Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 10"

Revision as of 15:07, 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.

$P_r=\begin{pmatrix} 2000-k \\ k \end{pmatrix}$

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

~Tomas Diaz. orders@tomasdiaz.com

@@ Line 15: / Line 15: @@
 <math>P_r=\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_{n_x}-(2000-k) \\ P_{n_y}-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>
 ~Tomas Diaz. orders@tomasdiaz.com

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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 10"

Revision as of 15:07, 25 November 2023

Problem

Solution