Mock AIME 6 2006-2007 Problems/Problem 10

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 $R(\theta)$ be the rotational matrix for a point to rotate $(\theta)$ around the origin:

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

For $\theta = 90^\circ$

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

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

Let's write $P_n$ in matrix form as: $P_n=\begin{bmatrix} P_{x_n} \\ P_{y_n} \end{bmatrix}$ , 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:

$P_{n+1}=(R)(P_n-P_r)+P_r=\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} P_{x_n}-(2000-n) \\ P_{y_n}-n \end{bmatrix}+\begin{bmatrix} 2000-n \\ n \end{bmatrix}=\begin{bmatrix} 2000-P_{y_n} \\ P_{x_n}+2n-2000 \end{bmatrix}$

Now we find $P_{n+2}$ :

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

For this problem, we're only interested in the $y$ -coordinate. So,

$P_{y_{n+1}}=P_{x_n}+2n-2000$

$P_{y_{n+2}}=-P_{y_n}+2(n+1)$

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

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

Therefore we need to calculate $P_{y_1}$ :

$P_{y_1}=P_{x_0}+(2)(0)-2000=2007-2000=7$

$P_{y_m} \equiv 1\; (mod\;2)\equiv P_{y_1}\; (mod\;2)$

So, we use the recursive formula for $P_{y_{n+2}}$ starting with $P_{y_1}$ :

$P_{y_3}=-P_{y_1}+(2)(1)+2$

$P_{y_5}=-P_{y_3}+(2)(3)+2$

$P_{y_7}=-P_{y_5}+(2)(5)+2$

and we can find $P_{y_{2q+1}}$ for $q \in \mathbb{Z}^{+}$ as:

$P_{y_{2q+1}}=-P_{y_{2q-1}}+2(2q-1)+2=-P_{y_{2q-1}}+4q$

$P_{y_{2q+1}}=(-1)^q P_{y_1}+4(q-(q-1)+(q-2)-(q-3)+....+0)$

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

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