2004 AMC 10A Problem 20

Problem 20 of 25HarderGeometry

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?

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Solution

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.

More practice

Concepts: equilateral triangle · area ratio · Pythagorean Theorem

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.