AMC10 2023 B

AMC10 2023 B · Q20

AMC10 2023 B · Q20. It mainly tests Circle theorems, Geometry misc.

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

在半径为$2$的球体表面上绘制了四个全等的半圆，如图所示，形成一个闭合曲线，将表面分成两个全等区域。该曲线的长度为$\pi\sqrt{n}$。求$n$？

(A) 32 32

(B) 12 12

(D) 36 36

(E) 27 27

Answer

Correct choice: (A)

正确答案：（A）

Solution

Focus on 2 of the points. Let the center of the Sphere be A. Label two points that form the diameter of one of the four semicircles M and C respectively. Triangle AMC is a right triangle through the inscribed right triangle theorem, with AM=AC=2. This is a 45-45-90 triangle, so the length of the diameter MC is just the hypotenuse of the triangle AMC, which is 22. This means the radius is 2. The circumference of the semicircle is 12⋅22⋅π=π2, and because there are four congruent semicircles, the length of the semicircular region is 42π=π⋅42. The solution is in the form πn, so we convert π⋅42 to π162 to get π32. n=$\boxed{\textbf{(A) 32}}$.

关注球体中心的$2$个点。设球心为$A$。标记形成四个半圆之一的直径的两个点分别为$M$和$C$。三角形$AMC$根据内接直角三角定理是直角三角形，且$AM=AC=2$。这是一个$45-45-90$三角形，因此直径$MC$的长度是三角形$AMC$的斜边，为$2\sqrt{2}$。这意味着半径为$\sqrt{2}$。半圆的周长为$\frac{1}{2}\cdot2\pi\sqrt{2}=\pi\sqrt{2}$，因为有四个全等的半圆，半圆区域的长度为$4\pi\sqrt{2}=\pi\cdot4\sqrt{2}$。解的形式为$\pi\sqrt{n}$，所以将$\pi\cdot4\sqrt{2}$转换为$\pi\sqrt{32}$。$n=\boxed{\textbf{(A) 32}}$。

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