Difference between revisions of "1997 AIME Problems/Problem 14"
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:<math>z^{1997}=1=1(\cos 0 + i \sin 0)</math> | :<math>z^{1997}=1=1(\cos 0 + i \sin 0)</math> | ||
− | By [[De Moivre's Theorem]] | + | Define <math>\theta = 2\pi/1997</math>. By [[De Moivre's Theorem]] the roots are given by |
− | :<math>z=\cos | + | :<math>z=\cos (k\theta) +i\sin(k\theta), \qquad k \in \{0,1,\ldots,1996\}</math> |
− | Now, let <math>v</math> be the root corresponding to <math>\theta= | + | Now, let <math>v</math> be the root corresponding to <math>m\theta=2m\pi/1997</math>, and let <math>w</math> be the root corresponding to <math>n\theta=2n\pi/ 1997</math>. Then |
− | + | <cmath>\begin{align*} | |
− | + | |v+w|^2 &= \left(\cos(m\theta) + \cos(n\theta)\right)^2 + \left(\sin(m\theta) + \sin(n\theta)\right)^2 \\ | |
+ | &= 2 + 2\cos\left(m\theta\right)\cos\left(n\theta\right) + 2\sin\left(m\theta\right)\sin\left(n\theta\right) | ||
+ | \end{align*}</cmath> | ||
+ | The [[Trigonometric identities|cosine difference identity]] simplifies that to | ||
+ | <cmath>|v+w|^2 = 2+2\cos((m-n)\theta) </cmath> | ||
− | We need <math> | + | We need <math>|v+w|^2 \ge 2+\sqrt{3}</math>, which simplifies to <cmath>\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}</cmath>Thus, <cmath>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166</cmath>. |
Therefore, <math>m</math> and <math>n</math> cannot be more than <math>166</math> away from each other. This means that for a given value of <math>m</math>, there are <math>332</math> values for <math>n</math> that satisfy the inequality; <math>166</math> of them <math>> m</math>, and <math>166</math> of them <math>< m</math>. Since <math>m</math> and <math>n</math> must be distinct, <math>n</math> can have <math>1996</math> possible values. Therefore, the probability is <math>\frac{332}{1996}=\frac{83}{499}</math>. The answer is then <math>499+83=\boxed{582}</math>. | Therefore, <math>m</math> and <math>n</math> cannot be more than <math>166</math> away from each other. This means that for a given value of <math>m</math>, there are <math>332</math> values for <math>n</math> that satisfy the inequality; <math>166</math> of them <math>> m</math>, and <math>166</math> of them <math>< m</math>. Since <math>m</math> and <math>n</math> must be distinct, <math>n</math> can have <math>1996</math> possible values. Therefore, the probability is <math>\frac{332}{1996}=\frac{83}{499}</math>. The answer is then <math>499+83=\boxed{582}</math>. | ||
=== Solution 2 === | === Solution 2 === | ||
− | The solutions of the equation <math>z^{1997} = 1</math> are the <math>1997</math>th [[roots of unity]] and are equal to <math>\ | + | The solutions of the equation <math>z^{1997} = 1</math> are the <math>1997</math>th [[roots of unity]] and are equal to <math>\text{cis}(\theta_k)</math>, where <math>\theta_k = \tfrac {2\pi k}{1997}</math> for <math>k = 0,1,\ldots,1996.</math> Thus, they are located at uniform intervals on the unit circle in the complex plane. |
− | + | The quantity <math>|v+w|</math> is unchanged upon rotation around the origin, so, WLOG, we can assume <math>v=1</math> after rotating the axis till <math>v</math> lies on the real axis. Let <math>w=\text{cis}(\theta_k)</math>. Since <math>w\cdot \overline{w}=|w|^2=1</math> and <math>w+\overline{w}=2\text{Re}(w) = 2\cos\theta_k</math>, we have <cmath>|v + w|^2 = (1+w)(1+\overline{w}) = 2+2\cos\theta_k</cmath> | |
− | < | + | We want <math>|v + w|^2\ge 2 + \sqrt {3}.</math> From what we just obtained, this is equivalent to <cmath>\cos\theta_k\ge \frac {\sqrt {3}}2 \qquad \Leftrightarrow \qquad -\frac {\pi}6\le \theta_k \le \frac {\pi}6</cmath> which is satisfied by <math>k = 166,165,\ldots, - 165, - 166</math> (we don't include 0 because that corresponds to <math>v</math>). So out of the <math>1996</math> possible <math>k</math>, <math>332</math> work. Thus, <math>m/n = 332/1996 = 83/499.</math> So our answer is <math>83 + 499 = \boxed{582}.</math> |
− | |||
− | |||
− | |||
− | |||
− | </cmath> | ||
− | |||
− | We want <math>|v + w|^2\ge 2 + \sqrt {3}.</math> From what we just obtained, this is equivalent to < | ||
=== Solution 3 === | === Solution 3 === | ||
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We can solve a geometrical interpretation of this problem. | We can solve a geometrical interpretation of this problem. | ||
