Question

任意一个圆与函数 $y=\sin x$ 的图像的交点数可以是

An arbitrary circle can intersect the graph of $y=\sin x$ in

至多 $2$ 个点

at most $2$ points

至多 $4$ 个点

at most $4$ points

至多 $6$ 个点

at most $6$ points

至多 $8$ 个点

at most $8$ points

超过 $16$ 个点

more than $16$ points

答案：E

知识点：圆交点计数三角学

难度评级：2240

小提示：

考虑一个在原点与 $x$ 轴相切的圆

Consider a circle tangent to the $x$ -axis at the origin

大提示：

半径非常大时，在许多个正弦波范围内，圆的下弧都靠近坐标轴

A very large radius keeps the lower arc close to the axis across many sine waves

解答：

取一个在原点与 $x$ 轴相切、圆心位于坐标轴上方很高处的圆。随着半径增大，在任意长的区间内，它的下弧都保持为正且可以任意接近 $x$ 轴。在 $y=\sin x$ 的每个正弧上，正弦曲线在两端为 $0$ ，而在中间远高于这段圆弧，从而产生两个交点。因此，选取足够大的半径便可得到超过 $16$ 个交点。

所以正确答案是 E。

Take a circle tangent to the $x$ -axis at the origin with its center high above the axis. As its radius grows, its lower arc stays positive but arbitrarily close to the $x$ -axis over an arbitrarily long interval. On each positive arch of $y=\sin x,$ the sine curve is $0$ at the endpoints and rises well above this circular arc in between, producing two crossings. Choosing a sufficiently large radius therefore produces more than $16$ intersections.

Thus the correct answer is E.

1985 AMC 12 第 10 题

答案：E

难度评级：2240

小提示：

大提示：

解答：

其他年份的第 10 题