Difference between revisions of "2008 AIME II Problems/Problem 9"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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− | 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 "a" be the inclination of OP to the x-axis. If | + | 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 "a" be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then <math>x'=rcos(\pi/4+a)+10 = \sqrt{2}(x - y)/2 + 10</math> and y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2. |
Let (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then x(n+1) + y(n+1) = \sqrt{2}(xn)+10, x(n+1) - y(n+1) = -\sqrt{2}(yn)+10. This implies | Let (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then x(n+1) + y(n+1) = \sqrt{2}(xn)+10, x(n+1) - y(n+1) = -\sqrt{2}(yn)+10. This implies | ||
x(n+2) = -yn + 5\sqrt{2}+ 10, y(n+2)=xn + 5\sqrt{2}. | x(n+2) = -yn + 5\sqrt{2}+ 10, y(n+2)=xn + 5\sqrt{2}. |
Revision as of 01:36, 7 May 2009
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 "a" 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 y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2.
Let (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then x(n+1) + y(n+1) = \sqrt{2}(xn)+10, x(n+1) - y(n+1) = -\sqrt{2}(yn)+10. This implies
x(n+2) = -yn + 5\sqrt{2}+ 10, y(n+2)=xn + 5\sqrt{2}.
Substituting x0 = 5 and y0 = 0, we have x8 = 5 and y8 = 0 again for the first time. p = x150 = x6 = -5\sqrt{2} and q = y150 = y6 = 5 + 5\sqrt{2}. 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
![$a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{149}+ \ldots + 10$](http://latex.artofproblemsolving.com/2/1/6/216deeef71518a6e2e0de2625fa1053ea4d085f8.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/d/b/1dbc8f4f274808eb43ba2f70ee16d88e5f6c701d.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 |