9. The answer is $\mathbf{( A )}$. Given $A B C$ as the triangle forming the base, the safety zone is a triangle $A^{\prime} B^{\prime} C^{\prime}$ (with $A^{\prime}$ lying on the angle bisector of angle $A$ and cyclic permutations) inside the triangle $A B C$. Let $H$ and $K$ be the projections of $A^{\prime}$ and $B^{\prime}$, respectively, on the side $A B$. We have $A^{\prime} H=1$ meter. Since the triangle $A^{\prime} A H$ is a half equilateral triangle (the angles at $A, A^{\prime}$, and $H$ are 30, 60, and 90 degrees, respectively), the side $A H$ is $2 \cdot 1 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}$, from which $A^{\prime} B^{\prime}=H K=A B-A H-B K=A B-2 \cdot A H=$ $8-2 \cdot \sqrt{3}$ meters. Therefore, the area of the triangle $A^{\prime} B^{\prime} C^{\prime}$ is given by

![](https://cdn.mathpix.com/cropped/2024_04_17_74150396f6f11420587fg-14.jpg?height=420&width=422&top_left_y=1663&top_left_x=1431)
$(8-2 \cdot \sqrt{3})^{2} \cdot \frac{\sqrt{3}}{4}=(64+12-32 \sqrt{3}) \cdot \frac{\sqrt{3}}{4}=19 \sqrt{3}-24$, which is the desired area (in square meters).
9. The reds and the greens are having a snowball fight. The reds' base is an area in the shape of an equilateral triangle with a side length of 8 meters. The greens cannot enter the reds' base, but they can throw their projectiles into the base while still being outside the perimeter. Knowing that the greens can hit a target up to a maximum distance of 1 meter, how large (in square meters) is the area inside the reds' base that is safe from the greens' range?
(A) $19 \sqrt{3}-24$
(B) $4 \sqrt{3}$
(C) $3 \sqrt{3}$
(D) $19-8 \sqrt{3}$
(E) every point in the red area is within the greens' range.