Difference between revisions of "2010 AIME II Problems/Problem 9"
m (→Solution) |
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label('M',M); | label('M',M); | ||
label('N',N); | label('N',N); | ||
+ | label('O',(0,0)); | ||
</asy></center> | </asy></center> | ||
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and <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>. | and <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>. | ||
+ | |||
+ | Let <math>O</math> be the center. | ||
and let <math>BC=2</math> | and let <math>BC=2</math> | ||
+ | |||
+ | ===Solution 1=== | ||
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>. | ||
Line 79: | Line 84: | ||
Thus, answer is <math>\boxed{011}</math> | Thus, answer is <math>\boxed{011}</math> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Let's coordinate bash this out. | ||
+ | |||
+ | 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>, | ||
+ | |||
+ | <math>C</math> is at <math>(\cos(120^\circ),\sin(120^\circ))=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)</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> | ||
+ | |||
+ | <math>I=\frac{C+D}{2}=\left(-\frac{3}{4},\frac{\sqrt{3}}{4}\right)</math> | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | Line <math>AH</math> has the slope of <math>-\frac{\sqrt{3}}{2}</math> | ||
+ | |||
+ | and the equation of <math>y=-\frac{\sqrt{3}}{2}(x-1)</math> | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | Line <math>BI</math> has the slope of <math>\frac{\sqrt{3}}{5}</math> | ||
+ | |||
+ | and the equation <math>y-\frac{3}{2}=\frac{\sqrt{3}}{5}\left(x-\frac{\sqrt{1}}{2}\right)</math> | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | 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{\sqrt{1}}{2}\right)</math> | ||
+ | |||
+ | <math>-5\sqrt{3}x=2\sqrt{3}x-\sqrt{3}</math> | ||
+ | |||
+ | <math>x=\frac{1}{7}</math> | ||
+ | |||
+ | <math>y=-\frac{\sqrt{3}}{2}(x-1)=\frac{3\sqrt{3}}{7}</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}} |
Revision as of 20:01, 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.
and let
Solution 1
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 |