View source for Mock AIME 6 2006-2007 Problems/Problem 10

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

==Solution==
Let <math>R</math> be the rotational matrix for a point along the origin:

<math>R=\begin{bmatrix} cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta) \end{bmatrix}</math>

For <math>\theta = 90^\circ</math>

<math>R=\begin{bmatrix} cos(90^\circ) & -sin(90^\circ)\\ sin(90^\circ) & cos(90^\circ) \end{bmatrix}=\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}</math>

Let <math>P_r</math> be the point of rotation, then <math>P_r=\begin{bmatrix} 2000-k \\ k \end{bmatrix}</math>

Let's write <math>P_n</math> in matrix form as: <math>P_n=\begin{bmatrix} P_{x_n} \\ P_{y_n} \end{bmatrix}</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.

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+1}=R(P_n-P_r)+P_r=\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} P_{x_n}-(2000-k) \\ P_{y_n}-k \end{bmatrix}+\begin{bmatrix} 2000-k \\ k \end{bmatrix}=\begin{bmatrix} 2000-P_{y_n} \\ P_{x_n}+2n-2000 \end{bmatrix}</math>

Now we find <math>P_{n+2}</math>:

<math>P_{n+2}=\begin{bmatrix} 2000-P_{y_{n+1}} \\ P_{x_{n+1}}+2n-2000 \end{bmatrix}=\begin{bmatrix} 2000-(P_{x_n}+2n-2000) \\ (2000-P_{y_n})+2(n+1)-2000 \end{bmatrix}=\begin{bmatrix} 4000-P_{x_n}-2n \\ -P_{y_n}+2(n+1) \end{bmatrix}</math>

For this problem, we're only interested in the <math>y</math>-coordinate.  So,

<math>P_{y_{n+1}}=P_{x_n}+2n-2000</math>

<math>P_{y_{n+2}}=-P_{y_n}+2(n+1)</math>

Notice that since <math>2(n+1) \equiv 0\; (mod\;2)</math>, then <math>P_{y_{n+2}} \equiv P_{y_n}\; (mod\;2)</math>.

Since <math>P_{y_0}=0</math> and we're looking at a <math>y</math>-coordinate of <math>433</math>, we notice that <math>433 \equiv 1\; (mod\;2) \not\equiv P_{y_0}\; (mod\;2)</math>.  

Therefore we need to calculate <math>P_{y_1}</math>:

<math>P_{y_1}=P_{x_0}+(2)(0)-2000=2007+2-2000=7</math>

<math>P_{y_m} \equiv 1\; (mod\;2)\equiv P_{y_1}\; (mod\;2)</math> 

So, we use the recursive formula for <math>P_{y_{n+2}}</math> starting with <math>P_{y_1}</math>:

<math>P_{y_3}=-P_{y_1}+(2)(1)+2</math>

<math>P_{y_5}=-P_{y_3}+(2)(3)+2</math>

<math>P_{y_7}=-P_{y_5}+(2)(5)+2</math>

and we can find <math>P_{y_{2k+1}}</math> for <math>k \in \mathbb{Z}^{+}</math> as:

<math>P_{y_{2k+1}}=-P_{y_{2k-1}}+2(2k-1)+2=-P_{y_{2k-1}}+4k</math>

<math>P_{y_{2k+1}}=(-1)^k P_{y_1}+4(k-(k-1)+(k-2)-(k-3)+....+0)</math>

<math>P_{y_{2k+1}}=(-1)^k P_{y_1}+4 \left( \begin{cases} k, & k\;is\;odd \\ 0, & k\;is\;even\end{cases}+\begin{cases} -\frac{k-1}{2}, & k\;is\;odd \\ \frac{k}{2}, & k\;is\;even\end{cases} \right) </math>

<math>P_{y_{2k+1}}=(-1)^k P_{y_1}+ \begin{cases} 2k+2, & k\;is\;odd \\ 2k, & k\;is\;even\end{cases}=(-1)^k P_{y_1}+ 2k+\begin{cases} 2, & k\;is\;odd \\ 0, & k\;is\;even\end{cases}</math>

<math>P_{y_{2k+1}}=(-1)^kP_{y_1}+2k+(1-(-1)^k)=(-1)^k(P_{y_1}-1)+2k+1</math>

Since <math>P_{y_{2k+1}}=433</math>, and <math>P_{y_1}=7</math> then we solve for <math>k</math> as follows:

<math>(-1)^k(7-1)+2k+1=433</math>

<math>k=\frac{433-(-1)^k}{2}=216-(-1)^k</math>



~Tomas Diaz. orders@tomasdiaz.com