AMC10 2016 B

AMC10 2016 B · Q23

AMC10 2016 B · Q23. It mainly tests Area & perimeter, Polygons.

In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$, respectively, so that lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?

在正六边形 $ABCDEF$ 中，点 $W$、$X$、$Y$、$Z$ 分别取在边 $\overline{BC}$、$\overline{CD}$、$\overline{EF}$、$\overline{FA}$ 上，使得直线 $AB$、$ZW$、$YX$ 和 $ED$ 互相平行且等距。求六边形 $WCXYFZ$ 的面积与六边形 $ABCDEF$ 的面积之比。

(A) $\frac{1}{3}$ $\frac{1}{3}$

(B) $\frac{10}{27}$ $\frac{10}{27}$

(D) $\frac{4}{9}$ $\frac{4}{9}$

(E) $\frac{13}{27}$ $\frac{13}{27}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Extend sides $\overline{CB}$ and $\overline{FA}$ to meet at $G$. Note that $FC=2AB$ and $ZW=\frac{5}{3}AB$. Then the areas of $\triangle BAG$, $\triangle WZG$, and $\triangle CFG$ are in the ratio $1^2:\left(\frac{5}{3}\right)^2:2^2=9:25:36$. Thus $\frac{[ZWCF]}{[ABCF]}=\frac{36-25}{36-9}=\frac{11}{27}$, and by symmetry, $\frac{[WCXYFZ]}{[ABCDEF]}=\frac{11}{27}$ also.

答案（C）：延长边 $\overline{CB}$ 和 $\overline{FA}$，使其相交于 $G$。注意 $FC=2AB$ 且 $ZW=\frac{5}{3}AB$。则 $\triangle BAG$、$\triangle WZG$ 与 $\triangle CFG$ 的面积之比为 $1^2:\left(\frac{5}{3}\right)^2:2^2=9:25:36$。因此 $\frac{[ZWCF]}{[ABCF]}=\frac{36-25}{36-9}=\frac{11}{27}$，并且由对称性可知，$\frac{[WCXYFZ]}{[ABCDEF]}=\frac{11}{27}$ 也成立。

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