An equilateral triangle $ABC$ is inscribed in a square with side 1 (each vertex of the triangle is on a side of the square and no two are on the same side). Determine the greatest and smallest value of the side of $\Delta ABC$.
The maximum occurs when we put one of the vertices of the equilateral triangle in the vertex of the square.
$$2(1-x)^2=1^2+x^2$$$$x=2-\sqrt{3}$$$$\sqrt{1+(2+\sqrt{3})^2}=\boxed{\sqrt{6}-\sqrt{2}}$$The minimum is just $\boxed{1}$ since that's the minimum distance between points on opposite sides of the square.