Difference between revisions of "2008 AIME II Problems/Problem 9"
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Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is | Furthermore, <math>5a^{150} = - 5i</math>. Thus, the final answer is | ||
<center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | <center><math>5\sqrt {2} + 5(\sqrt {2} + 1) \approx 19.1 \Longrightarrow \boxed{019}</math></center> | ||
− | === | + | === 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). | 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). | ||
Revision as of 13:06, 10 December 2018
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 P(x, y) be the position of the particle on the xy-plane, r be the length OP where O is the origin, and be the inclination of OP to the x-axis. If (x', y') is the position of the particle after a move from P, then
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.