Fitting a tetrahedron through the smallest hole

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I'm designing a child's toy consisting of a closed box with a hole on top; a unit tetrahedron must fit through this hole.

What is the smallest possible area of the hole?

Currently my hole is an isosceles triangle with base $1$ and height $\frac{\sqrt{2}}{2}$ (the distance between the midpoints of two opposite edges), which gives area $\frac{\sqrt{2}}{4}$. With the correct orientation, the tetrahedron is dropped straight through the hole.

Is it possible to do better considering rotations of the tetrahedra during insertion?