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

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(→‎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 "a" 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+a)+10 = \sqrt{2}(x - y)/2 + 10</math> and <math>y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2</math>.
 
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 "a" 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+a)+10 = \sqrt{2}(x - y)/2 + 10</math> and <math>y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2</math>.
−
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. <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. <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

Revision as of 02:33, 7 May 2009

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 "a" be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then $x'=rcos(\pi/4+a)+10 = \sqrt{2}(x - y)/2 + 10$ and $y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2$. 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. $p = x_{150} = x_6 = -5\sqrt{2}$ and $q = y_{150} = y_6 = 5 + 5\sqrt{2}$. Hence the final answer is

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

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

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

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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