Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | Let <math>R</math> be the rotational matrix for a point | + | Let <math>R(\theta)</math> be the rotational matrix for a point to rotate <math>(\theta)</math> around the origin: |
− | <math>R=\begin{ | + | <math>R(\theta)=\begin{bmatrix} cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta) \end{bmatrix}</math> |
For <math>\theta = 90^\circ</math> | For <math>\theta = 90^\circ</math> | ||
− | <math>R=\begin{ | + | <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 | + | Let <math>P_r</math> be the point of rotation, then <math>P_r=\begin{bmatrix} 2000-n \\ n \end{bmatrix}</math> |
− | <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. |
− | <math>P_{n+1} | + | 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-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}</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=2007-2000=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_{2q+1}}</math> for <math>q \in \mathbb{Z}^{+}</math> as: | ||
+ | |||
+ | <math>P_{y_{2q+1}}=-P_{y_{2q-1}}+2(2q-1)+2=-P_{y_{2q-1}}+4q</math> | ||
+ | |||
+ | <math>P_{y_{2q+1}}=(-1)^q P_{y_1}+4(q-(q-1)+(q-2)-(q-3)+....+0)</math> | ||
+ | |||
+ | <math>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) </math> | ||
+ | |||
+ | <math>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}</math> | ||
+ | |||
+ | <math>P_{y_{2q+1}}=(-1)^qP_{y_1}+2q+(1-(-1)^q)=(-1)^q(P_{y_1}-1)+2q+1</math> | ||
+ | |||
+ | Since <math>P_{y_{2q+1}}=433</math>, and <math>P_{y_1}=7</math> then we solve for <math>q</math> as follows: | ||
+ | |||
+ | <math>(-1)^q(7-1)+2q+1=433</math> | ||
+ | |||
+ | <math>q=\frac{433-1-(-1)^q6}{2}=216-(-1)^q3</math> | ||
+ | |||
+ | Since <math>\pm 1 \equiv 1\; (mod\;2)</math> then <math>\pm 3 \equiv 1\; (mod\;2)</math> and <math>(-1)^q3 \equiv 1\; (mod\;2)</math> | ||
+ | |||
+ | Therefore <math>(216-(-1)^q3) \equiv 1\; (mod\;2)</math> and <math>q</math> is odd. | ||
+ | |||
+ | So, <math>q=216-(-1)(3)=219</math> and <math>m=2q+1=(2)(219)+1=\boxed{439}</math> | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
+ | |||
+ | {{alternate solutions}} |
Latest revision as of 15:20, 25 November 2023
Problem
Given a point in the coordinate plane, let be the rotation of around the point . Let be the point and for all integers . If has a -coordinate of , what is ?
Solution
Let be the rotational matrix for a point to rotate around the origin:
For
Let be the point of rotation, then
Let's write in matrix form as: , where and are the and coordinates of respectively.
We can write the equation of by translating the to the origin, multiply it by the rotation matrix and then add the point subtracted:
Now we find :
For this problem, we're only interested in the -coordinate. So,
Notice that since , then .
Since and we're looking at a -coordinate of , we notice that .
Therefore we need to calculate :
So, we use the recursive formula for starting with :
and we can find for as:
Since , and then we solve for as follows:
Since then and
Therefore and is odd.
So, and
~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.