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, the answer is 4 + 7 = | + | Thus, the answer is <math>4 + 7 = \boxed{011}.</math> |
===Solution 2 (Coordinate Bash)=== | ===Solution 2 (Coordinate Bash)=== |
Revision as of 19:27, 2 June 2023
Problem
Let be a regular hexagon. Let
,
,
,
,
, and
be the midpoints of sides
,
,
,
,
, and
, respectively. The segments
,
,
,
,
, and
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
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
Without loss of generality, let
Note that is the vertical angle to an angle of the regular hexagon, so it has a measure of
.
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
Solution 2 (Coordinate Bash)
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 .
Solution 3
Use the diagram. Now notice that all of the "overlapping triangles" are congruent. You can use the AA similarity to see that the small triangles are similar to the large triangles. Now you can proceed as in Solution 1.
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.