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

Revision as of 00:38, 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 (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then $x(n+1) + y(n+1) = \sqrt{2}(xn)+10$ , $x(n+1) - y(n+1) = -\sqrt{2}(yn)+10$ . This implies $x(n+2) = -yn + 5\sqrt{2}+ 10$ , $y(n+2)=xn + 5\sqrt{2}$ . Substituting x0 = 5 and y0 = 0, we have x8 = 5 and y8 = 0 again for the first time. $p = x150 = x6 = -5\sqrt{2}$ and $q = y150 = y6 = 5 + 5\sqrt{2}$ . 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

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

@@ Line 6: / Line 6: @@
 === 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 <math>x'=rcos(\pi/4+a)+10 = \sqrt{2}(x - y)/2 + 10</math> and y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2.
+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 (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then x(n+1) + y(n+1) =  \sqrt{2}(xn)+10, x(n+1) - y(n+1) = -\sqrt{2}(yn)+10. This implies
+Let (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then <math>x(n+1) + y(n+1) =  \sqrt{2}(xn)+10</math>, <math>x(n+1) - y(n+1) = -\sqrt{2}(yn)+10</math>. This implies
-x(n+2) = -yn + 5\sqrt{2}+ 10, y(n+2)=xn + 5\sqrt{2}.
+<math>x(n+2) = -yn + 5\sqrt{2}+ 10</math>, <math>y(n+2)=xn + 5\sqrt{2}</math>.
-Substituting x0 = 5 and y0 = 0, we have x8 = 5 and y8 = 0 again for the first time. p = x150 = x6 = -5\sqrt{2} and q = y150 = y6 = 5 + 5\sqrt{2}. Hence the final answer is
+Substituting x0 = 5 and y0 = 0, we have x8 = 5 and y8 = 0 again for the first time. <math>p = x150 = x6 = -5\sqrt{2}</math> and <math>q = y150 = y6 = 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>
 === Solution 2 ===

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

Page

Toolbox

Search