2025 AMC 12B 第 16 题

Question

一个指针钟从午夜开始走了

2025

分钟后停止。停止时，时针和分针之间的锐角的正切值是多少？

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?

0

\sqrt{2} - 1

2 - \sqrt{2}

\dfrac{\sqrt{2}}{2}

3 - \sqrt{2}

将

2025

分钟按

12

小时取余，找出钟面显示的时间。

Reduce

2025

minutes modulo

12

hours to find where the hands point

分别求时针和分针的角度，取锐角差，再计算

\tan 22.5^\circ

。

Find each hand’s angle, take the acute difference, then evaluate

\tan 22.5^\circ

2025

分钟是

33

小时

45

分钟，按

12

小时取余后为

9{:}45

。分针指向

270^\circ

，时针指向

9.75 \times 30^\circ = 292.5^\circ

，所以两针之间的锐角是

22.5^\circ

。由半角值，

\tan 22.5^\circ = \sqrt{2} - 1

。

所以正确答案是 B。

2025

minutes is

33

hours

45

minutes, which modulo

12

hours reads

9{:}45.

The minute hand points at

270^\circ

and the hour hand at

9.75 \times 30^\circ = 292.5^\circ,

so the acute angle between them is

22.5^\circ.

Using the half-angle value,

\tan 22.5^\circ = \sqrt{2} - 1.

Thus, the correct answer is B.

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