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Difference between revisions of "2008 AIME II Problems/Problem 9"

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=== Solution 1 ===
 
=== Solution 1 ===
  
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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'=rcos(\pi/4+\theta)+10 = \sqrt{2}(x - y)/2 + 10</math> and <math>y' = rsin(\pi/4+\theta) = \sqrt{2}(x + y)/2</math>.
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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>.
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https://www.desmos.com/calculator/febtiheosz
 
https://www.desmos.com/calculator/febtiheosz
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=== Solution 2 ===
 
=== Solution 2 ===
 
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
 
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

Revision as of 12:53, 5 November 2020

Problem

A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer less than or equal to $|p| + |q|$.

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 $\theta$ be the inclination of OP to the x-axis. If $(x', y')$ is the position of the particle after a move from $P$, then we have two equations for $x'$ and $y'$: \[x'=r\cos(\pi/4+\theta)+10 = \frac{\sqrt{2}(x - y)}{2} + 10\] \[y' = r\sin(\pi/4+\theta) = \frac{\sqrt{2}(x + y)}{2}\]. Let $(x_n, y_n)$ be the position of the particle after the nth move, where $x_0 = 5$ and $y_0 = 0$. Then $x_{n+1} + y_{n+1} = \sqrt{2}x_n+10$, $x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10$. This implies $x_{n+2} = -y_n + 5\sqrt{2}+ 10$, $y_{n+2}=x_n + 5\sqrt{2}$. Substituting $x_0 = 5$ and $y_0 = 0$, we have $x_8 = 5$ and $y_8 = 0$ again for the first time. Thus, $p = x_{150} = x_6 = -5\sqrt{2}$ and $q = y_{150} = y_6 = 5 + 5\sqrt{2}$. Hence, the final answer is

$5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$

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$\left( \theta \right)$ rotates the object in the complex plane by $\theta$ counterclockwise. In this case, we use $a = cis(\frac{\pi}{4})$. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to

$a_{150} = (((5a + 10)a + 10)a + 10 \ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \ldots + 10$

where a is cis$\left( \theta \right)$. By De-Moivre's theorem, $\left(cis( \theta \right)^n )$=cis$\left(n \theta \right)$. Therefore,

$10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})$

Furthermore, $5a^{150} = - 5i$. Thus, the final answer is

$5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}$

Solution 3

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).

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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