2016 AMC 10B Problem 23

Problem 23 of 25HarderGeometry

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?

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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

ZW

2d.

Let the altitude from

P

ED

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

PED

\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 .