2006 AMC 10B Problem 15

Let the side of $ABCD$ be $s$. Since $\angle BAD = 60^\circ$ and $AB = AD$, triangle $ABD$ is equilateral, and so is triangle $CBD$. Hence the short diagonal is $BD = s$, and the long diagonal $AC$ is twice the altitude of an equilateral triangle: $AC = 2 \cdot \frac{s\sqrt3}{2} = s\sqrt3$.

Now look at rhombus $BFDE$. Its angles are also $60^\circ$ and $120^\circ$ (it is similar to $ABCD$). In the figure $E$ and $F$ sit close to segment $BD$, so the angles at $E$ and $F$ are the obtuse $120^\circ$ ones and the angles at $B$ and $D$ are $60^\circ$. The diagonal joining the two acute vertices of a rhombus is the longer one, so $BD$ is the long diagonal of $BFDE$.

Matching long diagonal to long diagonal, the similarity ratio is
$$\frac{BD}{AC} = \frac{s}{s\sqrt3} = \frac{1}{\sqrt3},$$
so the areas are in ratio $\left(\frac{1}{\sqrt3}\right)^2 = \frac{1}{3}$. The area of $BFDE$ is $\frac{24}{3} = 8$.

The answer is $\boxed{\textbf{(C)}\ 8}$.

A $60^\circ$ rhombus is two equilateral triangles glued along the short diagonal, which fixes the diagonal ratio $1 : \sqrt3$ with no coordinates. Similar figures have areas in the square of any pair of corresponding lengths; the only care needed is to identify which diagonal of the small rhombus corresponds to which of the large one.

Triangle $EBD$ is isosceles with apex angle $120^\circ$ on base $BD$, so its apex $E$ is the center of equilateral triangle $ABD$ and $[EBD] = \frac{1}{3}[ABD] = \frac{1}{3} \cdot 12 = 4$. Likewise $[FBD] = 4$, so $[BFDE] = 8$.

• Comparing BD with itself or with a side, instead of matching corresponding diagonals (long to long) of the two similar rhombi.
• Using the ratio $1/\sqrt3$ directly on the area (answer $8\sqrt3$, not offered) instead of squaring it.