2004 AMC 10A Problem 20

Question

Points

E

and

F

are located on square

ABCD

so that

\triangle BEF

is equilateral. What is the ratio of the area of

\triangle DEF

to that of

\triangle ABE?

\dfrac{4}{3}

\dfrac{3}{2}

\sqrt{3}

2

1 + \sqrt{3}

Let the square have side

1

and set

ED = DF = x

From

EF^2 = EB^2,

get

2x^2 = 1 + (1 - x)^2

Let the square have side

1,

and by symmetry let

ED = DF = x,

so

AE = 1 - x.

Since

\triangle BEF

is equilateral,

EF^2 = EB^2,

giving

2x^2 = 1 + (1 - x)^2,

which simplifies to

x^2 = 2(1 - x).

The right triangles have areas

[DEF] = \tfrac12 x^2

and

[ABE] = \tfrac12(1 - x),

so

\begin{aligned} \dfrac{[DEF]}{[ABE]} &= \dfrac{x^2}{1 - x} \\ &= \dfrac{2(1 - x)}{1 - x} = 2. \end{aligned}

Thus, the correct answer is D.

Problem 20 in other years