AMC12 2008 B

AMC12 2008 B · Q15

AMC12 2008 B · Q15. It mainly tests Triangles (properties), Area & perimeter.

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 $S$ be the smallest convex polygon that contains $R$. What is the area of the region that is inside $S$ but outside $R$?

在单位正方形的每条边上构造一个边长为1的正三角形。在每个正三角形的新边上再构造另一个边长为1的正三角形。正方形和12个三角形的内部没有公共点。令 $R$ 为正方形与所有三角形的并集所形成的区域，$S$ 为包含 $R$ 的最小凸多边形。$S$ 内而 $R$ 外的区域面积是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

(B) $\frac{\sqrt{2}}{4}$ $\frac{\sqrt{2}}{4}$

(D) $\sqrt{3}$ $\sqrt{3}$

(E) $2\sqrt{3}$ $2\sqrt{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is the area of $S - R$) is, $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$ which makes the answer $\boxed{C}$.

这些正三角形在单位正方形的每一边上形成边长为 $1, 1, 1, 2$ 的梯形（一个单位正六边形的一半）。在这些梯形“之间”的四个三角形是等腰三角形，两腰长均为 $1$，夹角为 $30^{\circ}$，因此这些三角形的总面积（即 $S - R$ 的面积）为 $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$，故答案为 $\boxed{C}$。

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