Difference between revisions of "2017 AMC 12A Problems/Problem 25"
m (→Solution) |
m |
||
Line 66: | Line 66: | ||
</cmath> | </cmath> | ||
Let <math>\zeta=e^{\frac{i\pi}{4}}</math>. We are looking for <math>\frac{P(1)+P(\zeta)+P(\zeta^2)+P(\zeta^3)+P(\zeta^4)+P(\zeta^5)+P(\zeta^6)+P(\zeta^7)}{8}</math>. <math>P(1)=P(-1)=2^{16}</math>, and all of the rest are equal to <math>0</math>. So, we get an answer of <math>\frac{2^{16}+2^{16}}{8\cdot 6^{12}}=\frac{2^{14}}{6^{12}}=\frac{2^2}{3^{10}}</math>. But wait! We need to multiply by <math>{12}\choose {4}</math>=<math>\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot3\cdot1}=5\cdot9\cdot11</math>. So, the answer is <math>\boxed{\text{\textbf{(E) }} \frac{2^2\cdot5\cdot11}{3^{10}} }</math> | Let <math>\zeta=e^{\frac{i\pi}{4}}</math>. We are looking for <math>\frac{P(1)+P(\zeta)+P(\zeta^2)+P(\zeta^3)+P(\zeta^4)+P(\zeta^5)+P(\zeta^6)+P(\zeta^7)}{8}</math>. <math>P(1)=P(-1)=2^{16}</math>, and all of the rest are equal to <math>0</math>. So, we get an answer of <math>\frac{2^{16}+2^{16}}{8\cdot 6^{12}}=\frac{2^{14}}{6^{12}}=\frac{2^2}{3^{10}}</math>. But wait! We need to multiply by <math>{12}\choose {4}</math>=<math>\frac{12\cdot11\cdot10\cdot9}{4\cdot3\cdot3\cdot1}=5\cdot9\cdot11</math>. So, the answer is <math>\boxed{\text{\textbf{(E) }} \frac{2^2\cdot5\cdot11}{3^{10}} }</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | We can write the points in polar form <math>(r, \theta)</math> as | ||
+ | <cmath>V = \{ (\sqrt{2}, \frac{\pi}{2}), (\sqrt{2}, \frac{3\pi}{2}), (\frac{1}{2}, \frac{\pi}{4}), (\frac{1}{2}, \frac{3\pi}{4}), (\frac{1}{2}, \frac{5\pi}{4}), (\frac{1}{2}, \frac{7\pi}{4}).</cmath>Note that when multiplying complex numbers, the <math>r</math>s multiply and the <math>\theta</math>s add, and since <math>-1 = (1, \pi),</math> we need <math>8</math> complex numbers with <math>r = \sqrt{2} = 2^{0.5}</math> and <math>4</math> with <math>r = \frac{1}{2} = 2^{-1}.</math> By binomial distribution, the probability of this occurring is <math>\dbinom{12}{4} (\frac{1}{3})^8 (\frac{2}{3})^4.</math> For the <math>\theta</math> part, note that <math>\frac{4\theta}{\pi}</math> must be congruent to <math>4 \mod 8, </math> and by using simple symmetry, the probability of the aforementioned occurring is <math>\frac{1}{4}. </math> This is since <math>2a + 6b</math> is even for all <math>a + b =8, </math> and the number of ordered quadruples <math>(a_1, a_2, a_3, a_4)</math> such that <math>a_i \in \{1, 3, 5, 7\}</math> for all <math>1 \leq i \leq 4</math> and <math>a_1 + a_2 + a_3 + a_4 \equiv 2k \mod 8</math> is the same for all <math>1 \leq k \leq 4,</math> again by using symmetry. Thus, our probability is <cmath>(\frac{1}{4}) (\frac{1}{3})^8 (\frac{2}{3})^4 \dbinom{12}{4} = \frac{2^2}{3^{12}} \dbinom{12}{4} = \boxed{\frac{2^2 \cdot 11 \cdot 5}{3^{10}}}.</cmath> | ||
+ | |||
+ | -fidgetboss_4000 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2017|ab=A|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:28, 3 June 2020
Problem
The vertices of a centrally symmetric hexagon in the complex plane are given by For each , , an element is chosen from at random, independently of the other choices. Let be the product of the numbers selected. What is the probability that ?
Solution 1
It is possible to solve this problem using elementary counting methods. This solution proceeds by a cleaner generating function.
We note that both lie on the imaginary axis and each of the have length and angle of odd multiples of , i.e. . When we draw these 6 complex numbers out on the complex plane, we get a crystal-looking thing. Note that the total number of ways to choose 12 complex numbers is . Now we count the number of good combinations.
We first consider the lengths. When we multiply 12 complex numbers together, their magnitudes multiply. Suppose we have of the numbers ; then we must have . Having will take care of the length of the product; now we need to deal with the angle.
We require . Letting be , we see that the angles we have available are , where we must choose exactly 8 angles from the set and exactly 4 from the set . If we found a good combination where we had of each angle , then the amount this would contribute to our count would be . We want to add these all up. We proceed by generating functions.
Consider The expansion will be of the form . Note that if we reduced the powers of mod and fished out the coefficient of and plugged in (and then multiplied by ) then we would be done. Since plugging in doesn't affect the 's, we do that right away. The expression then becomes where the last equality is true because we are taking the powers of mod . Let denote the coefficient of in . Note . We use the roots of unity filter, which states where . In our case , so we only need to find the average of the 's. We plug in and take the average to find the sum of all coefficients of . Plugging in makes all of the above zero except for and . Averaging, we get . Now the answer is simply
Solution 2
By changing to , we can give a bijection between cases where and cases where , so we'll just find the probability that and divide by in the end. Multiplying the hexagon's vertices by doesn't change , and switching any with doesn't change the property , so the probability that remains the same if we only select our 's at random from Since and , we must choose exactly times to make . To ensure is real, we must either choose times, times, or both and times. This gives us a total of good sequences , and hence the final result is
Solution 3
We use generating functions and a roots of unity filter. Notice that all values in are eighth roots of unity multiplied by a constant. Let be a primitive eighth root of unity (). The numbers in are then . To have , we must have that , so eight of the must belong to and the other four must belong to So, we write the generating function to describe the product. Note that this assumes that the that belong to come first, so we will need to multiply by at the end. We now apply a roots of unity filter to find the sum of the coefficients of the exponents that are , or equivalently the coefficients of the powers that are multiples of of the following function: Let . We are looking for . , and all of the rest are equal to . So, we get an answer of . But wait! We need to multiply by =. So, the answer is
Solution 4
We can write the points in polar form as Note that when multiplying complex numbers, the s multiply and the s add, and since we need complex numbers with and with By binomial distribution, the probability of this occurring is For the part, note that must be congruent to and by using simple symmetry, the probability of the aforementioned occurring is This is since is even for all and the number of ordered quadruples such that for all and is the same for all again by using symmetry. Thus, our probability is
-fidgetboss_4000
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.