Difference between revisions of "2008 AMC 12A Problems/Problem 25"
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<cmath> = \frac{1}{2^{99}}e^{-i\pi/2}z_{100} = -\frac{1}{2^{99}}i(2 + 4i) = \frac{1}{2^{97}} - \frac{1}{2^{98}}i.</cmath> | <cmath> = \frac{1}{2^{99}}e^{-i\pi/2}z_{100} = -\frac{1}{2^{99}}i(2 + 4i) = \frac{1}{2^{97}} - \frac{1}{2^{98}}i.</cmath> | ||
− | Hence, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{( | + | Hence, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{(D)}\frac{1}{2^{98}}}</math> ~ brainfertilzer. |
==Video Solution== | ==Video Solution== |
Revision as of 10:58, 9 December 2022
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, .
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 ..
Solution 3
The ordered pairs and 's makes us think to use complex numbers. We have
, so
. Letting
(so
), we have
. Letting
, we have
, so
. This is the reverse transformation. We have
Hence, ~ brainfertilzer.
Video Solution
https://www.youtube.com/watch?v=_4UJzyBslFA
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.