2008 AMC 12A Problems/Problem 25
Problem
A sequence ,
,
,
of points in the coordinate plane satisfies
for
.
Suppose that . What is
?
Solution
This sequence can also be expressed using matrix multiplication as follows:
.
Thus, is formed by rotating
counter-clockwise about the origin by
and dilating the point's position with respect to the origin by a factor of
.
So, starting with and performing the above operations
times in reverse yields
.
Rotating clockwise by
yields
. A dilation by a factor of
yields the point
.
Therefore, .
Shortcut: no answer has in the denominator. So the point cannot have orientation
or
. Also there are no negative answers. Any other non-multiple of
rotation of
would result in the need of radicals. So either it has orientation
or
. Both answers add up to
. Thus,
.
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.