2004 AMC 10A Problem 20

1. Let the square have side $1$. Point $E$ is on side $AD$ and $F$ is on side $CD$. Call $AE = t$.
2. The picture is symmetric. Triangles $ABE$ and $CBF$ both have a right angle, a leg of length $1$ (sides $AB$ and $CB$), and equal long sides ($BE = BF$, since the triangle $BEF$ is equilateral). Two right triangles with a matching leg and matching long side are the same size and shape. So $CF = t$ too.
3. That means $DE = 1 - t$ and $DF = 1 - t$, so triangle $DEF$ is a right triangle with two equal legs.
4. Use the equilateral triangle. Its sides $BE$ and $EF$ are equal, so their squares are equal. Use the Pythagorean theorem on each:
- In triangle $ABE$: $BE^2 = 1^2 + t^2 = 1 + t^2$.
- In triangle $DEF$: $EF^2 = (1-t)^2 + (1-t)^2 = 2(1-t)^2$.
5. Set them equal and simplify:
$$1 + t^2 = 2(1 - 2t + t^2) = 2 - 4t + 2t^2 \quad\Longrightarrow\quad t^2 + 1 = 4t.$$
6. We want the areas, and area of $DEF$ uses $(1-t)^2$. Expand it and use step 5:
$$(1 - t)^2 = t^2 + 1 - 2t = 4t - 2t = 2t.$$
7. Now the two areas are
$$\text{area of } DEF = \tfrac12 (1-t)^2 = \tfrac12 \times 2t = t, \qquad \text{area of } ABE = \tfrac12 \times 1 \times t = \tfrac{t}{2}.$$
8. Their ratio is $t \div \tfrac{t}{2} = 2$.

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

The figure is symmetric about the diagonal $BD$, which leaves just one unknown length. The equal sides of the equilateral triangle give one equation. Then, instead of solving for $t$ (which is $2 - \sqrt3$), use the equation to simplify the ratio directly. When a problem asks for a ratio, the messy numbers often cancel if you let them.

Use angles. The angle at $B$ in the square is $90^\circ$ and the equilateral triangle uses $60^\circ$ of it, leaving $30^\circ$, split equally by symmetry: $15^\circ$ on each side. So $t = \tan 15^\circ = 2 - \sqrt3$. Then $1 - t = \sqrt3 - 1$ and
$$(\sqrt3 - 1)^2 = 3 - 2\sqrt3 + 1 = 4 - 2\sqrt3 = 2(2 - \sqrt3) = 2t,$$
which is the same fact as step 6, and the ratio is again $2$.

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 $\boxed{\textbf{(D)}\ 2}$.