2008 AIME II Problems/Problem 9: Difference between revisions

Revision as of 13:41, 19 April 2008

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

Show periodic with $8$ steps, then invert twice. Template:Incomplete

Solution 2

Let the particle's position be represented by a complex number. The transformation takes $z$ to $f(z) = az + b$ where $a = e^{i\pi/4} = \frac {\sqrt {2}}{2} + i\frac {\sqrt {2}}{2}$ and $b = 10$ . We let $a_0 = 5$ and $a_{n + 1} = f(a_n)$ so that we want to find $a_{150}$ .

Basically, the thing comes out to

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

Notice that

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

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

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

@@ Line 14: / Line 14: @@
 Notice that
 <center><math>10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10( - \sqrt {2}/2 - i\sqrt {2}/2)</math></center>
-Furthermore, <math>5a^{150} = - 5i</math>. So our 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>

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