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Difference between revisions of "2003 AMC 12A Problems/Problem 14"
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==Solution==
 
==Solution==
 
===Solution 1===
 
===Solution 1===
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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 <math>\text{ABCD}</math> is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of <math>\text{AKB}</math>, and thus <math>\text{DMC}</math>, is <math>2\sqrt{3}</math>.  
  
−
The diagonal of the square KNML will then be <math>4+4\sqrt{3}</math>. From here there are 2 ways to proceed:
+
The diagonal of the square <math>\text{KNML}</math> will then be <math>4+4\sqrt{3}</math>. From here there are 2 ways to proceed:
  
−
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>.
+
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{\textbf{(D) } 32+16\sqrt{3}}</math>.
  
−
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}=\boxed{\mathrm{(D)}\ 32+16\sqrt{3}}</math>.
+
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}=\boxed{\textbf{(D) } 32+16\sqrt{3}}</math>.
  
 
===Solution 2===
 
===Solution 2===
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Using Law of Cosines, we find that side <math>(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sqrt{3}</math>.
 
Using Law of Cosines, we find that side <math>(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sqrt{3}</math>.
  
−
However, the area of <math>KLMN</math> is simply <math>(KL)^2</math>, hence the answer is <math>32+16\sqrt{3}\implies {D}</math>.
+
However, the area of <math>KLMN</math> is simply <math>(KL)^2</math>, hence the answer is <math>\boxed{\textbf{(D) } 32+16\sqrt{3}}</math>
  
 
===Solution 3===
 
===Solution 3===
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First we show that <math>KLMN</math> is a square. How? Show that it is a rhombus that has a right angle.
 
First we show that <math>KLMN</math> is a square. How? Show that it is a rhombus that has a right angle.
  
−
<math>\angle{NAK}=\angle{KBL}=\angle{LCM}=\angle{DNM}=150^\circ</math>. All the sides of the equilateral triangles are equal, so the triangles are congruent. Notice that <math>\angle{KNL}=45^\circ</math>, etc, so <math>\angle{KNM}=90^circ. So we have a square.
+
<math>\angle{NAK}=\angle{KBL}=\angle{LCM}=\angle{DNM}=150^\circ</math>. All the sides of the equilateral triangles are equal, so the triangles are congruent. Notice that <math>\angle{KNL}=45^\circ</math>, etc, so <math>\angle{KNM}=90^\circ</math>. So we have a square.
  
−
Instead of going to find </math>KN<math> using Law of Cosines, we can inscribe the square in another bigger one and then subtract the four right triangles in the corners, so after all of this, we find that the answer is </math>32+16\sqrt{3}<math>. Select </math>\boxed{D}$.
+
Instead of going to find <math>KN</math> using Law of Cosines, we can inscribe the square in another bigger one and then subtract the four right triangles in the corners, so after all of this, we find that the answer is <math>\boxed{\textbf{(D) } 32+16\sqrt{3}}</math>
  
−
~hastapasta
+
~hastapasta (edited)
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:26, 22 June 2023

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

The diagonal of the square $\text{KNML}$ 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 $\frac{16+48+32\sqrt{3}}{2}=\boxed{\textbf{(D) } 32+16\sqrt{3}}$.

Second: 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 $\frac{(4+4\sqrt{3})^2}{2}=\boxed{\textbf{(D) } 32+16\sqrt{3}}$.

Solution 2

Because $ABCD$ has area $16$, its side length is simply $\sqrt{16}\implies 4$.

Angle chasing, we find that the angle of $KBL=360-(90+2(60))=360-(210)=150$.

We also know that $KB=4, BL=4$.

Using Law of Cosines, we find that side $(KL)^2=16+16-2(16)cos{150}\implies (KL)^2=32+\frac{32\sqrt{3}}{2}\implies (KL)^2=32+16\sqrt{3}$.

However, the area of $KLMN$ is simply $(KL)^2$, hence the answer is $\boxed{\textbf{(D) } 32+16\sqrt{3}}$

Solution 3

First we show that $KLMN$ is a square. How? Show that it is a rhombus that has a right angle.

$\angle{NAK}=\angle{KBL}=\angle{LCM}=\angle{DNM}=150^\circ$. All the sides of the equilateral triangles are equal, so the triangles are congruent. Notice that $\angle{KNL}=45^\circ$, etc, so $\angle{KNM}=90^\circ$. So we have a square.

Instead of going to find $KN$ using Law of Cosines, we can inscribe the square in another bigger one and then subtract the four right triangles in the corners, so after all of this, we find that the answer is $\boxed{\textbf{(D) } 32+16\sqrt{3}}$

~hastapasta (edited)

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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