2013 AMC 12A Problem 11

Question

Triangle

ABC

is equilateral with

AB = 1.

Points

E

and

G

are on

\overline{AC}

and points

D

and

F

are on

\overline{AB}

such that both

\overline{DE}

and

\overline{FG}

are parallel to

\overline{BC}.

Furthermore, triangle

ADE

and trapezoids

DFGE

and

FBCG

all have the same perimeter. What is

DE + FG?

1

\dfrac{3}{2}

\dfrac{21}{13}

\dfrac{13}{8}

\dfrac{5}{3}

Let

x = DE

and

y = FG,

and write each of the three perimeters in terms of

x

and

y

The three equal perimeters give

3x = 3y - x = 3 - y

Let

x = DE

and

y = FG.

The parallel cuts make the small regions equilateral or isosceles trapezoids, so the perimeters are

\begin{gathered} \triangle ADE: 3x, \\ \quad DFGE: 3y - x, \\ \quad FBCG: 3 - y. \end{gathered}

Setting them equal,

3x = 3y - x

gives

4x = 3y,

and

3x = 3 - y.

Solving yields

x = \tfrac{9}{13}

and

y = \tfrac{12}{13},

so

DE + FG = \tfrac{21}{13}.

Thus, the correct answer is C.

Problem 11 in other years