[Geometry]
Problem: Regular hexagon $A B C D E F$ and regular hexagon $G H I J K L$ both have side length 24 . The hexagons overlap, so that $G$ is on $\overline{A B}, B$ is on $\overline{G H}, K$ is on $\overline{D E}$, and $D$ is on $\overline{J K}$. If $[G B C D K L]=\frac{1}{2}[A B C D E F]$, compute $L F$.

Solution: The diagram below shows the hexagons.

<img_3234>

The area of hexagon $G B C D K L$ can be computed as $[G B C D K L]=[A B C D E F]-[A G L K E F]$, and $[A G L K E F]$ can be computed by dividing concave hexagon $A G L K E F$ into two parallelograms sharing $\overline{F L}$. If $A B=s$, then the height $A E$ is $s \sqrt{3}$, so the height of parallelogram $A G L F$ is $\frac{s \sqrt{3}}{2}$. Thus $[A G L F]=L F \cdot \frac{s \sqrt{3}}{2}$ and $[A G L K E F]=L F \cdot s \sqrt{3}$. On the other hand, the area of a regular hexagon of side length $s$ is $\frac{3 s^{2} \sqrt{3}}{2}$. Because $[G B C D K L]=\frac{1}{2}[A B C D E F]$, it follows that $[A G L K E F]=\frac{1}{2}[A B C D E F]$, and

$$
L F \cdot s \sqrt{3}=\frac{1}{2}\left(\frac{3 s^{2} \sqrt{3}}{2}\right)=\frac{3 s^{2} \sqrt{3}}{4}
$$

whence $L F=\frac{3}{4} s$. With $s=24$, the answer is $\mathbf{1 8}$.
Compute $[B C D K L G]$ as twice the area of trapezoid $B C L G$. If $A B=s$, then $B G=s-L F$ and $C L=2 s-L F$, while the height of the trapezoid is $\frac{s \sqrt{3}}{2}$.[^0]


[^0]:    ${ }^{1}$ The answer 115 was also accepted for this problem because of an alternate (and unintended) reasonable interpretation of the problem statement. Some students also counted portions that contained the "hole", with the hole being strictly inside the portion, and not along its edges.



Thus the area of the trapezoid is:

$$
\frac{1}{2}\left(\frac{s \sqrt{3}}{2}\right)((s-L F)+(2 s-L F))=\frac{s \sqrt{3}(3 s-2 L F)}{4}
$$

Double that area to obtain

$$
[B C D K L G]=\frac{s \sqrt{3}(3 s-2 L F)}{2}
$$

On the other hand, $[A B C D E F]=\frac{3 s^{2} \sqrt{3}}{2}$, so

$$
\begin{aligned}
\frac{s \sqrt{3}(3 s-2 L F)}{2} & =\frac{3 s^{2} \sqrt{3}}{4} \\
3 s-2 L F & =\frac{3 s}{2} \\
L F & =\frac{3}{4} s .
\end{aligned}
$$

Substituting $s=24$ yields $L F=\mathbf{1 8}$.

Final answer: 18