Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 10"
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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>. | 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> | + | 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> | ||
+ | |||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com |
Revision as of 14:32, 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 along 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
~Tomas Diaz. orders@tomasdiaz.com