AMC10 2011 A

AMC10 2011 A · Q20

AMC10 2011 A · Q20. It mainly tests Probability (basic), Geometry misc.

Two points on the circumference of a circle of radius $r$ are selected independently and at random. From each point a chord of length $r$ is drawn in a clockwise direction. What is the probability that the two chords intersect?

在一个半径为 $r$ 的圆周上独立随机选取两点。从每点沿顺时针方向画一条长度为 $r$ 的弦。两条弦相交的概率是多少？

(A) $\frac{1}{6}$ $\frac{1}{6}$

(B) $\frac{1}{5}$ $\frac{1}{5}$

(D) $\frac{1}{3}$ $\frac{1}{3}$

(E) $\frac{1}{2}$ $\frac{1}{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let point $A$ be the first point chosen, and let point $B$ be the opposite endpoint of the corresponding chord. Drawing a radius to each endpoint of this chord of length $r$ results in an equilateral triangle. Hence a chord of length $r$ subtends an arc $\frac{1}{6}$ the circumference of the circle. Let diameter $FC$ be parallel to $AB$, and divide the circle into six equal portions as shown. The second point chosen will result in a chord that intersects $AB$ if and only if the point is chosen from minor $\widehat{FB}$. Hence the probability is $\frac{1}{3}$.

答案（D）：设点 $A$ 为第一个选取的点，点 $B$ 为对应弦的另一个端点（对端点）。向这条长度为 $r$ 的弦的两个端点分别作半径，可得到一个等边三角形。因此，长度为 $r$ 的弦所对的弧长是圆周的 $\frac{1}{6}$。令直径 $FC$ 与 $AB$ 平行，并如图将圆分成六个相等部分。第二个选取的点所确定的弦与 $AB$ 相交当且仅当该点取自小弧 $\widehat{FB}$。因此概率为 $\frac{1}{3}$。

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