Difference between revisions of "2008 AIME II Problems/Problem 9"
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<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 ==== |
− | + | As before, consider <math>z</math> as a complex number. Consider the transformation <math>z \to (z-\omega)e^{i\theta} + \omega</math>. This is a clockwise rotation of <math>z</math> by <math>\theta</math> radians about the points <math>\omega</math>. Let <math>f(z)</math> denote one move of <math>z</math>. Then | |
− | === Solution | + | [[File:2008AIMEII9Sol3.png|center|300px]] |
+ | Therefore, <math>z</math> rotates along a circle with center <math>\omega = 5+(5+5\sqrt2)i</math>. Since <math>8 \cdot \frac{\pi}{4} = 2\pi</math>, <math>f^9(z) = f(z) \implies f^{150} = f(6) \implies p+q = \boxed{019}</math>, as desired (the final algebra bash isn't bad). | ||
+ | |||
+ | === Solution 4 === | ||
Let <math>T: | Let <math>T: | ||
<center><math>T^{150} | <center><math>T^{150} |
Revision as of 22:43, 12 January 2023
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
.
Contents
[hide]Solutions
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
As before, consider as a complex number. Consider the transformation
. This is a clockwise rotation of
by
radians about the points
. Let
denote one move of
. Then
Therefore, rotates along a circle with center
. Since
,
, as desired (the final algebra bash isn't bad).
Solution 4
Let . We assume that the rotation matrix
here. Then we have

This simplifies to

Since , so we have
, giving
. The answer is yet
.
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.