2004 AMC 10A Problem 20

Let the square have side $1$, with $E$ on $\overline{AD}$ and $F$ on $\overline{CD}$, and put $AE = t$.

Right triangles $ABE$ and $CBF$ have equal legs $AB = CB = 1$ and equal hypotenuses $BE = BF$, so they are congruent (hypotenuse-leg). Hence $CF = t$ as well, and $DE = DF = 1 - t$.

Now express the equilateral side two ways. From right triangle $ABE$: $BE^2 = 1 + t^2$. From right isosceles triangle $DEF$: $EF^2 = 2(1-t)^2$. Setting them equal,
$$1 + t^2 = 2(1 - t)^2 = 2 - 4t + 2t^2 \quad\Longrightarrow\quad t^2 - 4t + 1 = 0.$$

We do not need $t$ itself. Rewrite the relation as $t^2 + 1 = 4t$. Then
$$(1 - t)^2 = t^2 + 1 - 2t = 4t - 2t = 2t.$$

The areas are $[DEF] = \tfrac12 (1-t)^2 = \tfrac12 \cdot 2t = t$ and $[ABE] = \tfrac12 \cdot 1 \cdot t = \tfrac{t}{2}$. Their ratio is
$$\frac{t}{t/2} = 2.$$

The answer is $\boxed{\textbf{(D)}\ 2}$.

The figure is symmetric about diagonal $BD$, which makes $\triangle DEF$ an isosceles right triangle and reduces everything to one unknown. The equilateral condition then becomes a single Pythagorean equation. The final trick is common on the AMC: when a ratio is requested, manipulate the defining relation to get the ratio directly rather than solving for $t = 2 - \sqrt3$ and substituting radicals.

Angle chase: symmetry gives $\angle ABE = \angle CBF = \tfrac{90^\circ - 60^\circ}{2} = 15^\circ$, so $t = \tan 15^\circ = 2 - \sqrt3$. Then $1 - t = \sqrt3 - 1$ and $(\sqrt3 - 1)^2 = 4 - 2\sqrt3 = 2(2 - \sqrt3) = 2t$, giving the same ratio $\frac{(1-t)^2/2}{t/2} = 2$.

• Expecting an answer involving sqrt(3) because the triangle is equilateral, and picking (C) or (E) without computing.
• Solving the quadratic and taking the root $t = 2 + \sqrt3$, which exceeds the side of the square.