Difference between revisions of "2010 AIME II Problems/Problem 9"
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\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> | \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, the answer is 4 + 7 = <math>\boxed{011}</math>. |
===Solution 2=== | ===Solution 2=== |
Revision as of 14:04, 13 October 2014
Problem
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
.
Contents
[hide]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]](http://latex.artofproblemsolving.com/f/9/f/f9fa5842a4caeeff7427aa1c43da018886b49731.png)
Let be the intersection of
and
and be the intersection of
and
.
Let be the center.
Solution 1
Let (without loss of generality).
Note that is the vertical angle to an angle of regular hexagon, and so has degree
.
Because and
are rotational images of one another, we get that
and hence
.
Using a similar argument, , and
Applying the Law of cosines on ,
Thus, the answer is 4 + 7 = .
Solution 2
We can use coordinates. Let be at
with
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
Finally,
Thus, the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.