Question

圆 $A$ 的半径为 $100$ 。圆 $B$ 的半径 $r \lt 100$ 是整数，并且在圆 $A$ 内部相切地沿圆 $A$ 的圆周滚动一圈。圆 $B$ 旅程开始和结束时，两圆有相同的切点。 $r$ 可能有多少个值？

Circle $A$ has radius $100.$ Circle $B$ has an integer radius $r \lt 100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A.$ The two circles have the same points of tangency at the beginning and end of circle $B$ 's trip. How many possible values can $r$ have?

$4$

$8$

$9$

$50$

$90$

答案：B

知识点：圆周长整除性因数个数

难度评级：1630

解答：

两个圆的周长分别为 $200\pi$ 和 $2\pi r$ ，所以初始切点在 $\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}$ 圈后回到原处。

为使它是大于 $1$ 的整数， $r$ 必须是小于 $100$ 的 $100$ 的因数： $1, 2, 4, 5, 10, 20, 25$ ，和 $50$ ，共有 $8$ 个值。

所以正确答案是 B。

The circumferences are $200\pi$ and $2\pi r,$ so the initial point of tangency returns after $\dfrac{200\pi}{2\pi r} = \dfrac{100}{r}$ rolls.

For this to be an integer greater than $1,$ $r$ must be a divisor of $100$ less than $100:$ namely $1, 2, 4, 5, 10, 20, 25,$ and $50.$ That is $8$ values.

Thus, the correct answer is B.

2009 AMC 10A 第 19 题

答案：B

难度评级：1630

解答：

其他年份的第 19 题