2008 AMC 12A Problems/Problem 25
Problem
A sequence , , , of points in the coordinate plane satisfies
for .
Suppose that . What is ?
Solution 1
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, .
[s]Shortcut: no answer has in the numerator. 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, .[/s] Does not work as there are negative answers.
Solution 2 (algebra)
Let . Then, we can begin to list out terms as follows:
We notice that the sequence follows the rule
We can now start listing out every third point, getting:
We can make two observations from this:
(1) In , the coefficient of and is
(2) The positioning of and , and their signs, cycle with every terms.
We know then that from (1), the coefficients of and in are both
We can apply (2), finding , so the positions and signs of and are the same in as they are in .
From this, we can get . We know that , so we get the following:
The answer is ..
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
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