Question

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 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 $ABCDEF?$

$\ \dfrac{1}{3}$

$\ \dfrac{10}{27}$

$\ \dfrac{11}{27}$

$\ \dfrac{4}{9}$

$\ \dfrac{13}{27}$

Answer: C

Concepts:regular polygon area ratio similarity

Difficulty rating: 2300

Solution:

Extend $\overline{FE}$ and $\overline{CD}$ until they meet at $P.$

Let $d$ be the distance between $ZW$ and $FC.$ Because the four given lines are equally spaced and $FC$ lies halfway between $ZW$ and $YX,$ the distance from $ED$ to $ZW$ is $2d.$ Let the altitude from $P$ to $ED$ be $h.$ The equilateral triangles $PED$ and $PFC$ have side lengths in the ratio $1:2,$ so their altitudes satisfy $h+3d=2h,$ giving $h=3d.$ Therefore the side-length ratio of $PZW$ to $PED$ is $\frac{h+2d}{h}=\frac53.$

Taking $[PED]=1,$ similarity gives $[PZW]=\frac{25}{9}$ and $[PFC]=4.$ Hence $\begin{aligned} [ZWCF]&=4-\frac{25}{9}=\frac{11}{9}, \\ [EDCF]&=4-1=3. \end{aligned}$ The desired hexagon consists of two congruent copies of $ZWCF,$ while the regular hexagon consists of two congruent copies of $EDCF.$ Thus the requested ratio is $\frac{11/9}{3}=\frac{11}{27}.$

Thus, the correct answer is C .

2016 AMC 10B Problem 23

Answer: C

Difficulty rating: 2300

Solution:

Problem 23 in Other Years