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2010 AIME II Problems/Problem 9

Revision as of 19:04, 3 April 2010 by Moplam (talk | contribs) (→‎Solution 2)

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); label('O',(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 $BC=2$

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 $\triangle BCI$, $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))}=\sqrt{7}$

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

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

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

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

Thus, answer is $\boxed{011}$

Solution 2

Let's coordinate bash this out.

Let $O$ be at $(0,0)$ with $A$ be 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 $(\cos(180^\circ),\sin(180^\circ))=(-1,0)$,

$H=\frac{B+C}{2}=\left(0,\frac{\sqrt{3}}{2}\right)$

$I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)$


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 $M$

$-\frac{\sqrt{3}}{2}(x-1)-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{1}{2}\right)$

$-5\sqrt{3}x=2\sqrt{3}x-\sqrt{3}$

$x=\frac{1}{7}$

$y=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}$


$\sqrt{x^2+y^2}=OM=\frac{1}{7}\sqrt{1^2+(3\sqrt{3})^2}=\frac{1}{7}\sqrt{28}=\frac{2}{\sqrt{7}}$

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

Thus, answer is $\boxed{011}$

P.S: Not too bad, isn't it?

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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