2008 AIME II Problems/Problem 9
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
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
Notice that
Furthermore, . Thus, the final answer is
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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All AIME Problems and Solutions |