2006 AIME II Problems/Problem 12
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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
Notice that 
because 
. Also, 
because they both correspond to arc 
. So 
.
![[EAF] = \frac12 (AE)(EF)\sin \angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/3/1/a318bfce5e8506e3827a2d13ff3f88f3aeebbc63.gif)
Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, ![[GBC] = \frac {BC^2}{AF^2}\cdot[EAF] = \frac {12}{11^2 + 13^2 - 2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac {143\sqrt3}4 = \...](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/0/0/300556300f92f45fd1d8c3597598072f08e5c08c.gif)
. Therefore, the answer is 
.
See also
| 2006 AIME II (Problems • Resources) | ||
| Preceded by Problem 11 | Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
