2008 AIME II Problems/Problem 9: Difference between revisions
solution 2 by Phelphedo, solution 1 will be up soon |
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</cmath> | </cmath> | ||
Furthermore, <math>5a^{150} = - 5i</math>. So our final answer is | Furthermore, <math>5a^{150} = - 5i</math>. So our final answer is | ||
< | <cmath> | ||
5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}. | 5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}.</cmath> | ||
== See also == | == See also == | ||
Revision as of 19:32, 3 April 2008
Problem
A particle is located on the coordinate plane at 
. Define a move for the particle as a counterclockwise rotation of 
radians about the origin followed by a translation of 
units in the positive 
-direction. Given that the particle's position after 
moves is 
, find the greatest integer less than or equal to 
.
Solution
Solution 1
Show periodic with 
steps, then invert twice. Template:Incomplete
Solution 2
Let the particle's position be represented by a complex number. The transformation takes 
to 
where 
and 
. We let 
and 
so that we want to find 
.
Basically, the thing comes out to
![\[a_{150} = (((5a + 10)a + 10)a + 10...) = 5a^{150} + 10 a^{149} + 10a^{149} ... + 10\]](http://latex.artofproblemsolving.com/7/e/d/7ed122a02603e16779f665377cdee442019d9a18.png)
Notice that
![\[10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10( - \sqrt {2}/2 - i\sqrt {2}/2)\]](http://latex.artofproblemsolving.com/1/3/c/13cf83e85ad9854a2ebc057762265c80624426f6.png)
Furthermore, 
. So our final answer is
![\[5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}.\]](http://latex.artofproblemsolving.com/5/5/6/55620a665e752acdd4cbd9b88c8f4dd5defd6c5e.png)
See also
| 2008 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
