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>. |
+ | |||
+ | ==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== | ==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 17:09, 6 February 2023
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 .
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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