Difference between revisions of "2010 AIME II Problems/Problem 9"
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Let <math>O</math> be the center. | Let <math>O</math> be the center. | ||
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===Solution 1=== | ===Solution 1=== | ||
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+ | 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>. | Note that <math>\angle BMH</math> is the vertical angle to an angle of regular hexagon, thus, it is <math>120^\circ</math>. |
Revision as of 20:02, 3 April 2010
Problem 9
Let be a regular hexagon. Let
,
,
,
,
, and
be the midpoints of sides
,
,
,
,
, and
, 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
be expressed as a fraction
where
and
are relatively prime positive integers. Find
.
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]](http://latex.artofproblemsolving.com/2/5/6/2566ffedf32eee10dacd7da4088512e3c2e4eb7e.png)
Let be the intersection of
and
and be the intersection of
and
.
Let be the center.
Solution 1
Let
Note that is the vertical angle to an angle of regular hexagon, thus, it is
.
Because and
are rotational images of one another, we get that
and hence
.
Using a simlar argument, .
Applying law of cosine on ,
Thus, answer is
Solution 2
Let's coordinate bash this out.
Let be at
with
be at
,
then is at
,
is at
,
is at
,
Line has the slope of
and the equation of
Line has the slope of
and the equation
Let's solve the system of equation to find
Thus, answer is
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 |