Question

如图，在半径为 $2$ 的球面上画出四个全等半圆，形成一条闭合曲线，将球面分成两个全等区域。该曲线长度为 $\pi\sqrt{n}$ 。求 $n$ 。

Four congruent semicircles are drawn on the surface of a sphere with radius $2,$ as shown, creating a closed curve that divides the surface into two congruent regions. The length of the curve is $\pi\sqrt{n}.$ What is $n?$

$32$

$12$

$48$

$36$

$27$

答案：A

知识点：球弧立体几何

难度评级：2100

解答：

这条曲线由四段全等的半圆弧组成。若弧所在圆的半径为 $r$ ，则总长度是单段半圆弧长 $\pi r$ 的 $4$ 倍。四段弧相交于四个点，这四点构成一个内接于球面大圆的正方形；每段弧的直径就是该正方形的一条边。球半径为 $2$ ，所以正方形边长为 $2\sqrt2$ ，从而 $r = \sqrt2$ 。（也可以这样验证：小圆所在平面到球心的距离为 $\frac{2}{\sqrt2} = \sqrt2$ ，所以小圆半径为 $\sqrt{2^2 - (\sqrt2)^2} = \sqrt2$ 。）因此总长度为 $4 \cdot \pi\sqrt2 = \pi\sqrt{32}$ ，所以 $n = 32$ 。正确答案是 A。

The curve is four congruent semicircular arcs, so its length is $4$ times one semicircle, $\pi r,$ where $r$ is the arc radius. The arcs meet at four points that form a square inscribed in a great circle of the radius- $2$ sphere, and each arc's diameter is a side of that square, a chord of length $2\sqrt2.$ So $r = \sqrt2.$ (Check it another way: the small circle sits in a plane at distance $\frac{2}{\sqrt2} = \sqrt2$ from the center, giving radius $\sqrt{2^2 - (\sqrt2)^2} = \sqrt2.$ ) The total length is $4 \cdot \pi\sqrt2 = \pi\sqrt{32},$ so $n = 32.$ Therefore, the answer is A.

2023 AMC 10B 第 20 题

答案：A

难度评级：2100

解答：

其他年份的第 20 题