Difference between revisions of "2023 AMC 12A Problems/Problem 25"
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~lprado | ~lprado | ||
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+ | ==Solution== | ||
+ | |||
+ | 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) | ||
+ | |||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 17:44, 10 November 2023
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
For odd , we have
Thus, for , we have
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn
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.