AMC12 2025 B

AMC12 2025 B · Q16

AMC12 2025 B · Q16. It mainly tests Exponents & radicals, Trigonometry (basic).

An analog clock starts at midnight and runs for $2025$ minutes before stopping. What is the tangent of the acute angle between the hour hand and the minute hand when the clock stops?

一个指针式时钟从午夜开始运行，运行了$2025$分钟后停止。时钟停止时，时针和分针之间的锐角的正切值为多少？

(A) 0 0

(B) \sqrt{2}-1 \sqrt{2}-1

(D) \frac{\sqrt{2}}{2} \frac{\sqrt{2}}{2}

(E) 3-\sqrt{2} 3-\sqrt{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

When divided by $60$, $2025$ yields $33$, with a remainder of $45$. Since $33 \equiv 9 \pmod{24}$, the clock is now at $9:45$. At this time, the hour is $75\%$ over, so the hour hand is also $75\%$ of the way between $9$ and $10$. There are $360/12 = 30 ^\circ$ degrees between $9$ and $10$, and since the hour hand is $75\%$ of the way, it is $30^\circ \cdot \frac{3}{4} = \frac{45}{2}^\circ$ from the minute hand, which is pointing directly at $9$. We now just need to find $\tan{(45/2)}$. Let this be $x$. Using the tangent addition formula, we get that \[\tan{45} = \frac{x + x}{1 - x\cdot x}\] \[1 = \frac{2x}{1-x^2}\] \[x^2+2x-1 = 0.\] Using the Quadratic Formula, and knowing $x$ must be positive since $0 < x < 90$, we get $x = \tan{(45/2)}= \boxed{\sqrt{2}-1}$ Note: One could solve this quickly using the *tangent half angle identity*, which is $\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1+\cos(\theta)} = \frac{1-\cos(\theta)}{\sin(\theta)}$ ~math660

$2025$除以$60$得商$33$，余数$45$。因为$33\equiv 9\pmod{24}$，此时钟指向$9:45$。此时，时针已走过$75\%$的进度，即从$9$到$10$之间的$75\%$。$9$到$10$之间有$360/12 = 30 ^\circ$度，时针走了$30^\circ \cdot \frac{3}{4} = \frac{45}{2}^\circ$，而分针指向$9$。现在只需求$\tan{(45/2)}$。设此值为$x$。使用正切加法公式，得到 \[\tan{45} = \frac{x + x}{1 - x\cdot x}\] \[1 = \frac{2x}{1-x^2}\] \[x^2+2x-1 = 0。\] 使用二次公式，且知$x$为正（因为$0 < x < 90$），得$x = \tan{(45/2)}= \boxed{\sqrt{2}-1}$ 注：也可以使用*正切半角公式*快速求解，即$\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1+\cos(\theta)} = \frac{1-\cos(\theta)}{\sin(\theta)}$

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