Difference between revisions of "2008 AIME II Problems/Problem 9"
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Basically, the thing comes out to | Basically, the thing comes out to | ||
− | <center><math>a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{ | + | <center><math>a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10</math></center> |
Notice that | Notice that | ||
− | <center><math>10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10( | + | <center><math>10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10( \sqrt {2}/2 - i\sqrt {2}/2)</math></center> |
Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is | Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is | ||
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | <center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> |
Revision as of 00:58, 15 August 2014
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
Let P(x, y) be the position of the particle on the xy-plane, r be the length OP where O is the origin, and be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then
and
.
Let
be the position of the particle after the nth move, where
and
. Then
,
. This implies
,
.
Substituting
and
, we have
and
again for the first time. Thus,
and
. Hence, the final answer is
![$5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$](http://latex.artofproblemsolving.com/d/6/e/d6e93554c7f003c0766d1509e99aad3fbd7d9f19.png)
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
![$a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10$](http://latex.artofproblemsolving.com/b/d/b/bdba935af30d8eca5158fe6587d0c83bf1952d15.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/3/7/9/379a39d9221ecd166a626870fbff994aad79a5c5.png)
Furthermore, . Thus, the final answer is
![$5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$](http://latex.artofproblemsolving.com/d/6/e/d6e93554c7f003c0766d1509e99aad3fbd7d9f19.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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.