Difference between revisions of "2015 AIME II Problems/Problem 13"
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We are trying to make <math>n</math> so that the imaginary part of this expression is negative. | We are trying to make <math>n</math> so that the imaginary part of this expression is negative. | ||
− | The argument of <math>z</math> is <math>1</math>. The argument of <math>z-1</math>, however, is a little more tricky. <math>z-1</math> is on a circle centered on <math>(-1,0)</math> with radius <math>1</math>. The change in angle due to <math>z</math> is <math>1</math> with respect to the center, but the angle that <math>z-1</math> makes with the <math>y</math>-axis is <math>half</math> the change, due to | + | The argument of <math>z</math> is <math>1</math>. The argument of <math>z-1</math>, however, is a little more tricky. <math>z-1</math> is on a circle centered on <math>(-1,0)</math> with radius <math>1</math>. The change in angle due to <math>z</math> is <math>1</math> with respect to the center, but the angle that <math>z-1</math> makes with the <math>y</math>-axis is <math>half</math> the change, due to Circle Theorems (this intercepted arc is the argument of <math>z</math>), because the <math>y</math>- axis is tangent to the circle at the origin. So <math>\text{arg}(z-1)=\frac{\pi+1}{2}</math>. Dividing <math>z</math> by <math>z-1</math> subtracts the latter argument from the former, so the angle of the quotient with the <math>x</math>-axis is <math>\frac{1-\pi}{2}</math>. |
We want the argument of the whole expression <math>-\pi<\theta<0</math>. This translates into <math>\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}</math>. <math>z^n-1</math> also consists of points on the circle centered at <math>(-1,0)</math>, so we deal with this argument similarly: the argument of <math>z^n</math> is twice the angle <math>z^n-1</math> makes with the <math>y</math>-axis. Since <math>z^n-1</math> is always negative, <math>\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}</math>, and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a <math>\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}</math> angle with the <math>y</math>-axis both ways. | We want the argument of the whole expression <math>-\pi<\theta<0</math>. This translates into <math>\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}</math>. <math>z^n-1</math> also consists of points on the circle centered at <math>(-1,0)</math>, so we deal with this argument similarly: the argument of <math>z^n</math> is twice the angle <math>z^n-1</math> makes with the <math>y</math>-axis. Since <math>z^n-1</math> is always negative, <math>\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}</math>, and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a <math>\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}</math> angle with the <math>y</math>-axis both ways. |
Revision as of 20:36, 29 February 2016
Contents
[hide]Problem
Define the sequence by
, where
represents radian measure. Find the index of the 100th term for which
.
Solution 1
If ,
. Then if
satisfies
,
, and
Since
is positive, it does not affect the sign of
. Let
. Now since
and
,
is negative if and only if
, or when
. Since
is irrational, there is always only one integer in the range, so there are values of
such that
at
. Then the hundredth such value will be when
and
.
Solution 2
Notice that is the imaginary part of
, by Euler's formula. Using the geometric series formula, we find that this sum is equal to
We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have
We only need to look at the imaginary part, which is
Since
,
, so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have
. This only holds when
is between
and
for integer
[continuity proof here], and since this has exactly one integer solution for every such interval, the
th such
is
.
Solution 3
Similar to solution 2, we set a complex number . We start from
instead of
because
starts from
: be careful.
The sum of .
We are trying to make so that the imaginary part of this expression is negative.
The argument of is
. The argument of
, however, is a little more tricky.
is on a circle centered on
with radius
. The change in angle due to
is
with respect to the center, but the angle that
makes with the
-axis is
the change, due to Circle Theorems (this intercepted arc is the argument of
), because the
- axis is tangent to the circle at the origin. So
. Dividing
by
subtracts the latter argument from the former, so the angle of the quotient with the
-axis is
.
We want the argument of the whole expression . This translates into
.
also consists of points on the circle centered at
, so we deal with this argument similarly: the argument of
is twice the angle
makes with the
-axis. Since
is always negative,
, and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a
angle with the
-axis both ways.
So the argument of must be in the bound
by doubling, namely the last
negative before another rotation. Since there is always one
in this category per rotation because
is irrational,
and the answer is
.
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.