AMC10 2015 A

AMC10 2015 A · Q14

AMC10 2015 A · Q14. It mainly tests Circle theorems, Transformations.

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?

下图显示一个半径为 $20$ cm 的钟面，以及一个半径为 $10$ cm 的圆盘，该圆盘在 12 点位置与钟面外切。圆盘上画有一个箭头，初始时指向竖直向上的方向。让圆盘沿钟面顺时针滚动。问当箭头下一次再次指向竖直向上时，圆盘将与钟面在什么位置相切？

(A) 2 o'clock 2点

(B) 3 o'clock 3点

(D) 6 o'clock 6点

(E) 8 o'clock 8点

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The circumference of the disk is half the circumference of the clock face. As the disk rolls $\frac{1}{4}$ of the way around the circumference of the clock face (from 12 o’clock to 3 o’clock), the disk rolls through $\frac{1}{2}$ of its own circumference. At that point, the arrow of the disk is pointing at the point of tangency, so the arrow on the disk will have turned $\frac{3}{4}$ of one revolution. In general, as the disk rolls through an angle $\alpha$ around the clock face, the arrow on the disk turns through an angle $3\alpha$ on the disk. The arrow will again be pointing in the upward vertical direction when the disk has turned through $1$ complete revolution, and the angle traversed on the clock face is $\frac{1}{3}$ of the way around the face. The point of tangency will be at 4 o’clock.

答案（C）：圆盘的周长是钟面周长的一半。当圆盘沿钟面圆周滚动 $\frac{1}{4}$ 圈（从 12 点到 3 点）时，圆盘自身滚过了其周长的 $\frac{1}{2}$。此时，圆盘上的箭头指向切点，因此圆盘上的箭头已经转过了 $\frac{3}{4}$ 圈。一般地，当圆盘沿钟面转过角度 $\alpha$ 时，圆盘上的箭头在圆盘上转过角度 $3\alpha$。当圆盘转过整整 $1$ 圈时，箭头会再次指向竖直向上的方向，而在钟面上走过的角度是整圈的 $\frac{1}{3}$。切点将位于 4 点钟方向。

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