Question

方程的所有正实数解 $x$ 之和是多少？ $\begin{gathered} \small 2\cos(2x)\left(\cos(2x) - \cos\left(\dfrac{2014\pi^2}{x}\right)\right) \\ = \cos(4x) - 1? \end{gathered}$

What is the sum of all positive real solutions $x$ to the equation $\begin{gathered} \small 2\cos(2x)\left(\cos(2x) - \cos\left(\dfrac{2014\pi^2}{x}\right)\right) \\ = \cos(4x) - 1? \end{gathered}$

$\pi$

$810\pi$

$1008\pi$

$1080\pi$

$1800\pi$

答案：D

知识点：三角恒等式因数之和

难度评级：2890

解答：

令 $x = \tfrac{\pi y}{2}$ 。两边除以 $2$ ，并使用 $\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y)$ ，方程化简为 $\cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1.$

两个余弦必须都等于 $1$ ，或都等于 $-1$ ，因此 $y$ 和 $\tfrac{4028}{y}$ 是同奇偶性的整数。因为 $4028 = 2^2 \cdot 19 \cdot 53$ 是偶数，二者都必须为偶数，所以 $y = 2a$ ，其中 $a$ 是 $2014 = 2 \cdot 19 \cdot 53$ 的正奇因数，故 $a \in \{1, 19, 53, 19 \cdot 53\}$ 。

每个这样的 $a$ 给出 $x = \tfrac{\pi y}{2} = \pi a$ ，所以解的和为 $\begin{gathered} \pi(1 + 19 + 53 + 19\cdot53) \\ = \pi(19+1)(53+1) \\ = 1080\pi. \end{gathered}$

所以正确答案是 D。

Let $x = \tfrac{\pi y}{2}.$ Dividing by $2$ and using $\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y),$ the equation simplifies to $\cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1.$

Both cosines must equal $1$ or both equal $-1,$ so $y$ and $\tfrac{4028}{y}$ are integers of the same parity. Since $4028 = 2^2 \cdot 19 \cdot 53$ is even, both must be even, so $y = 2a$ with $a$ a positive odd divisor of $2014 = 2 \cdot 19 \cdot 53,$ giving $a \in \{1, 19, 53, 19 \cdot 53\}.$

Each such $a$ gives $x = \tfrac{\pi y}{2} = \pi a,$ so the sum of solutions is $\begin{gathered} \pi(1 + 19 + 53 + 19\cdot53) \\ = \pi(19+1)(53+1) \\ = 1080\pi. \end{gathered}$

Thus, the correct answer is D.

2014 AMC 12B 第 25 题

答案：D

难度评级：2890

解答：

其他年份的第 25 题