Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 12A Problems/Problem 14"

(→‎Solution 2)
(→‎Solution 3)
Line 47: Line 47:
 
<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.
 
<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}</math>.
+
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

Revision as of 19:52, 11 May 2022

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_14&oldid=173889"