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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 10"
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<math>R=\begin{pmatrix} cos(90^\circ) & -sin(90^\circ)\\ sin(90^\circ) & cos(90^\circ) \end{pmatrix}=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}</math>
 
<math>R=\begin{pmatrix} cos(90^\circ) & -sin(90^\circ)\\ sin(90^\circ) & cos(90^\circ) \end{pmatrix}=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}</math>
  
−
Let <math>P_r</math> be the point of rotation.
+
Let <math>P_r</math> be the point of rotation, then <math>P_r=\begin{pmatrix} 2000-k \\ k \end{pmatrix}</math>
  
−
<math>P_r=\begin{pmatrix} 2000-k \\ k \end{pmatrix}</math>
+
Let's write <math>P_n</math> in matrix form as: <math>P_n=\begin{pmatrix} P_{x_n} \\ P_{y_n} \end{pmatrix}</math>, where <math>P_{x_n}</math> and <math>P_{y_n}</math> are the <math>x</math> and <math>y</math> coordinates of <math>P_n</math> respectively.
  
−
Let's write <math>P_n</math> in matrix form as:
+
We can write the equation of <math>P_{n+1}</math> by translating the <math>P_n</math> to the origin, multiply it by the rotation matrix <math>R</math> and then add the point subtracted:
−
 
 
−
<math>P_n=\begin{pmatrix} P_{x_n} \\ P_{y_n} \end{pmatrix}</math>
 
  
 
<math>P_{n+1}=R(P_n-P_r)+P_r=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} P_{x_n}-(2000-k) \\ P_{y_n}-k \end{pmatrix}+\begin{pmatrix} 2000-k \\ k \end{pmatrix}</math>
 
<math>P_{n+1}=R(P_n-P_r)+P_r=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} P_{x_n}-(2000-k) \\ P_{y_n}-k \end{pmatrix}+\begin{pmatrix} 2000-k \\ k \end{pmatrix}</math>

Revision as of 14:10, 25 November 2023

Problem

Given a point $P$ in the coordinate plane, let $T_k(P)$ be the $90^\circ$ rotation of $P$ around the point $(2000-k,k)$. Let $P_0$ be the point $(2007,0)$ and $P_{n+1}=T_n(P_n)$ for all integers $n\ge 0$. If $P_m$ has a $y$-coordinate of $433$, what is $m$?

Solution

Let $R$ be the rotational matrix for a point along the origin:

$R=\begin{pmatrix} cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta) \end{pmatrix}$

For $\theta = 90^\circ$

$R=\begin{pmatrix} cos(90^\circ) & -sin(90^\circ)\\ sin(90^\circ) & cos(90^\circ) \end{pmatrix}=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$

Let $P_r$ be the point of rotation, then $P_r=\begin{pmatrix} 2000-k \\ k \end{pmatrix}$

Let's write $P_n$ in matrix form as: $P_n=\begin{pmatrix} P_{x_n} \\ P_{y_n} \end{pmatrix}$, where $P_{x_n}$ and $P_{y_n}$ are the $x$ and $y$ coordinates of $P_n$ respectively.

We can write the equation of $P_{n+1}$ by translating the $P_n$ to the origin, multiply it by the rotation matrix $R$ and then add the point subtracted:

$P_{n+1}=R(P_n-P_r)+P_r=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}\begin{pmatrix} P_{x_n}-(2000-k) \\ P_{y_n}-k \end{pmatrix}+\begin{pmatrix} 2000-k \\ k \end{pmatrix}$


~Tomas Diaz. orders@tomasdiaz.com

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