2023 AMC 12A Problems/Problem 25

Problem

There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that $\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}$ whenever $\tan 2023x$ is defined. What is $a_{2023}?$

$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$

Solution

Note that $\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}$ , 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 $\tan{2x}, \tan{3x},$ and $\tan{4x}$ , and can notice the pattern from that. The expression given essentially matches the formula of $\tan{kx}$ exactly. $a_{2023}$ is evidently equivalent to $\pm\binom{2023}{2023}$ , 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 $\binom{k}{k}\tan^{k}{x}$ is $\boxed{\textbf{(C) } -1}$ .

Notice: If you have time and don't know $\tan{3x}$ and $\tan{4x}$ , you'd have to keep deriving $\tan{kx}$ until you see the pattern.

~lprado

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2023 AMC 12A Problems/Problem 25

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