2008 AIME II Problems/Problem 13
Problem
A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let be the region outside the hexagon, and let
. Then the area of
has the form
, where
and
are positive integers. Find
.
Solution
If a point is in
, then the point
is in
(where cis denotes
). Since
is symmetric about the origin, it suffices to consider the result of the transformation when
, and then to multiply by
to account for the entire area.
We note that the region , where
is the region outside the circle of radius
centered at the origin, then
is simply the region inside a circle of radius
centered at the origin. It now suffices to find what happens to the mapping of the region
.
The equation of the hexagon side in that region is , which is transformed to
. Let
where
; then
, so the equation becomes
. Hence the side is sent to a unit circle centered at
.
Then is the union of six unit circles centered at $\cis \frac{k\pi}{6}$ (Error compiling LaTeX. Unknown error_msg),
, and the region
. That is show below.
![[asy] picture p; draw(p,expi(pi/6)--expi(-pi/6)--(0,0)--cycle);draw(p,arc(1/3^.5,1/3^.5,-60,60)); add(p);add(rotate(60)*p);add(rotate(120)*p);add(rotate(180)*p);add(rotate(240)*p);add(rotate(300)*p); [/asy]](http://latex.artofproblemsolving.com/2/b/c/2bcffc613cf155df3a52c36e16a36a29bc1e93b7.png)
The area of the regular hexagon is just . The area of each of the
sectors is
. Their sum is
, and
.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |