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Difference between revisions of "2010 AIME II Problems/Problem 9"

m (→‎Solution)
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== Problem 9 ==
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== Problem ==
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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 midpoints of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The segments <math>\overbar{AH}</math>, <math>\overbar{BI}</math>, <math>\overbar{CJ}</math>, <math>\overbar{DK}</math>, <math>\overbar{EL}</math>, and <math>\overbar{FG}</math> bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of <math>ABCDEF</math> be expressed as a fraction <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+
Let <math>ABCDEF</math> be a [[regular polygon|regular]] [[hexagon]]. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the [[midpoint]]s of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The [[segment]]s <math>\overbar{AH}</math>, <math>\overbar{BI}</math>, <math>\overbar{CJ}</math>, <math>\overbar{DK}</math>, <math>\overbar{EL}</math>, and <math>\overbar{FG}</math> bound a smaller regular hexagon. Let the [[ratio]] of the area of the smaller hexagon to the area of <math>ABCDEF</math> be expressed as a fraction <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
  
 +
__TOC__
 
==Solution==
 
==Solution==
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===diagram===
 
 
<center><asy>
 
<center><asy>
 
defaultpen(0.8pt+fontsize(12pt));
 
defaultpen(0.8pt+fontsize(12pt));
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label('A',A, E);
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label('$A$',A,(1,0));
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label('B',B,NE);
+
label('$B$',B,NE);
−
label('C',C,NW);
+
label('$C$',C,NW);
−
label('D',D, W);
+
label('$D$',D, W);
−
label('E',E,SW);
+
label('$E$',E,SW);
−
label('F',F,SE);
+
label('$F$',F,SE);
−
label('G',G,NE);
+
label('$G$',G,NE);
−
label('H',H, N);
+
label('$H$',H, (0,1));
−
label('I',I,NW);
+
label('$I$',I,NW);
−
label('J',J,SW);
+
label('$J$',J,SW);
−
label('K',K, S);
+
label('$K$',K, S);
−
label('L',L,SE);
+
label('$L$',L,SE);
−
label('M',M);
+
label('$M$',M);
−
label('N',N);
+
label('$N$',N);
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label('O',(0,0));
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label('$O$',(0,0),NE); dot((0,0));
 
</asy></center>
 
</asy></center>
  
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Let <math>O</math> be the center.
 
Let <math>O</math> be the center.
−
 
−
 
  
 
===Solution 1===
 
===Solution 1===
 +
Let <math>BC=2</math> (without loss of generality).
  
−
Let <math>BC=2</math>
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Note that <math>\angle BMH</math> is the vertical angle to an angle of regular hexagon, and so has degree <math>120^\circ</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>.
 
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>.
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Using a similar argument, <math>NI=MH</math>, and
−
 
 
−
<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>
 
  
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<math>BM+MH=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}}</math>
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<cmath>MN=BI-NI-BM=BI-(BM+MH).</cmath>
  
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<math>MN=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}}</math>
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Applying the [[Law of cosines]] on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\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>
+
<cmath>\begin{align*}\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}\end{align*}</cmath>
  
−
Thus, answer is <math>\boxed{011}</math>
+
Thus, answer is <math>\boxed{011}</math>.
  
 
===Solution 2===
 
===Solution 2===
−
Let's coordinate bash this out.
+
We can use coordinates. Let <math>O</math> be at <math>(0,0)</math> with <math>A</math> at <math>(1,0)</math>,  
−
 
 
−
Let <math>O</math> be at <math>(0,0)</math> with <math>A</math> be at <math>(1,0)</math>,  
 
  
 
then <math>B</math> is at <math>(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>,  
 
then <math>B</math> is at <math>(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)</math>,  
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<math>D</math> is at <math>(\cos(180^\circ),\sin(180^\circ))=(-1,0)</math>,  
 
<math>D</math> is at <math>(\cos(180^\circ),\sin(180^\circ))=(-1,0)</math>,  
  
−
<math>H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right)</math>
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<cmath>\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\
−
 
+
&I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}</cmath>
−
<math>I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)</math>
 
  
 
<br/>
 
<br/>
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Let's solve the system of equation to find <math>M</math>
 
Let's solve the system of equation to find <math>M</math>
  
−
<math>-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)</math>
+
<cmath>\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\
 +
-5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\
 +
x&=\frac{1}{7} \\
 +
y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}</cmath>
  
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<math>-5\sqrt{3}x=2\sqrt{3}x-\sqrt{3}</math>
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Finally,
  
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<math>x=\frac{1}{7}</math>
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<cmath>\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\
 +
&\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}</cmath>
  
−
<math>y=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}</math>
+
Thus, the answer is <math>\boxed{011}</math>.
−
 
 
−
<br/>
 
−
 
 
−
<math>\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}}</math>
 
−
 
 
−
<math>\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}</math>
 
−
 
 
−
Thus, answer is <math>\boxed{011}</math>
 
−
 
 
−
P.S: Not too bad, isn't it?
 
  
 
== 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}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 10:53, 6 April 2010

Problem

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,(1,0)); 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, (0,1)); label('$I$',I,NW); label('$J$',J,SW); label('$K$',K, S); label('$L$',L,SE); label('$M$',M); label('$N$',N); label('$O$',(0,0),NE); dot((0,0)); [/asy]

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

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

Let $O$ be the center.

Solution 1

Let $BC=2$ (without loss of generality).

Note that $\angle BMH$ is the vertical angle to an angle of regular hexagon, and so has degree $120^\circ$.

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 similar argument, $NI=MH$, and

\[MN=BI-NI-BM=BI-(BM+MH).\]

Applying the Law of cosines on $\triangle BCI$, $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}$

\begin{align*}\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}\end{align*}

Thus, answer is $\boxed{011}$.

Solution 2

We can use coordinates. Let $O$ be at $(0,0)$ with $A$ at $(1,0)$,

then $B$ is at $(\cos(60^\circ),\sin(60^\circ))=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$,

$C$ is at $(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$,

$D$ is at $(\cos(180^\circ),\sin(180^\circ))=(-1,0)$,

\begin{align*}&H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right) \\ &I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)\end{align*}


Line $AH$ has the slope of $-\frac{\sqrt{3}}{2}$ and the equation of $y=-\frac{\sqrt{3}}{2}(x-1)$


Line $BI$ has the slope of $\frac{\sqrt{3}}{5}$ and the equation $y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)$


Let's solve the system of equation to find $M$

\begin{align*}-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}&=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right) \\ -5\sqrt{3}x&=2\sqrt{3}x-\sqrt{3} \\ x&=\frac{1}{7} \\ y&=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}\end{align*}

Finally,

\begin{align*}&\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}} \\ &\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{OM}{OA}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}\end{align*}

Thus, the 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
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