Difference between revisions of "2008 AIME II Problems/Problem 9"
Sophieli666 (talk | contribs) (→Solution 2) |
(→Solution) |
||
Line 29: | Line 29: | ||
==== Unclear/Unfinished Solution ==== | ==== Unclear/Unfinished Solution ==== | ||
Each move has two parts, a rotation and a translation. But consider what happens to a translation after four rotations: it is cancelled out by another translation in the opposite direction. Thus the particle repeats position every 8 moves. So we only have to move backwards two steps from move 152 = 8(19). | Each move has two parts, a rotation and a translation. But consider what happens to a translation after four rotations: it is cancelled out by another translation in the opposite direction. Thus the particle repeats position every 8 moves. So we only have to move backwards two steps from move 152 = 8(19). | ||
+ | |||
+ | === Solution 3 === | ||
+ | Let <math>T:\begin{pmatrix}x\\y\end{pmatrix}\rightarrow R(\frac{\pi}{4})\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix}</math>. We assume that the rotation matrix <math>R(\frac{\pi}{4}) = R</math> here. Then we have | ||
+ | <center><math>T^{150}\begin{pmatrix}5\\0\end{pmatrix}=R(R(...R(R\begin{pmatrix}5\\0\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}...)+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}</math></center> | ||
+ | This simplifies to | ||
+ | <center><math>R^{150}\begin{pmatrix}5\\0\end{pmatrix}+(I+R^2+R^3+...+R^{149})\begin{pmatrix}10\\0\end{pmatrix}</math></center> | ||
+ | Since <math>R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O</math>, so we have <math>R^6\begin{pmatrix}5\\0\end{pmatrix}+(-R^6-R^7)\begin{pmatrix}10\\0\end{pmatrix}</math>, giving <math>p=-5\sqrt{2}, q=5\sqrt{2}+5</math>. The answer is yet <math>\lfloor10\sqrt{2}+5\rfloor=\boxed{019}</math>. | ||
== See also == | == See also == |
Revision as of 23:54, 2 June 2022
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
Solution
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
Unclear/Unfinished Solution
Each move has two parts, a rotation and a translation. But consider what happens to a translation after four rotations: it is cancelled out by another translation in the opposite direction. Thus the particle repeats position every 8 moves. So we only have to move backwards two steps from move 152 = 8(19).
Solution 3
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.