Question

两根直管（圆柱体）的半径分别为 $1$ 和 $\tfrac14$ ，它们平行放置并在平坦地面上相切。下图是正面视图。第三根平行管也放在同一地面上，并且同时与前两根相切。它的可能半径之和是多少？

Two straight pipes (circular cylinders), with radii $1$ and $\tfrac14,$ lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?

$\dfrac{1}{9}$

$1$

$\dfrac{10}{9}$

$\dfrac{11}{9}$

$\dfrac{19}{9}$

答案：C

知识点：相切圆勾股定理分类讨论

难度评级：2120

解答：

两个半径为 $R$ 和 $r$ 的圆靠在地面上并互相相切时，它们与地面的接触点水平距离为 $2\sqrt{Rr}$ 。所以半径 $1$ 和 $\tfrac14$ 的两根管接触地面的点相距 $2\sqrt{1 \cdot \tfrac14} = 1$ 。设第三根管半径为 $r$ 。它到大管接触点的距离为 $2\sqrt{r}$ ，到小管接触点的距离为 $2\sqrt{\tfrac14 r} = \sqrt{r}$ 。若夹在两者之间， $2\sqrt r + \sqrt r = 1$ ，所以 $\sqrt r = \tfrac13$ ， $r = \tfrac19$ 。若在小管外侧， $2\sqrt r - \sqrt r = 1$ ，所以 $r = 1$ 。（在大管外侧不可能。）半径之和为 $\tfrac19 + 1 = \tfrac{10}{9}$ 。因此正确答案是 C。

Two circles of radii $R$ and $r$ resting on the floor and touching each other have contact points a horizontal distance $2\sqrt{Rr}$ apart. So the radius- $1$ and radius- $\tfrac14$ pipes touch the floor $2\sqrt{1 \cdot \tfrac14} = 1$ apart. A third pipe of radius $r$ sits $2\sqrt{r}$ from the big pipe's contact point and $2\sqrt{\tfrac14 r} = \sqrt{r}$ from the small pipe's. Nestled between them, $2\sqrt r + \sqrt r = 1,$ so $\sqrt r = \tfrac13$ and $r = \tfrac19.$ Sitting past the small pipe, $2\sqrt r - \sqrt r = 1,$ so $r = 1.$ (Past the big pipe can't happen.) The sum is $\tfrac19 + 1 = \tfrac{10}{9}.$ Thus, C is the correct answer.

2024 AMC 10B 第 21 题

答案：C

难度评级：2120

解答：

其他年份的第 21 题