Difference between revisions of "2008 AMC 12A Problems/Problem 25"
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<math>\mathrm{(A)}\ -\frac{1}{2^{97}}\qquad\mathrm{(B)}\ -\frac{1}{2^{99}}\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ \frac{1}{2^{98}}\qquad\mathrm{(E)}\ \frac{1}{2^{96}}</math> | <math>\mathrm{(A)}\ -\frac{1}{2^{97}}\qquad\mathrm{(B)}\ -\frac{1}{2^{99}}\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ \frac{1}{2^{98}}\qquad\mathrm{(E)}\ \frac{1}{2^{96}}</math> | ||
− | ==Solution== | + | ==Solution 1== |
This sequence can also be expressed using matrix multiplication as follows: | This sequence can also be expressed using matrix multiplication as follows: | ||
Line 19: | Line 19: | ||
Rotating <math>(2,4)</math> clockwise by <math>99 \cdot 30^\circ \equiv 90^\circ</math> yields <math>(4,-2)</math>. A dilation by a factor of <math>\frac{1}{2^{99}}</math> yields the point <math>(a_1,b_1) = \left(\frac{4}{2^{99}}, -\frac{2}{2^{99}} \right) = \left(\frac{1}{2^{97}}, -\frac{1}{2^{98}} \right)</math>. | Rotating <math>(2,4)</math> clockwise by <math>99 \cdot 30^\circ \equiv 90^\circ</math> yields <math>(4,-2)</math>. A dilation by a factor of <math>\frac{1}{2^{99}}</math> yields the point <math>(a_1,b_1) = \left(\frac{4}{2^{99}}, -\frac{2}{2^{99}} \right) = \left(\frac{1}{2^{97}}, -\frac{1}{2^{98}} \right)</math>. | ||
− | 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>. |
+ | |||
+ | ==Solution 2 (algebra)== | ||
+ | Let <math>(x,y)=(a_1,b_1)</math>. Then, we can begin to list out terms as follows: | ||
+ | |||
+ | |||
+ | |||
+ | <math>(a_2,b_2)=(x\sqrt{3}-y,y\sqrt{3}+x)</math> | ||
+ | |||
+ | <math>(a_3,b_3)=(2x-2y\sqrt{3},2y+2x\sqrt{3})</math> | ||
+ | |||
+ | <math>(a_4,b_4)=(-8y,8x)</math> | ||
+ | |||
+ | |||
+ | |||
+ | We notice that the sequence follows the rule <math>(a_{n+3},b_{n+3})=(-2^3b_n,2^3a_n)</math> | ||
+ | |||
+ | We can now start listing out every third point, getting: | ||
+ | |||
+ | |||
+ | |||
+ | <math>(a_1,b_1)=(x,y)</math> | ||
+ | |||
+ | <math>(a_4,b_4)=(-2^3y,2^3x)</math> | ||
+ | |||
+ | <math>(a_7,b_7)=(-2^6x,-2^6y)</math> | ||
+ | |||
+ | <math>(a_{10},b_{10})=(2^9y,-2^9x)</math> | ||
+ | |||
+ | <math>(a_{13},b_{13})=(2^{12}x,2^{12}y)</math> | ||
+ | |||
+ | |||
+ | |||
+ | We can make two observations from this: | ||
+ | |||
+ | (1) In <math>a_n</math>, the coefficient of <math>x</math> and <math>y</math> is <math>2^{n-1}</math> | ||
+ | |||
+ | (2) The positioning of <math>x</math> and <math>y</math>, and their signs, cycle with every <math>12</math> terms. | ||
+ | |||
+ | |||
+ | |||
+ | We know then that from (1), the coefficients of <math>x</math> and <math>y</math> in <math>(a_{100},b_{100})</math> are both <math>2^{99}</math> | ||
+ | |||
+ | We can apply (2), finding <math>100 \text{(mod }12)=4</math>, so the positions and signs of <math>x</math> and <math>y</math> are the same in <math>(a_{100},b_{100})</math> as they are in <math>(a_{4},b_{4})</math>. | ||
+ | |||
+ | From this, we can get <math>(a_{100},b_{100})=(-2^{99}y,2^{99}x)</math>. We know that <math>(a_{100},b_{100})=(2,4)</math>, so we get the following: | ||
+ | |||
+ | |||
+ | |||
+ | <math>-2^{99}y=2 \Rightarrow y=-\frac{1}{2^{98}}</math> | ||
+ | |||
+ | <math>2^{99}x=4 \Rightarrow x=\frac{1}{2^{97}}</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}} |
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 |
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All AMC 12 Problems and Solutions |
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