Difference between revisions of "2008 AIME II Problems/Problem 9"
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− | Therefore, <math>z</math> rotates along a circle with center <math>\omega = 5+(5+5\sqrt2)i</math>. Since <math>8 \cdot \frac{\pi}{4} = 2\pi</math>, <math>f^9(z) = f(z) \implies f^{150} = f( | + | Therefore, <math>z</math> rotates along a circle with center <math>\omega = 5+(5+5\sqrt2)i</math>. Since <math>8 \cdot \frac{\pi}{4} = 2\pi</math>, <math>f^9(z) = f(z) \implies f^{150}(z) = f^6(z) \implies p+q = \boxed{019}</math>, as desired (the final algebra bash isn't bad). |
=== Solution 4 === | === Solution 4 === |
Revision as of 22:44, 12 January 2023
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 .
Contents
[hide]Solutions
Solution 1
Let be the position of the particle on the -plane, be the length where is the origin, and be the inclination of OP to the x-axis. If is the position of the particle after a move from , then we have two equations for 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
If you're curious, the points do eventually form an octagon and repeat. Seems counterintuitive, but believe it or not, it happens.
https://www.desmos.com/calculator/febtiheosz
Solution 2
Let the particle's position be represented by a complex number. Recall that multiplying a number by cis rotates the object in the complex plane by counterclockwise. In this case, we use . Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to
where a is cis. By De-Moivre's theorem, =cis. Therefore,
Furthermore, . Thus, the final answer is
Solution 3
As before, consider as a complex number. Consider the transformation . This is a clockwise rotation of by radians about the points . Let denote one move of . Then
Therefore, rotates along a circle with center . Since , , as desired (the final algebra bash isn't bad).
Solution 4
Let . We assume that the rotation matrix here. Then we have
This simplifies to
Since , so we have , giving . The answer is yet .
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.