Difference between revisions of "2008 AIME II Problems/Problem 9"
Cosmoalpern (talk | contribs) (→solution 3) |
(→Solution 1) |
||
Line 6: | Line 6: | ||
=== Solution 1 === | === Solution 1 === | ||
− | Let P(x, y) be the position of the particle on the xy-plane, r be the length OP where O is the origin, and <math>\theta</math> be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then <math>x'= | + | Let <math>P(x, y)</math> be the position of the particle on the <math>xy</math>-plane, <math>r</math> be the length <math>OP</math> where <math>O</math> is the origin, and <math>\theta</math> be the inclination of OP to the x-axis. If <math>(x', y')</math> is the position of the particle after a move from <math>P</math>, then we have two equations for <math>x'</math> and <math>y'</math>: |
+ | <cmath>x'=r\cos(\pi/4+\theta)+10 = \frac{\sqrt{2}(x - y)}{2} + 10</cmath> | ||
+ | <cmath>y' = r\sin(\pi/4+\theta) = \frac{\sqrt{2}(x + y)}{2}</cmath>. | ||
Let <math>(x_n, y_n)</math> be the position of the particle after the nth move, where <math>x_0 = 5</math> and <math>y_0 = 0</math>. Then <math>x_{n+1} + y_{n+1} = \sqrt{2}x_n+10</math>, <math>x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10</math>. This implies | Let <math>(x_n, y_n)</math> be the position of the particle after the nth move, where <math>x_0 = 5</math> and <math>y_0 = 0</math>. Then <math>x_{n+1} + y_{n+1} = \sqrt{2}x_n+10</math>, <math>x_{n+1} - y_{n+1} = -\sqrt{2}y_n+10</math>. This implies | ||
<math>x_{n+2} = -y_n + 5\sqrt{2}+ 10</math>, <math>y_{n+2}=x_n + 5\sqrt{2}</math>. | <math>x_{n+2} = -y_n + 5\sqrt{2}+ 10</math>, <math>y_{n+2}=x_n + 5\sqrt{2}</math>. | ||
Line 15: | Line 17: | ||
https://www.desmos.com/calculator/febtiheosz | https://www.desmos.com/calculator/febtiheosz | ||
+ | |||
=== Solution 2 === | === Solution 2 === | ||
Let the particle's position be represented by a complex number. Recall that multiplying a number by cis<math>\left( \theta \right)</math> rotates the object in the complex plane by <math>\theta</math> counterclockwise. In this case, we use <math>a = cis(\frac{\pi}{4})</math>. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to | Let the particle's position be represented by a complex number. Recall that multiplying a number by cis<math>\left( \theta \right)</math> rotates the object in the complex plane by <math>\theta</math> counterclockwise. In this case, we use <math>a = cis(\frac{\pi}{4})</math>. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to |
Revision as of 12:53, 5 November 2020
Problem
A particle is located on the coordinate plane at . Define a move for the particle as a counterclockwise rotation of radians about the origin followed by a translation of units in the positive -direction. Given that the particle's position after moves is , find the greatest integer less than or equal to .
Solution
Solution 1
Let be the position of the particle on the -plane, be the length where is the origin, and be the inclination of OP to the x-axis. If is the position of the particle after a move from , then we have two equations for and : . Let be the position of the particle after the nth move, where and . Then , . This implies , . Substituting and , we have and again for the first time. Thus, and . Hence, the final answer is
If you're curious, the points do eventually form an octagon and repeat. Seems counterintuitive, but believe it or not, it happens.
https://www.desmos.com/calculator/febtiheosz
Solution 2
Let the particle's position be represented by a complex number. Recall that multiplying a number by cis rotates the object in the complex plane by counterclockwise. In this case, we use . Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to
where a is cis. By De-Moivre's theorem, =cis. Therefore,
Furthermore, . Thus, the final answer is
Solution 3
Each move has two parts, a rotation and a translation. But consider what happens to a translation after four rotations: it is cancelled out by another translation in the opposite direction. Thus the particle repeats position every 8 moves. So we only have to move backwards two steps from move 152 = 8(19).
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.