AMC12 2018 B

AMC12 2018 B · Q20

AMC12 2018 B · Q20. It mainly tests Triangles (properties), Area & perimeter.

Let $ABCDEF$ be a regular hexagon with side length 1. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline{AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?

设 $ABCDEF$ 是一个边长为 1 的正六边形。分别用 $X$、$Y$ 和 $Z$ 表示边 $\overline{AB}$、$\overline{CD}$ 和 $\overline{EF}$ 的中点。$ riangle ACE$ 和 $ riangle XYZ$ 的内部交集内部形成的凸六边形的面积是多少？

(A) $\frac{3}{8}\sqrt{3}$ $\frac{3}{8}\sqrt{3}$

(B) $\frac{7}{16}\sqrt{3}$ $\frac{7}{16}\sqrt{3}$

(D) $\frac{1}{2}\sqrt{3}$ $\frac{1}{2}\sqrt{3}$

(E) $\frac{9}{16}\sqrt{3}$ $\frac{9}{16}\sqrt{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $O$ be the center of the regular hexagon. Points $B,O,E$ are collinear and $BE=BO+OE=2$. Trapezoid $FABE$ is isosceles, and $\overline{XZ}$ is its midline. Hence $XZ=\frac{3}{2}$ and analogously $XY=ZY=\frac{3}{2}$. Denote by $U_1$ the intersection of $\overline{AC}$ and $\overline{XZ}$ and by $U_2$ the intersection of $\overline{AC}$ and $\overline{XY}$. It is easy to see that $\triangle AXU_1$ and $\triangle U_2XU_1$ are congruent $30-60-90^\circ$ right triangles. By symmetry the area of the convex hexagon enclosed by the intersection of $\triangle ACE$ and $\triangle XYZ$, shaded in the figure, is equal to the area of $\triangle XYZ$ minus $3$ times the area of $\triangle U_2XU_1$. The hypotenuse of $\triangle U_2XU_1$ is $XU_2=AX=\frac{1}{2}$, so the area of $\triangle U_2XU_1$ is $$\frac{1}{2}\cdot\frac{\sqrt{3}}{4}\cdot\left(\frac{1}{2}\right)^2=\frac{1}{32}\sqrt{3}.$$ The area of the equilateral triangle $XYZ$ with side length $\frac{3}{2}$ is equal to $\frac{1}{4}\sqrt{3}\cdot\left(\frac{3}{2}\right)^2=\frac{9}{16}\sqrt{3}$. Hence the area of the shaded hexagon is $$\frac{9}{16}\sqrt{3}-3\cdot\frac{1}{32}\sqrt{3}=3\sqrt{3}\left(\frac{3}{16}-\frac{1}{32}\right)=\frac{15}{32}\sqrt{3}.$$

答案（C）：设 $O$ 为正六边形的中心。点 $B,O,E$ 共线，且 $BE=BO+OE=2$。梯形 $FABE$ 为等腰梯形，$\overline{XZ}$ 是其中位线。因此 $XZ=\frac{3}{2}$，同理 $XY=ZY=\frac{3}{2}$。记 $U_1$ 为 $\overline{AC}$ 与 $\overline{XZ}$ 的交点，$U_2$ 为 $\overline{AC}$ 与 $\overline{XY}$ 的交点。不难看出 $\triangle AXU_1$ 与 $\triangle U_2XU_1$ 是全等的 $30-60-90^\circ$ 直角三角形。由对称性，图中阴影部分（由 $\triangle ACE$ 与 $\triangle XYZ$ 的交集所围成的凸六边形）的面积，等于 $\triangle XYZ$ 的面积减去 $3$ 个 $\triangle U_2XU_1$ 的面积。$\triangle U_2XU_1$ 的斜边为 $XU_2=AX=\frac{1}{2}$，因此 $$\triangle U_2XU_1\text{ 的面积 }=\frac{1}{2}\cdot\frac{\sqrt{3}}{4}\cdot\left(\frac{1}{2}\right)^2=\frac{1}{32}\sqrt{3}.$$ 边长为 $\frac{3}{2}$ 的等边三角形 $XYZ$ 的面积为 $\frac{1}{4}\sqrt{3}\cdot\left(\frac{3}{2}\right)^2=\frac{9}{16}\sqrt{3}$。因此阴影六边形的面积为 $$\frac{9}{16}\sqrt{3}-3\cdot\frac{1}{32}\sqrt{3}=3\sqrt{3}\left(\frac{3}{16}-\frac{1}{32}\right)=\frac{15}{32}\sqrt{3}.$$

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