Olimpiadi di Matematica

Solution:

The answer is $\mathbf{(A)}$. Let $ABC$ be the triangle forming the base; the safety zone is a triangle $A'B'C'$ (with $A'$ belonging to the bisector of the angle at $A$, and cyclically for the others) inside the triangle $ABC$. Letting $H$ and $K$ be the projections of $A'$ and $B'$ respectively onto the side $AB$, we have $A'H=1$ meter. Since the triangle $A'AH$ is half of an equilateral triangle (the angles at $A$, $A'$, and $H$ are respectively $30^\circ$, $60^\circ$, and $90^\circ$), the side $AH$ has length $2 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$, from which $A'B' = HK = AB - AH - BK = AB - 2 \cdot AH = 8 - 2 \cdot \sqrt{3}$ meters. Therefore the area of triangle $A'B'C'$ is given by

$$(8-2 \cdot \sqrt{3})^2 \cdot \frac{\sqrt{3}}{4} = (64 + 12 - 32 \sqrt{3}) \cdot \frac{\sqrt{3}}{4} = 19 \sqrt{3} - 24,$$

which is the required area (in square meters).