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 "2010 AIME II Problems/Problem 9"

(Created page with '== Problem 9 == Let <math>ABCDEF</math> be a regular hexagon. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the midpoi…')
 
(→‎Solution)
Line 54: Line 54:
 
</asy></center>
 
</asy></center>
  
−
Let <math>M</math> be the intersection of <math>\overbar{AG}</math> and <math>\overbar{BI}</math>
+
Let <math>M</math> be the intersection of <math>\overline{AG}</math> and <math>\overline{BI}</math>
  
−
and <math>N</math> be the intersection of <math>\overbar{BI}</math> and <math>\overbar{CJ}</math>.
+
and <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>.
 +
 
 +
and let <math>BC=2</math>
 +
 
 +
Note that <math>\angle BMH</math> is the vertical angle to an angle of regular hexagon, thus, it is <math>120^\circ</math>.
 +
 
 +
Because <math>\triangle ABH</math> and <math>\triangle BCI</math> are rotational images of one another, we get that <math>\angle{MBH}=\angle{HAB}</math> and hence <math>\triangle ABH \sim \triangle BMH \sim \triangle BCI</math>.
 +
 
 +
Using a simlar argument, <math>NI=MH</math>.
 +
 
 +
<math>MN=BI-NI-BM=BI-(BM+MH)</math>
 +
 
 +
Applying law of cosine on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))=\sqrt{7}</math>
 +
 
 +
<math>\frac{BC+CI}{BI}=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH}</math>
 +
 
 +
<math>BM+MH=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}}</math>
 +
 
 +
<math>MN=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}}</math>
 +
 
 +
<math>\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}</math>
 +
 
 +
Thus, answer is <math>\boxed{011}</math>
  
−
More coming^v^
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=8|num-a=10|n=II}}
 
{{AIME box|year=2010|num-b=8|num-a=10|n=II}}

Revision as of 19:15, 3 April 2010

Problem 9

Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\overbar{AH}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{BI}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{CJ}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{DK}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{EL}$ (Error compiling LaTeX. Unknown error_msg), and $\overbar{FG}$ (Error compiling LaTeX. Unknown error_msg) bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of $ABCDEF$ be expressed as a fraction $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

[asy] defaultpen(0.8pt+fontsize(12pt)); pair A,B,C,D,E,F; pair G,H,I,J,K,L; A=dir(0); B=dir(60); C=dir(120); D=dir(180); E=dir(240); F=dir(300); draw(A--B--C--D--E--F--cycle,blue); G=(A+B)/2; H=(B+C)/2; I=(C+D)/2; J=(D+E)/2; K=(E+F)/2; L=(F+A)/2; int i; for (i=0; i<6; i+=1) { draw(rotate(60*i)*(A--H),dotted); } pair M,N,O,P,Q,R; M=extension(A,H,B,I); N=extension(B,I,C,J); O=extension(C,J,D,K); P=extension(D,K,E,L); Q=extension(E,L,F,G); R=extension(F,G,A,H); draw(M--N--O--P--Q--R--cycle,red); label('A',A, E); label('B',B,NE); label('C',C,NW); label('D',D, W); label('E',E,SW); label('F',F,SE); label('G',G,NE); label('H',H, N); label('I',I,NW); label('J',J,SW); label('K',K, S); label('L',L,SE); label('M',M); label('N',N); [/asy]

Let $M$ be the intersection of $\overline{AG}$ and $\overline{BI}$

and $N$ be the intersection of $\overline{BI}$ and $\overline{CJ}$.

and let $BC=2$

Note that $\angle BMH$ is the vertical angle to an angle of regular hexagon, thus, it is $120^\circ$.

Because $\triangle ABH$ and $\triangle BCI$ are rotational images of one another, we get that $\angle{MBH}=\angle{HAB}$ and hence $\triangle ABH \sim \triangle BMH \sim \triangle BCI$.

Using a simlar argument, $NI=MH$.

$MN=BI-NI-BM=BI-(BM+MH)$

Applying law of cosine on $\triangle BCI$, $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))=\sqrt{7}$ (Error compiling LaTeX. Unknown error_msg)

$\frac{BC+CI}{BI}=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH}$

$BM+MH=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}}$

$MN=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}}$

$\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}$

Thus, answer is $\boxed{011}$

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_9&oldid=34208"