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

Revision as of 17:45, 28 October 2017

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 $x'=rcos(\pi/4+\theta)+10 = \sqrt{2}(x - y)/2 + 10$ and $y' = rsin(\pi/4+\theta) = \sqrt{2}(x + y)/2$ . 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, [url="https://www.desmos.com/calculator/febtiheosz"]the points[/url] do eventually form an octagon and repeat. Seems counterintuitive, but believe it or not, it happens.

Solution 2

Let the particle's position be represented by a complex number. Recall that multiplying a number by where a is cis $\left( \theta \right)$ . rotates the object in the complex plane by $\theta$ counterclockwise. 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

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

@@ Line 11: / Line 11: @@
 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>
+If you're curious, [url="https://www.desmos.com/calculator/febtiheosz"]the points[/url] do eventually form an octagon and repeat. Seems counterintuitive, but believe it or not, it happens.
 === Solution 2 ===

2008 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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