Question

在一个单位正方形的每条边上，向外作一个边长为 $1$ 的等边三角形。在每个等边三角形的新边上，再作一个边长为 $1$ 的等边三角形。正方形和这 $12$ 个三角形的内部没有公共点。令 $R$ 为正方形和所有三角形的并集，令 $S$ 为包含 $R$ 的最小凸多边形。位于 $S$ 内部但在 $R$ 外部的区域面积是多少？

On each side of a unit square, an equilateral triangle of side length $1$ is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length $1$ is constructed. The interiors of the square and the $12$ triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and let $S$ be the smallest convex polygon that contains $R.$ What is the area of the region that is inside $S$ but outside $R?$

$\dfrac{1}{4}$

$\dfrac{\sqrt{2}}{4}$

$1$

$\sqrt{3}$

$2\sqrt{3}$

答案：C

知识点：三角形面积导角三角学

难度评级：1660

解答：

凸包 $S$ 与 $R$ 的差只出现在正方形的四个角附近，每处形成一个小三角形空隙。每个空隙三角形有两条长度为 $1$ 的边，也就是相邻三角形的外边。

这两条边之间的夹角为 $360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ$ ，所以每个空隙面积为 $\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14.$

总面积为 $4 \cdot \tfrac14 = 1$ 。

因此，正确答案是 C。

The convex hull $S$ differs from $R$ only near the four corners of the square, where a small triangular gap forms. Each gap triangle has two sides of length $1$ (outer edges of adjacent triangles).

The angle between those two sides is $360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ,$ so each gap has area $\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14.$

The total area is $4 \cdot \tfrac14 = 1.$

Thus, the correct answer is C.

2008 AMC 12B 第 15 题

答案：C

难度评级：1660

解答：

其他年份的第 15 题