2016 AMC 10B Problem 23

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

Small hint:

Extend

\overline{FE}

and

\overline{CD}

until they meet

Big hint:

Compare the areas of three similar triangles with vertex at the extension point

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{\frac{11}{9}}{3}=\frac{11}{27}.

Thus, the correct answer is C .