE);
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label("$B$",B,SW);
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label("$C$",C,NW);
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label("$D$",D,NE);
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label("$K$",K,NNW);
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label("$L$",L,E);
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label("$M$",M,S);
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label("$N$",N,W);
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</asy>
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<math> \textrm{(A)}\ 32\qquad\textrm{(B)}\ 16+16\sqrt{3}\qquad\textrm{(C)}\ 48\qquad\textrm{(D)}\ 32+16\sqrt{3}\qquad\textrm{(E)}\ 64 </math>
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==Solution==
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===Solution 1===
 
Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is <math>2\sqrt{3}</math>.  
 
Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is <math>2\sqrt{3}</math>.  
  
 
The diagonal of the square KNMC will then be <math>4+4\sqrt{3}</math>. From here there are 2 ways to proceed:
 
The diagonal of the square KNMC will then be <math>4+4\sqrt{3}</math>. From here there are 2 ways to proceed:
  
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First: Since the diagonal is <math>4+4\sqrt{3}</math>, the side length is <math>\frac{4+4\sqrt{3}}{\sqrt{2}}</math>, and the area is thus <math>\frac{16+48+32\sqrt{3}}{2}=32+16\sqrt{3}</math>.
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First: Since the diagonal is <math>4+4\sqrt{3}</math>, the side length is <math>\frac{4+4\sqrt{3}}{\sqrt{2}}</math>, and the area is thus <math>\frac{16+48+32\sqrt{3}}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>.
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Second: Since a square is a rhombus, the area of the square is <math>\frac{d_1d_2}{2}</math>, where <math>d_1</math> and <math>d_2</math> are the diagonals of the rhombus. Since the diagonal is <math>4+4\sqrt{3}</math>, the area is <math>\frac{(4+4\sqrt{3})^2}{2}=32+16\sqrt{3}</math>.
 
  
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Either way, our answer is <math>\boxed{D}</math>.
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===Solution 2===
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Since a square is a rhombus, the area of the square is <math>\frac{d_1d_2}{2}</math>, where <math>d_1</math> and <math>d_2</math> are the diagonals of the rhombus. Since the diagonal is <math>4+4\sqrt{3}</math>, the area is <math>\frac{(4+4\sqrt{3})^2}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2003|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2003|ab=A|num-b=13|num-a=15}}

Revision as of 14:10, 1 August 2011

Problem

Points $K, L, M,$ and $N$ lie in the plane of the square $ABCD$ such that $AKB$, $BLC$, $CMD$, and $DNA$ are equilateral triangles. If $ABCD$ has an area of 16, find the area of $KLMN$.

[asy] unitsize(2cm); defaultpen(fontsize(8)+linewidth(0.8)); pair A=(-0.5,0.5), B=(0.5,0.5), C=(0.5,-0.5), D=(-0.5,-0.5); pair K=(0,1.366), L=(1.366,0), M=(0,-1.366), N=(-1.366,0); draw(A--N--K--A--B--K--L--B--C--L--M--C--D--M--N--D--A); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$K$",K,NNW); label("$L$",L,E); label("$M$",M,S); label("$N$",N,W); [/asy]

$\textrm{(A)}\ 32\qquad\textrm{(B)}\ 16+16\sqrt{3}\qquad\textrm{(C)}\ 48\qquad\textrm{(D)}\ 32+16\sqrt{3}\qquad\textrm{(E)}\ 64$

Solution

Solution 1

Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is $2\sqrt{3}$.

The diagonal of the square KNMC will then be $4+4\sqrt{3}$. From here there are 2 ways to proceed:

First: Since the diagonal is $4+4\sqrt{3}$, the side length is $\frac{4+4\sqrt{3}}{\sqrt{2}}$, and the area is thus $\frac{16+48+32\sqrt{3}}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}$.

Solution 2

Since a square is a rhombus, the area of the square is $\frac{d_1d_2}{2}$, where $d_1$ and $d_2$ are the diagonals of the rhombus. Since the diagonal is $4+4\sqrt{3}$, the area is $\frac{(4+4\sqrt{3})^2}{2}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}$.

See Also

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 12 Problems and Solutions
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