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

Revision as of 16:05, 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:

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

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

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:

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

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 $P_{y_1}$ :

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

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

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 $P_{y_{2k+1}}$ for $k \in \mathbb{Z}^{+}$ as:

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

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

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

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

$P_{y_{2k+1}}=(-1)^kP_{y_1}+2k+(1-(-1)^k)=(-1)^k(P_{y_1}-1)+2k+1$

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

$(-1)^k(7-1)+2k+1=433$

$k=\frac{433-(-1)^k6}{2}=216-(-1)^k3$

Since $\pm 1 \equiv 1\; (mod\;2)$ then $\pm 3 \equiv 1\; (mod\;2)$ and $(-1)^k3 \equiv 1\; (mod\;2)$

Therefore $(216-(-1)^k3) \equiv 1\; (mod\;2)$ and $k$ is odd.

So, $k=216-(-1)(3)=219$ and $m=2k+1=(2)(219)+1=\boxed{439}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 65: / Line 65: @@
 <math>k=\frac{433-(-1)^k6}{2}=216-(-1)^k3</math>
+Since <math>\pm 1 \equiv 1\; (mod\;2)</math> then <math>\pm 3 \equiv 1\; (mod\;2)</math> and <math>(-1)^k3 \equiv 1\; (mod\;2)</math>
+Therefore <math>(216-(-1)^k3) \equiv 1\; (mod\;2)</math> and <math>k</math> is odd.
+So, <math>k=216-(-1)(3)=219</math> and <math>m=2k+1=(2)(219)+1=\boxed{439}</math>
 ~Tomas Diaz. orders@tomasdiaz.com
+{{alternate solutions}}

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

Revision as of 16:05, 25 November 2023

Problem

Solution