AMC12 2021 B

AMC12 2021 B · Q8

AMC12 2021 B · Q8. It mainly tests Pythagorean theorem, Circle theorems.

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?

三条等间距的平行线与一个圆相交，形成三条弦，长分别为$38,38,$和$34$。两条相邻平行线之间的距离是多少？

(A) 5\frac12 5\frac12

(B) 6 6

(D) 7 7

(E) 7\frac12 7\frac12

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is \[2d + d = 3d\] The distance between each of the chords is $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we obtain two different right triangles: - One with base $\dfrac{38}{2} = 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ in the diagram) - Another with base $\dfrac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ in the diagram) By the Pythagorean theorem, we obtain the following system of equations: \begin{align*} 19^2 + d^2 &= r^2, \\ 17^2 + (3d)^2 &= r^2. \end{align*} That is, \begin{align*} 361 + d^2 &= r^2, \\ 289 + 9d^2 &= r^2. \end{align*} Set the right-hand sides equal: \[ 361 + d^2 = 289 + 9d^2 \] \[ 361 - 289 = 9d^2 - d^2 \] \[ 72 = 8d^2 \] \[ d^2 = 9 \implies d = 3. \] Thus, \[ 2d = 6. \]

由于两条平行弦长度相等（$38$），它们必须距离圆心等距。设每条弦距离圆心的垂直距离为$d$。因此，长度为$34$的弦距离圆心的距离为 \[2d + d = 3d\] 各弦之间的距离为$2d$。设圆的半径为$r$。画出到交点处的半径，得到两个不同的直角三角形： - 一个底边$\dfrac{38}{2} = 19$，高$d$，斜边$r$（图中$\triangle RAO$） - 另一个底边$\dfrac{34}{2} = 17$，高$3d$，斜边$r$（图中$\triangle LBO$）由勾股定理，得到方程组： \begin{align*} 19^2 + d^2 &= r^2, \\ 17^2 + (3d)^2 &= r^2. \end{align*} 即， \begin{align*} 361 + d^2 &= r^2, \\ 289 + 9d^2 &= r^2. \end{align*} 令右边相等： \[361 + d^2 = 289 + 9d^2\] \[361 - 289 = 9d^2 - d^2\] \[72 = 8d^2\] \[d^2 = 9 \implies d = 3.\] 因此， \[2d = 6.\]

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