− | Without loss of generality, let <math> | + | Without loss of generality, let <math>w = 1</math>. We are now looking for a point exactly one unit away from <math>w</math> such that the point is at least <math>\sqrt{2 + \sqrt{3}}</math> units away from the origin. Note that the "boundary" condition is when the point will be exactly <math>\sqrt{2+\sqrt{3}}</math> units away from the origin; these points will be the intersections of the circle centered at <math>(1,0)</math> with radius <math>1</math> and the circle centered at <math>(0,0)</math> with radius <math>\sqrt{2+\sqrt{3}}</math>. The equations of these circles are <math>(x-1)^2 + y^2 = 1</math> and <math>x^2 + y^2 = 2 + \sqrt{3}</math>. Solving for <math>x</math> yields <math>x = 1 + \frac{\sqrt{3}}{2}</math>. Clearly, this means that the real part of <math>v</math> is greater than <math>\frac{\sqrt{3}}{2}</math>. Solving, we note that <math>332</math> possible <math>v</math>s exist, meaning that <math>\frac{m}{n} = \frac{332}{1996} = \frac{83}{499}</math>. Therefore, the answer is <math>83 + 499 = \boxed{582}</math>. |
+ | |||
+ | === Solution 4 === | ||
+ | Since <math>z^{1997}=1</math>, the roots will have magnitude <math>1</math>. Thus, the roots can be written as <math>\cos(\theta)+i\sin(\theta)</math> and <math>\cos(\omega)+i\sin(\omega)</math> for some angles <math>\theta</math> and <math>\omega</math>. We rewrite the requirement as <math>\sqrt{2+\sqrt3}\le|\cos(\theta)+\cos(\omega)+i\sin(\theta)+i\sin(\omega)|</math>, which can now be easily manipulated to <math>2+\sqrt{3}\le(\cos(\theta)+\cos(\omega))^2+(\sin(\theta)+\sin(\omega))^2</math>. | ||
+ | WLOG, let <math>\theta = 0</math>. Thus, our inequality becomes <math>2+\sqrt{3}\le(1+\cos(\omega))^2+(\sin(\omega))^2</math>, <math>2+\sqrt{3}\le2+2\cos(\omega)</math>, and finally <math>\cos(\omega)\ge\frac{\sqrt{3}}{2}</math>. Obviously, <math>\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}</math>, and thus it follows that, on either side of a given point, <math>\frac{1997}{12}\approx166</math> points will work. The probability is <math>\frac{166\times2}{1996} = \frac{83}{499}</math>, and thus our requested sum is <math>\boxed{582}</math> | ||
+ | ~SigmaPiE | ||
== See also == | == See also == | ||
{{AIME box|year=1997|num-b=13|num-a=15}} | {{AIME box|year=1997|num-b=13|num-a=15}} |
Latest revision as of 01:25, 19 June 2024
Problem
Let and
be distinct, randomly chosen roots of the equation
. Let
be the probability that
, where
and
are relatively prime positive integers. Find
.
Contents
Solution
Solution 1
Define . By De Moivre's Theorem the roots are given by
Now, let be the root corresponding to
, and let
be the root corresponding to
. Then
The cosine difference identity simplifies that to
We need , which simplifies to
Thus,
.
Therefore, and
cannot be more than
away from each other. This means that for a given value of
, there are
values for
that satisfy the inequality;
of them
, and
of them
. Since
and
must be distinct,
can have
possible values. Therefore, the probability is
. The answer is then
.
Solution 2
The solutions of the equation are the
th roots of unity and are equal to
, where
for
Thus, they are located at uniform intervals on the unit circle in the complex plane.
The quantity is unchanged upon rotation around the origin, so, WLOG, we can assume
after rotating the axis till
lies on the real axis. Let
. Since
and
, we have
We want
From what we just obtained, this is equivalent to
which is satisfied by
(we don't include 0 because that corresponds to
). So out of the
possible
,
work. Thus,
So our answer is
Solution 3
We can solve a geometrical interpretation of this problem.
Without loss of generality, let . We are now looking for a point exactly one unit away from
such that the point is at least
units away from the origin. Note that the "boundary" condition is when the point will be exactly
units away from the origin; these points will be the intersections of the circle centered at
with radius
and the circle centered at
with radius
. The equations of these circles are
and
. Solving for
yields
. Clearly, this means that the real part of
is greater than
. Solving, we note that
possible
s exist, meaning that
. Therefore, the answer is
.
Solution 4
Since , the roots will have magnitude
. Thus, the roots can be written as
and
for some angles
and
. We rewrite the requirement as
, which can now be easily manipulated to
.
WLOG, let . Thus, our inequality becomes
,
, and finally
. Obviously,
, and thus it follows that, on either side of a given point,
points will work. The probability is
, and thus our requested sum is
~SigmaPiE
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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