2023 AMC 12B 第 21 题

先试着解答 2023 AMC 12B 第 21 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2023 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

21.

一个灯罩的形状是直圆锥台的侧面。圆锥台的高为 333\sqrt{3} 英寸,上底直径为 66 英寸,下底直径为 1212 英寸。一只虫子在灯罩底边上,灯罩上边缘离虫子最远的位置有一团 蜂蜜。虫子想爬到蜂蜜处,但必须留在灯罩表面上。它到蜂蜜的最短路径长度是多少英寸?
A lampshade is made in the form of the lateral surface of the frustum of a right circular cone. The height of the frustum is 333\sqrt{3} inches, its top diameter is 66 inches, and its bottom diameter is 1212 inches. A bug is at the bottom of the lampshade and there is a glob of honey on the top edge of the lampshade at the spot farthest from the bug. The bug wants to crawl to the honey, but it must stay on the surface of the lampshade. What is the length in inches of its shortest path to the honey?
6+3π6+3\pi
6+6π6+6\pi
636\sqrt{3}
656\sqrt{5}
63+π6\sqrt{3}+\pi
答案:E
知识点:圆锥展开图(立体几何)切线
小提示:
将圆锥台延伸成完整圆锥(到顶点的斜距为 66 和 1212);展开后是角度为 π\pi 的扇形
Extend the frustum to a full cone (apex slant distances 66 and 1212); it unrolls to a sector of angle π\pi
大提示:
到蜂蜜的直线会穿过缺失的顶部小圆锥,因此路径先与半径 66 的内圆相切,再沿它的弧走
The straight line to the honey cuts across the missing top cone, so the path runs tangent to the inner circle of radius 6,6, then along its arc
解答:
将圆锥台延伸成完整圆锥。由于半径为 33 和 66,斜高带宽为 66,顶点到上边缘的斜距为 66,到下边缘的斜距为 1212。下底周长 12π12\pi 展开成半径 1212、角度 12π12=π\dfrac{12\pi}{12}=\pi 的扇形。在这个展开图中把虫子放在 (12,0)(12,0);蜂蜜位于沿底边走半圈的位置,在展开图中半径为 66、角度为 π2\tfrac{\pi}{2}。连接它们的直弦会进入半径 66 以内(不在表面上),所以测地线与半径 66 的圆相切:切线长为 122−62=63\sqrt{12^2-6^2}=6\sqrt3,切点角度为 π3\tfrac{\pi}{3},之后沿半径 66 的圆弧走过角度 π6\tfrac{\pi}{6},弧长为 6⋅π6=π6\cdot\tfrac{\pi}{6}=\pi。最短路径为 63+π6\sqrt3+\pi。
因此,正确答案是 E。
Extend the frustum to a full cone. Since the radii are 33 and 66 with slant band 6,6, the apex is slant distance 66 from the top rim and 1212 from the bottom rim. The bottom circumference 12π12\pi unrolls to a sector of radius 1212 and angle 12π12=π.\dfrac{12\pi}{12}=\pi. Place the bug at (12,0)(12,0) in this pattern; the honey, halfway around the base, is at radius 66 and angle π2.\tfrac{\pi}{2}. The straight chord between them passes within radius 66 (off the surface), so the geodesic goes tangent to the circle of radius 6:6: the tangent has length 122−62=63\sqrt{12^2-6^2}=6\sqrt3 and touches at angle π3,\tfrac{\pi}{3}, after which the path follows the arc of angle π6\tfrac{\pi}{6} on radius 6,6, of length 6⋅π6=π.6\cdot\tfrac{\pi}{6}=\pi. The shortest path is 63+π.6\sqrt3+\pi.
Thus, the correct answer is E.
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