Difference between revisions of "2008 AIME II Problems/Problem 9"
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==== 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 00:54, 3 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
![$5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$](http://latex.artofproblemsolving.com/d/6/e/d6e93554c7f003c0766d1509e99aad3fbd7d9f19.png)
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
![$a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10$](http://latex.artofproblemsolving.com/b/d/b/bdba935af30d8eca5158fe6587d0c83bf1952d15.png)
where a is cis. By De-Moivre's theorem,
=cis
.
Therefore,
![$10(a^{150} + \ldots + 1) 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})$](http://latex.artofproblemsolving.com/0/1/6/0168dc9c7fdea544ec38c70b55848433fed9b437.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)
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
![$T^{150}\begin{pmatrix}5\\0\end{pmatrix}=R(R(...R(R\begin{pmatrix}5\\0\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}...)+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}$](http://latex.artofproblemsolving.com/c/7/7/c779a76d8286dbe05981c0790cb34629e9605e30.png)
This simplifies to
![$R^{150}\begin{pmatrix}5\\0\end{pmatrix}+(I+R^2+R^3+...+R^{149})\begin{pmatrix}10\\0\end{pmatrix}$](http://latex.artofproblemsolving.com/c/6/f/c6f6b8137274ebe4c9479bdef91f1dad1f7d7633.png)
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.