Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(New Solution)
(→‎Solution 2)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 +
__TOC__
 +
 
== Problem ==
 
== 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>.
 
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 ==  
 
== Solutions ==  
 
=== Solution 1 ===
 
=== Solution 1 ===
Line 8: Line 9:
 
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>:
 
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>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
 
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>.
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>.
 
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,
 
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
 
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>
 
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center>
Line 31: Line 32:
  
 
[[File:2008AIMEII9Sol3.png|center|300px]]
 
[[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} = f(6) \implies p+q = \boxed{019}</math>, as desired (the final algebra bash isn't bad).
+
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 ===
 
=== Solution 4 ===

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

2008AIMEII9Sol3.png

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 $\lfloor10\sqrt{2}+5\rfloor=\boxed{019}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_9&oldid=213674"