Difference between revisions of "2008 AMC 12A Problems/Problem 25"
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We know then that from (1), the coefficients of <math>x</math> and <math>y</math> in <math>(a_{100},b_{100})</math> are both <math>2^{99}</math> | We know then that from (1), the coefficients of <math>x</math> and <math>y</math> in <math>(a_{100},b_{100})</math> are both <math>2^{99}</math> | ||
− | We can apply (2), finding <math>100 \text{(mod ) | + | We can apply (2), finding <math>100 \text{(mod }12)=4</math>, so the positions and signs of <math>x</math> and <math>y</math> are the same in <math>(a_{100},b_{100})</math> as they are in <math>(a_{4},b_{4})</math>. |
From this, we can get <math>(a_{100},b_{100})=(-2^{99}y,2^{99}x)</math>. We know that <math>(a_{100},b_{100})=(2,4)</math>, so we get the following: | From this, we can get <math>(a_{100},b_{100})=(-2^{99}y,2^{99}x)</math>. We know that <math>(a_{100},b_{100})=(2,4)</math>, so we get the following: |
Revision as of 22:13, 11 January 2018
Problem
A sequence ,
,
,
of points in the coordinate plane satisfies
for
.
Suppose that . What is
?
Solution 1
This sequence can also be expressed using matrix multiplication as follows:
.
Thus, is formed by rotating
counter-clockwise about the origin by
and dilating the point's position with respect to the origin by a factor of
.
So, starting with and performing the above operations
times in reverse yields
.
Rotating clockwise by
yields
. A dilation by a factor of
yields the point
.
Therefore, .
Shortcut: no answer has in the denominator. So the point cannot have orientation
or
. Also there are no negative answers. Any other non-multiple of
rotation of
would result in the need of radicals. So either it has orientation
or
. Both answers add up to
. Thus,
.
Solution 2 (algebra)
Let . Then, we can begin to list out terms as follows:
We notice that the sequence follows the rule
We can now start listing out every third point, getting:
We can make two observations from this:
(1) In , the coefficient of
and
is
(2) The positioning of and
, and their signs, cycle with every
terms.
We know then that from (1), the coefficients of and
in
are both
We can apply (2), finding , so the positions and signs of
and
are the same in
as they are in
.
From this, we can get . We know that
, so we get the following:
The answer is .
A.B.
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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