Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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\cos 2023 x + i \sin 2023 x | \cos 2023 x + i \sin 2023 x | ||
&= (\cos x + i \sin x)^{2023}\\ | &= (\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{2023}{3} \cos^{ | + | &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2023}{3} \cos^{2020} x (i \sin x)^{3}\\ |
&+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ | &+ \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\\ | &= \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\\ | ||
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==Solution 2 (Formula of tanx)== | ==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 | + | 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 \pm \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>. |
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~lprado | ~lprado | ||
− | + | ==Solution 3== | |
− | ==Solution== | ||
For odd <math>n</math>, we have | For odd <math>n</math>, we have | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~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== | ==Video Solution 1 by OmegaLearn== | ||
https://youtu.be/4KJR_1Kg4A4 | https://youtu.be/4KJR_1Kg4A4 | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/0KH554CLayE | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2023|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2023|ab=A|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:28, 2 August 2025
Contents
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.