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: | ||
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Shortcut: no answer has <math>3</math> in the denominator. So the point cannot have orientation <math>(2,4)</math> or <math>(-2,-4)</math>. Also there are no negative answers. Any other non-multiple of <math>90^\circ</math> rotation of <math>30n^\circ</math> would result in the need of radicals. So either it has orientation <math>(4,-2)</math> or <math>(-4,2)</math>. Both answers add up to <math>2</math>. Thus, <math>2/2^{99}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}</math>. | Shortcut: no answer has <math>3</math> in the denominator. So the point cannot have orientation <math>(2,4)</math> or <math>(-2,-4)</math>. Also there are no negative answers. Any other non-multiple of <math>90^\circ</math> rotation of <math>30n^\circ</math> would result in the need of radicals. So either it has orientation <math>(4,-2)</math> or <math>(-4,2)</math>. Both answers add up to <math>2</math>. Thus, <math>2/2^{99}=\boxed{\textbf{(D) }\frac{1}{2^{98}}}</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>. | ||
+ | |||
+ | A.B. | ||
==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}} |
Revision as of 21:12, 11 January 2018
Contents
[hide]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, .
Shortcut: no answer has in the denominator. 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, .
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 .
A.B.
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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