, 2003

Solution:

The answer is (B). The six isosceles triangles have two angles of $75^{\circ}$. The angle between two sides of two isosceles triangles having a common vertex is therefore $360^{\circ}-\left(75^{\circ}+75^{\circ}+120^{\circ}\right)=90^{\circ}$. With reference to the figure alongside, one observes that the required area $S$ can be obtained by subtracting from the area of the regular hexagon (which has side equal to the radius of the circumscribed circle) the area of 6 isosceles right triangles with side $\frac{1}{\sqrt{2}}$. We thus have
$$S=\frac{6 \cdot \frac{\sqrt{3}}{2}}{2}-6 \cdot \frac{\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{2}$$
from which
$$S=\frac{6 \sqrt{3}}{4}-\frac{6}{4}=\frac{3(\sqrt{3}-1)}{2}$$