Difference between revisions of "2008 AMC 12A Problems/Problem 25"
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Therefore, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \frac{1}{2^{98}} \Rightarrow D</math>. | Therefore, <math>a_1 + b_1 = \frac{1}{2^{97}} - \frac{1}{2^{98}} = \frac{1}{2^{98}} \Rightarrow D</math>. | ||
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==Solution 2 (algebra)== | ==Solution 2 (algebra)== | ||
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The answer is <math>x+y=\frac{1}{2^{97}}-\frac{1}{2^{98}}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}</math>.. | The answer is <math>x+y=\frac{1}{2^{97}}-\frac{1}{2^{98}}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}</math>.. | ||
+ | ==Solution 3== | ||
+ | |||
+ | The ordered pairs and <math>\sqrt{3}</math>'s makes us think to use complex numbers. We have <math>(a_{n+1},b_{n+1}) = 2\left(\frac{\sqrt{3}}{2}a_n - \frac{1}{2}b_n, \frac{\sqrt{3}}{2}b_n + \frac{1}{2}a_n\right)</math>, so <math>a_{n+1} + b_{n+1}i = 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)(a_n + b_ni) = \frac{1}{2}e^{i\pi/6}(a_n + b_ni)</math>. Letting <math>z_n = a_n + b_ni</math> (so <math>z_{n+1} = a_{n+1} + b_{n+1}i</math>), we have <math>z_{n+1} = 2e^{i\pi/6}z_n</math>. Letting <math>n\rightarrow n-1</math>, we have <math>z_n = 2e^{i\pi/6}z_{n-1}</math>, so <math>z_{n-1} = \frac{1}{2}e^{-i\pi/6}z_n</math>. This is the reverse transformation. We have | ||
+ | <cmath> z_{99} = \frac{1}{2}e^{-i\pi/6}z_{100}</cmath> | ||
+ | <cmath> z_{98} = \frac{1}{2^2}e^{2(-i\pi/6)}z_{100}</cmath> | ||
+ | <cmath> \vdots</cmath> | ||
+ | <cmath> z_{1} = \frac{1}{2^{99}}e^{99(-i\pi/6)}z_{100}</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{(D)}\frac{1}{2^{98}}}</math> ~ brainfertilzer. | ||
+ | |||
+ | |||
+ | ==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. | ||
+ | |||
+ | |||
+ | 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. | ||
+ | |||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | -skibbysiggy | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=_4UJzyBslFA | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=24|after=Last question}} | {{AMC12 box|year=2008|ab=A|num-b=24|after=Last question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:13, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.