Difference between revisions of "2008 AMC 12A Problems/Problem 25"
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Hence, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{(D)}\frac{1}{2^{98}}}</math> ~ brainfertilzer. | Hence, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \boxed{\mathbf{(D)}\frac{1}{2^{98}}}</math> ~ brainfertilzer. | ||
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+ | ==Solution 4 (Kinda braindead)== | ||
+ | Start by turning the two equations into <math>a_n = \frac{a_{n+1}\sqrt{3}+b_{n+1}}{4}</math> and <math>b_n = \frac{b_{n+1}\sqrt{3}-a_{n+1}}{4}</math>. Note that these are just obtained by solving the equations. | ||
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+ | This makes finding values of <math>a_n</math> and <math>b_n</math> much easier, and soon we notice that <math>a_{97} = \frac{1}{2}</math> and <math>b_{97} = -\frac{1}{4}</math>. After that, we get that <math>a_{94} = -\frac{1}{32}</math> and <math>b_{94} = -\frac{1}{16}</math>. Observe that <math>|a_n| = |\frac{b_{n+3}}{8}|</math> and <math>|b_n| = |\frac{a_{n+3}}{8}|</math>. This is basically just ignoring signs. | ||
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+ | Now, we proceed to find <math>|a_1| = \frac{1}{2^{97}}</math> while <math>|b_1| = \frac{1}{2^{98}}</math>. Despite there being 4 possible sign combinations for <math>(a_1, b_1)</math>, the only achievable answer choice is <math>\boxed{\textbf{(D)}\frac{1}{2^{98}}}</math> | ||
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+ | -skibbysiggy | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 22:13, 2 September 2024
Contents
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.
Solution 4 (Kinda braindead)
Start by turning the two equations into and . Note that these are just obtained by solving the equations.
This makes finding values of and much easier, and soon we notice that and . After that, we get that and . Observe that and . This is basically just ignoring signs.
Now, we proceed to find while . Despite there being 4 possible sign combinations for , the only achievable answer choice is
-skibbysiggy
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 |
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