Question

两根长度相同但直径不同的长圆柱管彼此平行地放在平面上。较大的圆柱管半径为 $72$ ，沿平面向半径为 $24$ 的较小圆柱管滚动。它滚过较小圆柱管并继续沿平面滚动，直到它以圆周上的同一点着地而停下，已完成一整圈旋转。如果较小圆柱管始终不动，且滚动无滑动，则较大圆柱管最终离起点的距离为 $x$ 。距离 $x$ 可表示为 $a\pi + b\sqrt{c}$ ，其中 $a$ 、 $b$ 、 $c$ 为整数，且 $c$ 不被任何质数的平方整除。求 $a + b + c$ 。

Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24.$ It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance $x$ from where it starts. The distance $x$ can be expressed in the form $a\pi + b\sqrt{c},$ where $a,$ $b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a + b + c.$

答案

答案：179

知识点：弧圆周长特殊直角三角形

难度评级：3060

小提示：

当两根管相切且都接触地面时，连接它们圆心的线段长为 $96$ ，竖直分量为 $48$ 。

When the tubes touch while both rest on the ground, the segment joining their centers has length $96$ and vertical component $48$

大提示：

总转角为 $360^\circ$ ：越过小管时，大管的自转角等于圆心扫过的 $120^\circ$ 加上接触弧所贡献的转角，其余转角来自平地滚动。

Total turning is $360^\circ:$ while crossing the small tube, the big tube’s rotation equals the center’s $120^\circ$ sweep plus the angular contribution from the contact arc; the rest is flat rolling

解答：

当滚动的大管同时接触地面和小管时，两圆心之间的距离为 $72 + 24 = 96$ ，竖直分量为 $72 - 24 = 48$ ，所以该线段与水平线成 $30^\circ$ 角。大管滚过小管时，它的圆心沿着以小管圆心为圆心、半径 $96$ 的圆弧运动，从一侧水平线上方 $30^\circ$ 到另一侧 $30^\circ$ ，扫过 $120^\circ$ 。圆心的水平位移为 $2 \cdot 96\cos 30^\circ = 96\sqrt{3}$ 。

在这段扫过过程中，小管上的接触弧对应大管圆周的 $120^\circ \cdot \frac{24}{72} = 40^\circ$ ，而扫角本身也使大管转过 $120^\circ$ ，所以越过小管时大管总共转过 $160^\circ$ 。为了恰好完成一整圈，剩下的 $360^\circ - 160^\circ = 200^\circ$ 转角发生在平地滚动中，此时圆心前进的滚动距离为 $\frac{200}{360} \cdot 2\pi \cdot 72 = 80\pi$ 。

因此 $x = 80\pi + 96\sqrt{3}$ ，所以 $a + b + c = 80 + 96 + 3 = 179$ 。

When the rolling tube touches both the ground and the small tube, the segment between centers has length $72 + 24 = 96$ and vertical component $72 - 24 = 48,$ so it makes a $30^\circ$ angle with the horizontal. As the big tube rolls over the small one, its center swings along an arc of radius $96$ about the small tube’s center, from $30^\circ$ above the horizontal on one side to $30^\circ$ on the other: a sweep of $120^\circ,$ advancing the center horizontally by $2 \cdot 96\cos 30^\circ = 96\sqrt{3}.$

During that sweep, the contact arc on the small tube is $120^\circ \cdot \frac{24}{72} = 40^\circ$ worth of the big tube’s circumference, and the sweep itself also rotates the big tube by $120^\circ,$ so crossing the small tube turns the big tube by $160^\circ$ in all. To complete exactly one revolution, the remaining $360^\circ - 160^\circ = 200^\circ$ of turning happens rolling on flat ground, where the center advances the rolled distance $\frac{200}{360} \cdot 2\pi \cdot 72 = 80\pi.$

Hence $x = 80\pi + 96\sqrt{3},$ and $a + b + c = 80 + 96 + 3 = 179.$

2007 AIME II 第 11 题

答案：179

难度评级：3060

小提示：

大提示：

解答：

其他年份的第 11 题