Question

下图显示一个半径为 $20$ 厘米的圆形钟面，以及一个半径为 $10$ 厘米的圆盘，它在 $12$ 点处与钟面外切。圆盘上画有一个箭头，初始时竖直向上。令圆盘沿钟面顺时针滚动。当箭头下一次竖直向上时，圆盘与钟面在哪个位置相切？

The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?

$2$ 点钟

$2$ o'clock

$3$ 点钟

$3$ o'clock

$4$ 点钟

$4$ o'clock

$6$ 点钟

$6$ o'clock

$8$ 点钟

$8$ o'clock

答案：C

知识点：圆周长圆比与比例

难度评级：1790

解答：

半径 $10$ 的圆盘在半径 $20$ 的钟面外侧滚动。若接触点绕钟面转过圆心角 $\theta$ ，圆盘相对于原方向转过 $\frac{20+10}{10}\theta=3\theta$ 。

箭头再次竖直向上时， $3\theta$ 等于 $2\pi$ ，所以 $\theta=\frac{2\pi}{3}$ ，对应四点钟方向。

所以正确答案是 C。

The disk of radius $10$ rolls externally around the clock face of radius $20$ . If the point of tangency moves through central angle $\theta$ around the clock, the disk rotates through $\frac{20+10}{10}\theta=3\theta$ relative to its original direction.

The arrow next points upward when $3\theta$ is a positive multiple of $2\pi$ . The first time this happens is $\theta=\frac{2\pi}{3}$ , which is one-third of the way around the clock from 12 o'clock, namely 4 o'clock.

Thus, C is the correct answer.

2015 AMC 10A 第 14 题

答案：C

难度评级：1790

解答：

其他年份的第 14 题