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

Latest revision as of 21:37, 28 January 2024

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 $|p| + |q|$ .

Solutions

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

As before, consider $z$ as a complex number. Consider the transformation $z \to (z-\omega)e^{i\theta} + \omega$ . This is a clockwise rotation of $z$ by $\theta$ radians about the points $\omega$ . Let $f(z)$ denote one move of $z$ . Then

Therefore, $z$ rotates along a circle with center $\omega = 5+(5+5\sqrt2)i$ . Since $8 \cdot \frac{\pi}{4} = 2\pi$ , $f^9(z) = f(z) \implies f^{150}(z) = f^6(z) \implies p+q = \boxed{019}$ , as desired (the final algebra bash isn't bad).

Solution 4

Let $T:\begin{pmatrix}x\\y\end{pmatrix}\rightarrow R(\frac{\pi}{4})\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix}$ . We assume that the rotation matrix $R(\frac{\pi}{4}) = R$ here. Then we have

$T^{150}\begin{pmatrix}5\\0\end{pmatrix}=R(R(...R(R\begin{pmatrix}5\\0\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}...)+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}$

This simplifies to

$R^{150}\begin{pmatrix}5\\0\end{pmatrix}+(I+R^2+R^3+...+R^{149})\begin{pmatrix}10\\0\end{pmatrix}$

Since $R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O$ , so we have $R^6\begin{pmatrix}5\\0\end{pmatrix}+(-R^6-R^7)\begin{pmatrix}10\\0\end{pmatrix}$ , giving $p=-5\sqrt{2}, q=5\sqrt{2}+5$ . The answer is yet $\lfloor10\sqrt{2}+5\rfloor=\boxed{019}$ .

@@ Line 1: / Line 1: @@
+__TOC__
 == Problem ==
 A particle is located on the coordinate plane at <math>(5,0)</math>. Define a ''move'' for the particle as a counterclockwise rotation of <math>\pi/4</math> radians about the origin followed by a translation of <math>10</math> units in the positive <math>x</math>-direction. Given that the particle's position after <math>150</math> moves is <math>(p,q)</math>, find the greatest integer less than or equal to <math>|p| + |q|</math>.
-__TOC__
+== Solutions ==
-== Solution ==
 === Solution 1 ===
 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>.
+<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
 <math>x_{n+2} = -y_n + 5\sqrt{2}+ 10</math>, <math>y_{n+2}=x_n + 5\sqrt{2}</math>.
@@ Line 23: / Line 24: @@
 where a is cis<math>\left( \theta \right)</math>. By De-Moivre's theorem, <math>\left(cis( \theta \right)^n )</math>=cis<math>\left(n \theta \right)</math>.
 Therefore,
-<center><math>10(a^{150} + \ldots + 1) = 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})</math></center>
+<center><math>10(a^{150} + \ldots + 1)= 10(1 + a + \ldots + a^6) = - 10(a^7) = - 10(\frac{ \sqrt {2} }{2} - \frac{i\sqrt {2}} {2})</math></center>
 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>
-==== Unclear/Unfinished Solution ====
+==== 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).
+As before, consider <math>z</math> as a complex number.  Consider the transformation <math>z \to (z-\omega)e^{i\theta} + \omega</math>.  This is a clockwise rotation of <math>z</math> by <math>\theta</math> radians about the points <math>\omega</math>.  Let <math>f(z)</math> denote one move of <math>z</math>.  Then
+[[File:2008AIMEII9Sol3.png|center|300px]]
+Therefore, <math>z</math> rotates along a circle with center <math>\omega = 5+(5+5\sqrt2)i</math>.  Since <math>8 \cdot \frac{\pi}{4} = 2\pi</math>, <math>f^9(z) = f(z) \implies f^{150}(z) = f^6(z) \implies p+q = \boxed{019}</math>, as desired (the final algebra bash isn't bad).
+=== Solution 4 ===
+Let <math>T:\begin{pmatrix}x\\y\end{pmatrix}\rightarrow R(\frac{\pi}{4})\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix}</math>. We assume that the rotation matrix <math>R(\frac{\pi}{4}) = R</math> here. Then we have
+<center><math>T^{150}\begin{pmatrix}5\\0\end{pmatrix}=R(R(...R(R\begin{pmatrix}5\\0\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}...)+\begin{pmatrix}10\\0\end{pmatrix})+\begin{pmatrix}10\\0\end{pmatrix}</math></center>
+This simplifies to
+<center><math>R^{150}\begin{pmatrix}5\\0\end{pmatrix}+(I+R^2+R^3+...+R^{149})\begin{pmatrix}10\\0\end{pmatrix}</math></center>
+Since <math>R+R^{7}=O, R^2+R^6=O, R^3+R^5=O, I+R^4=O</math>, so we have <math>R^6\begin{pmatrix}5\\0\end{pmatrix}+(-R^6-R^7)\begin{pmatrix}10\\0\end{pmatrix}</math>, giving <math>p=-5\sqrt{2}, q=5\sqrt{2}+5</math>. The answer is yet <math>\lfloor10\sqrt{2}+5\rfloor=\boxed{019}</math>.
 == 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

Page

Toolbox

Search