Difference between revisions of "1997 AIME Problems/Problem 14"
DeathLlama9 (talk | contribs) (→Solution) |
(→Solution 1) |
||
Line 12: | Line 12: | ||
Now, let <math>v</math> be the root corresponding to <math>\theta=\frac{2\pi m}{1997}</math>, and let <math>w</math> be the root corresponding to <math>\theta=\frac{2\pi n}{1997}</math>. The magnitude of <math>v+w</math> is therefore: | Now, let <math>v</math> be the root corresponding to <math>\theta=\frac{2\pi m}{1997}</math>, and let <math>w</math> be the root corresponding to <math>\theta=\frac{2\pi n}{1997}</math>. The magnitude of <math>v+w</math> is therefore: | ||
− | + | <cmath>\sqrt{\left(\cos\left(\frac{2\pi m}{1997}\right) + \cos\left(\frac{2\pi n}{1997}\right)\right)^2 + \left(\sin\left(\frac{2\pi m}{1997}\right) + \sin\left(\frac{2\pi n}{1997}\right)\right)^2}=\sqrt{2 + 2\cos\left(\frac{2\pi m}{1997}\right)\cos\left(\frac{2\pi n}{1997}\right) + 2\sin\left(\frac{2\pi m}{1997}\right)\sin\left(\frac{2\pi n}{1997}\right)}</cmath> | |
− | |||
− | We need < | + | We need <cmath>\cos \left(\frac{2\pi m}{1997}\right)\cos \left(\frac{2\pi n}{1997}\right) + \sin \left(\frac{2\pi m}{1997}\right)\sin \left(\frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</cmath>The [[Trigonometric identities|cosine difference identity]] simplifies that to <cmath>\cos\left(\frac{2\pi m}{1997} - \frac{2\pi n}{1997}\right) \ge \frac{\sqrt{3}}{2}</cmath>Thus, <cmath>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \lfloor \frac{1997}{12} \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>. |
Revision as of 19:53, 4 August 2021
Problem
Let and
be distinct, randomly chosen roots of the equation
. Let
be the probability that
, where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
By De Moivre's Theorem, we find that ()
Now, let be the root corresponding to
, and let
be the root corresponding to
. The magnitude of
is therefore:
We need The cosine difference identity simplifies that 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
for
They are also located at the vertices of a regular
-gon that is centered at the origin in the complex plane.
Without loss of generality, let Then
We want From what we just obtained, this is equivalent to
This occurs when
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.