Difference between revisions of "2018 AIME I Problems/Problem 6"
Line 21: | Line 21: | ||
Since <math>|z|=1</math>, let <math>z=\cos \theta + i\sin \theta</math>. For <math>z^{6!}-z^{5!}</math> to be real, the imaginary parts of <math>z^{6!}</math> and <math>z^{5!}</math> must be equal, so <math>\sin 720\theta=\sin 120\theta</math>. We need to find all solutions for <math>\theta</math> in the interval <math>[0,2\pi)</math>. This can be done by graphing <math>y=\sin 720\theta</math> and <math>y=\sin 120\theta</math> and finding their intersections. Since the period of <math>y=\sin 720\theta</math> is <math>\frac{\pi}{360}</math> and the period of <math>y=\sin 120\theta</math> is <math>\frac{\pi}{60}</math>, the common period of both graphs is <math>\frac{\pi}{60}</math>. Therefore, we only graph the functions in the domain <math>[0, \frac{\pi}{60})</math>. We can clearly see that there are twelve points of intersection. However, since we only graphed <math>\frac{1}{120}</math> of the interval <math>[0,2\pi)</math>, we need to multiply our answer by <math>120</math>. The answer is <math>12 \cdot 120 = 1440 = \boxed{440} \pmod{1000}</math>. | Since <math>|z|=1</math>, let <math>z=\cos \theta + i\sin \theta</math>. For <math>z^{6!}-z^{5!}</math> to be real, the imaginary parts of <math>z^{6!}</math> and <math>z^{5!}</math> must be equal, so <math>\sin 720\theta=\sin 120\theta</math>. We need to find all solutions for <math>\theta</math> in the interval <math>[0,2\pi)</math>. This can be done by graphing <math>y=\sin 720\theta</math> and <math>y=\sin 120\theta</math> and finding their intersections. Since the period of <math>y=\sin 720\theta</math> is <math>\frac{\pi}{360}</math> and the period of <math>y=\sin 120\theta</math> is <math>\frac{\pi}{60}</math>, the common period of both graphs is <math>\frac{\pi}{60}</math>. Therefore, we only graph the functions in the domain <math>[0, \frac{\pi}{60})</math>. We can clearly see that there are twelve points of intersection. However, since we only graphed <math>\frac{1}{120}</math> of the interval <math>[0,2\pi)</math>, we need to multiply our answer by <math>120</math>. The answer is <math>12 \cdot 120 = 1440 = \boxed{440} \pmod{1000}</math>. | ||
+ | ==Solution 6 (Official MAA)== | ||
+ | If <math>z</math> satisfies the given conditions, there is a <math>\theta \in [0,2\pi]</math> such that <math>z=e^{i\theta}</math> and <math>e^{720\theta i-120\theta i}</math> is real. This difference is real if and only if either the two numbers <math>720\theta</math> and <math>120\theta</math> represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer <math>k</math> such that <math>720\theta=120\theta+2k\pi,</math> which implies that <math>\theta</math> is a multiple of <math>\tfrac{\pi}{300}.</math> In the second case there is an integer <math>k</math> such that <math>720\theta=120\theta+(2k+1)\pi,</math> which implies that <math>\theta</math> is <math>\tfrac{\pi}{840}</math> plus a multiple of <math>\tfrac{\pi}{420}.</math> In the interval <math>[0,2\pi]</math> there are <math>600</math> values of <math>\theta</math> that are multiples of <math>\tfrac{\pi}{300},</math> there are <math>840</math> values that are <math>\tfrac{\pi}{840}</math> plus a multiple of <math>\tfrac{\pi}{420},</math> and there are no values of <math>\theta</math> that satisfy both of these conditions. Therefore there | ||
+ | must be <math>600+840=1440</math> complex numbers satisfying the given conditions. The requested remainder is <math>440.</math> | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 13:00, 3 March 2021
Contents
Problem
Let be the number of complex numbers with the properties that and is a real number. Find the remainder when is divided by .
Solution 1
Let . This simplifies the problem constraint to . This is true if . Let be the angle makes with the positive x-axis. Note that there is exactly one for each angle . This must be true for values of (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time ). For each of these solutions for , there are necessarily solutions for . Thus, there are solutions for , yielding an answer of .
Solution 2
The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to . Since , let , then we can write the imaginary part of . Using the sum-to-product formula, we get or . The former yields solutions, and the latter yields solutions, giving a total of solution, so our answer is .
Solution 3
As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let . We have two cases to consider. Either , or and are reflections across the imaginary axis. If , then . Thus, or , giving us 600 solutions. (Equalities are taken modulo ) For the second case, . This means , giving us 840 solutions. Our total count is thus , yielding a final answer of .
Solution 4
Because we know that Hence Because is real, it is equal to its complex conjugate. Hence Substituting the expression we that we derived earlier, we get This leaves us with a polynomial whose leading term is Hence our answer is .
Solution 5
Since , let . For to be real, the imaginary parts of and must be equal, so . We need to find all solutions for in the interval . This can be done by graphing and and finding their intersections. Since the period of is and the period of is , the common period of both graphs is . Therefore, we only graph the functions in the domain . We can clearly see that there are twelve points of intersection. However, since we only graphed of the interval , we need to multiply our answer by . The answer is .
Solution 6 (Official MAA)
If satisfies the given conditions, there is a such that and is real. This difference is real if and only if either the two numbers and represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer such that which implies that is a multiple of In the second case there is an integer such that which implies that is plus a multiple of In the interval there are values of that are multiples of there are values that are plus a multiple of and there are no values of that satisfy both of these conditions. Therefore there must be complex numbers satisfying the given conditions. The requested remainder is
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
See also
2018 AIME 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 AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.