2006 AMC 10B Problem 15

1. The big rhombus. Let each side be $s$. Since $AB = AD$ and the angle at $A$ is $60^\circ$, triangle $ABD$ has two equal sides with a $60^\circ$ angle between them, so it is equilateral. The same is true of triangle $CBD$. So the big rhombus is two equilateral triangles glued together along $BD$, and $BD = s$.
2. The other diagonal $AC$ goes across both triangles through their heights. The height of an equilateral triangle with side $s$ is $\frac{\sqrt3}{2}s$, so
$$AC = 2 \cdot \frac{\sqrt3}{2}s = \sqrt3\, s.$$
So $BD$ is the short diagonal and $AC$ is the long one.
3. The small rhombus. It is similar to the big one, so it also has angles $60^\circ$ and $120^\circ$. In the figure, $E$ and $F$ sit close to the segment $BD$, so the wide $120^\circ$ angles are at $E$ and $F$ and the narrow $60^\circ$ angles are at $B$ and $D$. The diagonal joining the two narrow corners is the long one, so $BD$ is the long diagonal of $BFDE$.
4. Compare matching lengths. Long diagonal to long diagonal:
$$\frac{BD}{AC} = \frac{s}{\sqrt3\, s} = \frac{1}{\sqrt3}.$$
5. Areas of similar shapes compare as the square of matching lengths: $\left(\frac{1}{\sqrt3}\right)^2 = \frac13$. So the small rhombus has area $\frac13 \cdot 24 = 8$.

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

A rhombus with a $60^\circ$ angle is two equilateral triangles, so its diagonals are always in the ratio $1 : \sqrt3$, with no measuring needed. For similar shapes, the area ratio is the square of the length ratio, but only if you compare lengths that play the same role in both shapes. The main work is deciding which diagonal of the small rhombus matches which diagonal of the big one.

Cut into equal pieces. Triangle $EBD$ has angles $30^\circ$, $30^\circ$, $120^\circ$ with the $120^\circ$ angle at $E$, so $E$ is the center of equilateral triangle $ABD$. Lines from the center to the three corners split $ABD$ into three equal triangles, so triangle $EBD$ is $\frac13$ of triangle $ABD$, which is half of $24$, so $\frac13 \cdot 12 = 4$. Triangle $FBD$ is also $4$ in the same way. The small rhombus has area $4 + 4 = 8$.

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 $\boxed{\textbf{(C)}\ 8}$.