Difference between revisions of "2006 AIME II Problems/Problem 12"
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Revision as of 23:25, 4 July 2013
Problem
Equilateral is inscribed in a circle of radius
. Extend
through
to point
so that
and extend
through
to point
so that
Through
draw a line
parallel to
and through
draw a line
parallel to
Let
be the intersection of
and
Let
be the point on the circle that is collinear with
and
and distinct from
Given that the area of
can be expressed in the form
where
and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime, find
Solution
![[asy] size(250); pointpen = black; pathpen = black + linewidth(0.65); pen s = fontsize(8); pair A=(0,0),B=(-3^.5,-3),C=(3^.5,-3),D=13*expi(-2*pi/3),E1=11*expi(-pi/3),F=E1+D; path O = CP((0,-2),A); pair G = OP(A--F,O); D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O); D(B--MP("D",D,W,s)--MP("F",F,s)--MP("E",E1,E,s)--C); D(A--F);D(B--MP("G",G,SW,s)--C); MP("11",(A+E1)/2,NE);MP("13",(A+D)/2,NW);MP("l_1",(D+F)/2,SW);MP("l_2",(E1+F)/2,SE); [/asy]](http://latex.artofproblemsolving.com/2/9/9/299c51b0ad858e4078795c0f549803abc5d0ec0d.png)
Notice that because
. Also,
because they both correspond to arc
. So
.
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, . Therefore, the answer is
.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.