Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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+ | ==Problem== | ||
+ | There is a unique sequence of integers <math>a_1, a_2, \cdots a_{2023}</math> such that | ||
+ | <cmath> | ||
+ | \tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} | ||
+ | </cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math> | ||
+ | <math>\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \cos 2023 x + i \sin 2023 x | ||
+ | &= (\cos x + i \sin x)^{2023}\ | ||
+ | &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2020}{3} \cos^{2020} x (i \sin x)^{3}\ | ||
+ | &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\ | ||
+ | &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\ | ||
+ | &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | By equating real and imaginary parts: | ||
+ | |||
+ | <cmath>\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x</cmath> | ||
+ | |||
+ | <cmath>\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x</cmath> | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \tan2023x | ||
+ | &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\ | ||
+ | &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x} | ||
+ | }\ | ||
+ | &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <cmath>a_{2023} = \boxed{\textbf{(C)}-1}</cmath> | ||
+ | |||
+ | This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove <math>\tan{nx} = \frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \dots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \dots}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 2 (Formula of tanx)== | ||
+ | Note that <math>\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}</math>, where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of <math>\tan{2x}, \tan{3x},</math> and <math>\tan{4x}</math>, and can notice the pattern from that. The expression given essentially matches the formula of <math>\tan{kx}</math> exactly. <math>a_{2023}</math> is evidently equivalent to <math>\pm\binom{2023}{2023}</math>, or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of <math>\binom{k}{k}\tan^{k}{x}</math> is <math>\boxed{\textbf{(C) } -1}</math>. | ||
+ | |||
+ | |||
+ | Notice: If you have time and don't know <math>\tan{3x}</math> and <math>\tan{4x}</math>, you'd have to keep deriving <math>\tan{kx}</math> until you see the pattern. | ||
+ | |||
+ | ~lprado | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | For odd <math>n</math>, we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \tan nx | ||
+ | & = \frac{\sin nx}{\cos nx} \ | ||
+ | & = \frac{\frac{1}{2i} \left( e^{i n x} - e^{-i n x} \right)} | ||
+ | {\frac{1}{2} \left( e^{i n x} + e^{-i n x} \right)} \ | ||
+ | & = - i \frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \ | ||
+ | & = - i \frac{\left( \cos x + i \sin x \right)^n - \left( \cos x - i \sin x \right)^n} | ||
+ | {\left( \cos x + i \sin x \right)^n + \left( \cos x - i \sin x \right)^n} \ | ||
+ | & = \frac{ - 2 i \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} | ||
+ | {2 \sum_{m=0}^{(n-1)/2} \binom{n}{2m} | ||
+ | \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \ | ||
+ | & = \frac{ \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} | ||
+ | {i \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m} | ||
+ | \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \ | ||
+ | & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( i \tan x \right)^{2m + 1}} | ||
+ | {i \sum_{m=0}^{(n-1)/2} \binom{n}{2m} | ||
+ | \left( i \tan x \right)^{2m}} \ | ||
+ | & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} | ||
+ | i^{2m + 1}} | ||
+ | {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} | ||
+ | \left( \tan x \right)^{2m} i^{2m + 1}} \ | ||
+ | & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} | ||
+ | \left( -1 \right)^m} | ||
+ | {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} | ||
+ | \left( \tan x \right)^{2m} \left( -1 \right)^m} | ||
+ | . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, for <math>n = 2023</math>, we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | a_{2023} & = \binom{2023}{2023} \left( -1 \right)^{(2023-1)/2} \ | ||
+ | & = \left( -1 \right)^{1011} \ | ||
+ | & = \boxed{\textbf{(C) -1}}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | |||
+ | We can use recursion to find a pattern for this problem. Notice that, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \tan x &= \tan x\ | ||
+ | \tan 2x &= \dfrac{\tan x + \tan x}{1 - \tan^2 x} = \dfrac{2\tan x}{1 - \tan^2 x}\ | ||
+ | \tan 3x &= \dfrac{\tan 2x + \tan x}{1 - \tan 2x \tan x} = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | The coefficient of the highest degree term seems to be always <math>\pm 1</math>. Now, we prove this by an imcomplete mathematical induction. | ||
+ | |||
+ | * Firstly, we suppose <math>n</math> is odd, and <math>\tan nx = \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x}</math>, then | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \tan(n + 1)x &= \dfrac{\tan nx + \tan x}{1 - \tan nx \tan x}\ | ||
+ | &= \dfrac{\dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x} + \tan x}{1 - \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x}{a_0 + a_2\tan^2 x + \cdots + a_{n - 1}\tan^{n - 1} x}\tan x}\ | ||
+ | &= \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_n\tan^n x + a_0\tan x + a_2\tan^3 x + \cdots + a_{n - 1}\tan^n x}{a_0 + a_2\tan^2x + \cdots + a_{n - 1}\tan^{n - 1}x - a_1\tan^2 x - a_3\tan^4 x - \cdots \boxed{-a_n\tan^{n + 1} x}} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | The coefficient of the highest degree term becomes <math>-a_n</math>. | ||
+ | |||
+ | * Secondly, we suppose <math>n</math> is even, and <math>\tan nx = \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x}</math>, then | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \tan(n + 1)x &= \dfrac{\tan nx + \tan x}{1 - \tan nx \tan x}\ | ||
+ | &= \dfrac{\dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x} + \tan x}{1 - \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x}\tan x}\ | ||
+ | &= \dfrac{a_1\tan x + a_3\tan^3 x + \cdots + a_{n - 1}\tan^{n - 1} x + a_0\tan x + a_2\tan^3 x + \cdots \boxed{+ a_n\tan^{n + 1} x}}{a_0 + a_2\tan^2 x + \cdots + a_n\tan^n x - a_1\tan^2 x - a_3\tan^4 x - \cdots - a_{n - 1}\tan^n x} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | The coefficient of the highest degree term remains <math>a_n</math>. | ||
+ | |||
+ | When <math>n = 1</math>, <math>a_n = 1</math>. During the process of <math>n</math> increasing to 2023, <math>a_n</math> changed its sign a total of <math>2022 \div 2 = 1011</math> times. | ||
+ | |||
+ | Hence, <math>a_{2023} = 1 \times (-1)^{1011} = \boxed{\textbf{(C)}-1}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath] | ||
+ | |||
+ | ==Video Solution 1 by OmegaLearn== | ||
+ | https://youtu.be/4KJR_1Kg4A4 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/0KH554CLayE | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2023|ab=A|num-b=24|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:38, 28 October 2024
Contents
[hide]Problem
There is a unique sequence of integers such that whenever is defined. What is
Solution 1
By equating real and imaginary parts:
This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove
Solution 2 (Formula of tanx)
Note that , where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of and , and can notice the pattern from that. The expression given essentially matches the formula of exactly. is evidently equivalent to , or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of is .
Notice: If you have time and don't know and , you'd have to keep deriving until you see the pattern.
~lprado
Solution 3
For odd , we have
Thus, for , we have
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4
We can use recursion to find a pattern for this problem. Notice that, The coefficient of the highest degree term seems to be always . Now, we prove this by an imcomplete mathematical induction.
- Firstly, we suppose is odd, and , then
The coefficient of the highest degree term becomes .
- Secondly, we suppose is even, and , then
The coefficient of the highest degree term remains .
When , . During the process of increasing to 2023, changed its sign a total of times.
Hence,
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 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.