Question

三条等距平行线与一个圆相交，形成三条长度分别为 $38,38$ , $34$ 的弦。相邻两条平行线之间的距离是多少？

Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34.$ What is the distance between two adjacent parallel lines?

$5\frac12$

$6$

$6\frac12$

$7$

$7\frac12$

答案：B

知识点：弦圆勾股定理

难度评级：1540

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文字解答：

两条长度为 $38$ 的弦到圆心的距离相等。它们不能位于两条最外侧的直线上：否则圆心会在中间直线上，该线所截的弦会是直径，长度大于 $38,$ 而不是给定的 $34.$ 因此两条相等弦位于相邻直线上，圆心在两线中间。设这个半间距为 $d.$ 则每条长 $38$ 的弦到圆心的距离是 $d$ ，长 $34$ 的弦到圆心的距离是 $3d$ 。

若圆的半径为 $r,$ 则

$r^2=19^2+d^2=17^2+(3d)^2.$

因此 $19^2-17^2=8d^2,$ 所以 $72=8d^2,$ 得到 $d=3.$ 相邻平行线的距离是 $2d=6.$

所以答案是 B。

The two chords of length $38$ are equally far from the center of the circle. They cannot lie on the two outer lines: then the center would lie on the middle line, whose chord would be a diameter longer than $38,$ not the given $34.$ Therefore, the equal chords lie on adjacent lines, with the center halfway between them. Let that half-distance be $d.$ Then each $38$ -chord is distance $d$ from the center, and the $34$ -chord is distance $3d$ from the center.

If the circle has radius $r,$ then

$r^2=19^2+d^2=17^2+(3d)^2.$

Thus $19^2-17^2=8d^2,$ so $72=8d^2,$ and $d=3.$ The distance between adjacent parallel lines is $2d=6.$

Thus, the answer is B .

2021 AMC 10B Spring 第 14 题

答案：B

难度评级：1540

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