Difference between revisions of "2020 CIME I Problems/Problem 6"
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Find the number of complex numbers <math>z</math> satisfying <math>|z|=1</math> and <math>z^{850}+z^{350}+1=0</math>. | Find the number of complex numbers <math>z</math> satisfying <math>|z|=1</math> and <math>z^{850}+z^{350}+1=0</math>. | ||
− | ==Solution== | + | ==Solution 1== |
− | We reduce the problem to <math>z^17+z^7+1</math>, remembering to multiply the final product by 50. We need the imaginary parts of the numbers <math>z^17,z^7</math> to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form <math> | + | We reduce the problem to <math>z^{17}+z^7+1</math>, remembering to multiply the final product by 50. We need the imaginary parts of the numbers <math>z^{17},z^7</math> to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form <math>cis(15x)</math> (this holds true because we are only looking for solutions with a magnitude of <math>1</math>). We also need the real parts to sum to <math>-1</math>. We check all the multiples of 15 that result in <math>cis(x)</math> being negative, and find that only two work(or alternatively, if you are good, you can guess that only <math>120</math> and <math>240</math> work). The answer is then <math>100</math>. |
− | |||
+ | ===Note=== | ||
+ | One definite way to find <math>cisa</math> and <math>cisb</math> such that they sum to <math>-1</math> is by setting up equations. | ||
+ | |||
+ | Let cos<math>a=x</math> and cos<math>b=y</math>. By definition | ||
+ | |||
+ | <cmath>x+y=-1</cmath> | ||
+ | <cmath>\sqrt{1-x^2}+\sqrt{1-y^2}=0</cmath> | ||
+ | |||
+ | We can ingnore plus or minus for the sins as this would still yield the same result. From the second equation <math>1-x^2=1-y^2</math> and <math>x= \pm y</math>. | ||
+ | |||
+ | Plugging into the first equation, we see that <math>x=-1/2, y=-1/2</math>. And we have <math>a=120</math> and <math>b=240</math> or the other way around. | ||
+ | |||
+ | (Note by Bigbrain_2009) | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>f(w)=w^{850}+w^{350}+1</math> and suppose <math>z</math> is such that <math>f(z)=0</math> and <math>|z|=1</math>. Note that <math>z</math> is not real (by descartes rule of signs), thus <math>\overline{z}=|z|^2/z=1/z</math> is also a root of <math>f</math>. It follows that <math>z^{850}f(1/z)=z^{850}+z^{500}+1=0</math>. Subtracting we have <cmath>0=z^{500}-z^{350}=z^{350}(z^{150}-1)=z^{350}(z^{50}-1)(z^{100}+z^{50}+1)</cmath> | ||
+ | Now <math>z^{350}\neq 0</math> else <math>f(z)=1\neq 0</math>, and <math>z^{50}\neq 1</math> else <math>f(z)=3\neq0</math>. Hence if <math>|z|=1</math> and <math>f(z)=0</math> then we must have <math>z^{100}+z^{50}+1=0</math>. Conversely, if <math>z</math> satisfies <math>z^{100}+z^{50}+1=0</math> then <math>z^{150}=1</math> so that <math>|z|=1</math> and <math>z^{850}+z^{350}+1=z^{100}+z^{50}+1=0</math>. Therefore <math>z</math> satisfies <math>f(z)=0</math> and <math>|z|=1</math> if and only if <math>z^{100}+z^{50}+1=0</math>. Note that this equation has 100 solutions (by the fundamental theorem of algebra) lying on the unit circle. Furthermore they are distinct since the solutions of <math>z^{150}=1</math> are distinct and <math>z^{100}+z^{50}+1</math> is a factor of <math>z^{150}-1</math>. Therefore the answer is 100. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We have <math>\frac{z^{850}+z^{350}+1}{3}=0</math>. Geometrically speaking, this means that the centroid with vertices on <math>z^{850}, z^{350}, 1</math> is the origin. However, because the circumcenter is also the origin, the only configuration that works is if all three points form an equilateral triangle. Thus, <math>z^{850}, z^{350}</math> are the primitive 3rd roots of unity in some order. Let <math>y=z^50</math>, then because <math>17</math> and <math>7</math> are relatively prime, there is one solution of <math>y</math> for each permutation, so <math>2</math> in total. Thus, there are <math>2*50=100</math> solutions. | ||
+ | |||
+ | ==See also== | ||
{{CIME box|year=2020|n=I|num-b=5|num-a=7}} | {{CIME box|year=2020|n=I|num-b=5|num-a=7}} | ||
{{MAC Notice}} | {{MAC Notice}} |
Latest revision as of 11:18, 1 February 2025
Problem 6
Find the number of complex numbers satisfying
and
.
Solution 1
We reduce the problem to , remembering to multiply the final product by 50. We need the imaginary parts of the numbers
to cancel, which by working modulo 360 we can easily determine only happens when the number is of the form
(this holds true because we are only looking for solutions with a magnitude of
). We also need the real parts to sum to
. We check all the multiples of 15 that result in
being negative, and find that only two work(or alternatively, if you are good, you can guess that only
and
work). The answer is then
.
Note
One definite way to find and
such that they sum to
is by setting up equations.
Let cos and cos
. By definition
We can ingnore plus or minus for the sins as this would still yield the same result. From the second equation and
.
Plugging into the first equation, we see that . And we have
and
or the other way around.
(Note by Bigbrain_2009)
Solution 2
Let and suppose
is such that
and
. Note that
is not real (by descartes rule of signs), thus
is also a root of
. It follows that
. Subtracting we have
Now
else
, and
else
. Hence if
and
then we must have
. Conversely, if
satisfies
then
so that
and
. Therefore
satisfies
and
if and only if
. Note that this equation has 100 solutions (by the fundamental theorem of algebra) lying on the unit circle. Furthermore they are distinct since the solutions of
are distinct and
is a factor of
. Therefore the answer is 100.
Solution 3
We have . Geometrically speaking, this means that the centroid with vertices on
is the origin. However, because the circumcenter is also the origin, the only configuration that works is if all three points form an equilateral triangle. Thus,
are the primitive 3rd roots of unity in some order. Let
, then because
and
are relatively prime, there is one solution of
for each permutation, so
in total. Thus, there are
solutions.
See also
2020 CIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All CIME Problems and Solutions |
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