Difference between revisions of "2008 AIME II Problems/Problem 9"
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=== Solution 1 === | === 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 <math>\theta</math> 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'= | + | Let <math>P(x, y)</math> be the position of the particle on the <math>xy</math>-plane, <math>r</math> be the length <math>OP</math> where <math>O</math> is the origin, and <math>\theta</math> be the inclination of OP to the x-axis. If <math>(x', y')</math> is the position of the particle after a move from <math>P</math>, then we have two equations for <math>x'</math> and <math>y'</math>: |
+ | <cmath>x'=r\cos(\pi/4+\theta)+10 = \frac{\sqrt{2}(x - y)}{2} + 10</cmath> | ||
+ | <cmath>y' = r\sin(\pi/4+\theta) = \frac{\sqrt{2}(x + y)}{2}</cmath>. | ||
Let <math>(x_n, y_n)</math> be the position of the particle after the nth move, where <math>x_0 = 5</math> and <math>y_0 = 0</math>. Then <math>x_{n+1} + y_{n+1} = \sqrt{2}x_n+10</math>, <math>x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10</math>. This implies | Let <math>(x_n, y_n)</math> be the position of the particle after the nth move, where <math>x_0 = 5</math> and <math>y_0 = 0</math>. Then <math>x_{n+1} + y_{n+1} = \sqrt{2}x_n+10</math>, <math>x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10</math>. This implies | ||
<math>x_{n+2} = -y_n + 5\sqrt{2}+ 10</math>, <math>y_{n+2}=x_n + 5\sqrt{2}</math>. | <math>x_{n+2} = -y_n + 5\sqrt{2}+ 10</math>, <math>y_{n+2}=x_n + 5\sqrt{2}</math>. | ||
Substituting <math>x_0 = 5</math> and <math>y_0 = 0</math>, we have <math>x_8 = 5</math> and <math>y_8 = 0</math> again for the first time. Thus, <math>p = x_{150} = x_6 = -5\sqrt{2}</math> and <math>q = y_{150} = y_6 = 5 + 5\sqrt{2}</math>. Hence, the final answer is | Substituting <math>x_0 = 5</math> and <math>y_0 = 0</math>, we have <math>x_8 = 5</math> and <math>y_8 = 0</math> again for the first time. Thus, <math>p = x_{150} = x_6 = -5\sqrt{2}</math> and <math>q = y_{150} = y_6 = 5 + 5\sqrt{2}</math>. Hence, the final answer is | ||
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | <center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | ||
+ | |||
+ | 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 === | === Solution 2 === | ||
− | Let the particle's position be represented by a complex number. Recall that multiplying a number by | + | Let the particle's position be represented by a complex number. Recall that multiplying a number by cis<math>\left( \theta \right)</math> rotates the object in the complex plane by <math>\theta</math> counterclockwise. In this case, we use <math>a = cis(\frac{\pi}{4})</math>. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to |
<center><math>a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10</math></center> | <center><math>a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10</math></center> | ||
where a is cis<math>\left( \theta \right)</math>. By De-Moivre's theorem, <math>\left(cis( \theta \right)^n )</math>=cis<math>\left(n \theta \right)</math>. | where a is cis<math>\left( \theta \right)</math>. By De-Moivre's theorem, <math>\left(cis( \theta \right)^n )</math>=cis<math>\left(n \theta \right)</math>. | ||
Therefore, | Therefore, | ||
− | <center><math>10(a^{150} + \ldots + 1) | + | <center><math>10(a^{150} + \ldots + 1) 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})</math></center> |
Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is | Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is | ||
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | <center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | ||
+ | |||
+ | ==== 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 <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
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 |
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