LIVE by Po-Shen Loh标识

LIVE

中文

2021 AMC 12B Fall 第 15 题

先试着解答 2021 AMC 12B Fall 第 15 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 2021 AMC 12B Fall 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。

15.

三张相同的边长为 66 的正方形纸片叠在一起。中间的纸片绕其中心顺时针旋转 30∘30^\circ，最上面的纸片绕其中心顺时针旋转 60∘60^\circ，得到下图所示的 2424 边形。该多边形面积可表示为 a−bca - b\sqrt{c}，其中 aa、bb、cc 为正整数，且 cc 不被任何质数的平方整除。求 a+b+ca + b + c。

Three identical square sheets of paper each with side length 66 are stacked on top of each other. The middle sheet is rotated clockwise 30∘30^\circ about its center and the top sheet is rotated clockwise 60∘60^\circ about its center, resulting in the 2424-sided polygon shown in the figure below. The area of this polygon can be expressed in the form a−bc,a - b\sqrt{c}, where a,a, b,b, and cc are positive integers, and cc is not divisible by the square of any prime. What is a+b+c?a + b + c?

7575

9393

9696

129129

147147

答案：E
知识点：面积分割三角学对称性
难度评级：2100
小提示：

由对称性，多边形的顶点每隔 15∘15^\circ 交替为外侧正方形顶点和内部边交点。

By symmetry the polygon’s vertices alternate between outer square-corners and inner edge-crossings, every 15∘15^\circ

大提示：

外侧顶点到中心的距离为 323\sqrt2，内部顶点到中心的距离为 232\sqrt3。

The outer vertices are at distance 323\sqrt2 from the center and the inner ones at distance 232\sqrt3

解答：

因为三个正方形分别旋转 0∘0^\circ，30∘30^\circ，60∘60^\circ，图形有 1212 重对称。它的 2424 个顶点每隔 15∘15^\circ 交替出现：外侧顶点到中心的距离为 323\sqrt2，内侧交点到中心的距离为 232\sqrt3。

连接中心与全部 2424 个顶点，把多边形分成 2424 个三角形，每个三角形的两边为 323\sqrt2 和 232\sqrt3，夹角为 15∘15^\circ。总面积是 24⋅12(32)(23)sin⁡15∘=726⋅6−24。 \begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4} \end{aligned}\text{。}

这化简为 186(6−2)=108−36318\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3，所以 a+b+c=108+36+3=147a + b + c = 108 + 36 + 3 = 147。

所以正确答案是 E。

Because the three squares are rotated by 0∘,0^\circ, 30∘,30^\circ, and 60∘,60^\circ, the figure has 1212-fold symmetry. Its 2424 vertices alternate every 15∘15^\circ: outer vertices are the square corners at distance 323\sqrt2 from the center, and inner vertices are edge crossings at distance 23.2\sqrt3.

Connecting the center to all 2424 vertices splits the polygon into 2424 triangles, each with sides 323\sqrt2 and 232\sqrt3 and included angle 15∘.15^\circ. The total area is 24⋅12(32)(23)sin⁡15∘=726⋅6−24. \begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}. \end{aligned}

This simplifies to 186(6−2)=108−363,18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3, so a+b+c=108+36+3=147.a + b + c = 108 + 36 + 3 = 147.

Thus, the correct answer is E.

第 14 题#14
完整试卷
第 16 题#16

其他年份的第 15 题

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12 · 1968 AMC 12 · 1969 AMC 12 · 1970 AMC 12 · 1971 AMC 12 · 1972 AMC 12 · 1973 AMC 12 · 1974 AMC 12 · 1975 AMC 12 · 1976 AMC 12 · 1977 AMC 12 · 1978 AMC 12 · 1979 AMC 12 · 1980 AMC 12 · 1981 AMC 12 · 1982 AMC 12 · 1983 AMC 12 · 1984 AMC 12 · 1985 AMC 12 · 1986 AMC 12 · 1987 AMC 12 · 1988 AMC 12 · 1989 AMC 12 · 1990 AMC 12 · 1991 AMC 12 · 1992 AMC 12 · 1993 AMC 12 · 1994 AMC 12 · 1995 AMC 12 · 1996 AMC 12 · 1997 AMC 12 · 1998 AMC 12 · 1999 AMC 12 · 2000 AMC 12 · 2001 AMC 12 · 2002 AMC 12A · 2002 AMC 12B · 2003 AMC 12A · 2003 AMC 12B · 2004 AMC 12A · 2004 AMC 12B · 2005 AMC 12A · 2005 AMC 12B · 2006 AMC 12A · 2006 AMC 12B · 2007 AMC 12A · 2007 AMC 12B · 2008 AMC 12A · 2008 AMC 12B · 2009 AMC 12A · 2009 AMC 12B · 2010 AMC 12A · 2010 AMC 12B · 2011 AMC 12A · 2011 AMC 12B · 2012 AMC 12A · 2012 AMC 12B · 2013 AMC 12A · 2013 AMC 12B · 2014 AMC 12A · 2014 AMC 12B · 2015 AMC 12A · 2015 AMC 12B · 2016 AMC 12A · 2016 AMC 12B · 2017 AMC 12A · 2017 AMC 12B · 2018 AMC 12A · 2018 AMC 12B · 2019 AMC 12A · 2019 AMC 12B · 2020 AMC 12A · 2020 AMC 12B · 2021 AMC 12A Spring · 2021 AMC 12B Spring · 2021 AMC 12A Fall · 2022 AMC 12A · 2022 AMC 12B · 2023 AMC 12A · 2023 AMC 12B · 2024 AMC 12A · 2024 AMC 12B · 2025 AMC 12A · 2025 AMC 12B