Origami in the 3D space

Let the coordinates of $M$ be $(x, y, z)$ . Then by the distance formula, $ME = \sqrt{2} = \sqrt{(x - 4)^2 + (y - \sqrt{2})^2 + z^2}$ , $MF = 2 = \sqrt{(x - 2)^2 + y^2 + z^2}$ , and $MA = 4 = \sqrt{x^2 + (y - 2\sqrt{2})^2 + z^2}$ , which solves to $M(x, y, z) = M(\frac{10}{3}, \frac{2\sqrt{2}}{3}, \frac{2\sqrt{3}}{3})$ for $z > 0$ .

The equation of the plane with $A$ , $E$ , and $M$ can be expressed generally as $x + by + cz = d$ . $A(0, 2\sqrt{2}, 0)$ gives $2\sqrt{2}b = d$ , $E(4, \sqrt{2}, 0)$ gives $4 + \sqrt{2}b = d$ and $M(\frac{10}{3}, \frac{2\sqrt{2}}{3}, \frac{2\sqrt{3}}{3})$ gives $\frac{10}{3} + \frac{2\sqrt{2}}{3}b + \frac{2\sqrt{3}}{3}c = d$ , which solves to $x + 2\sqrt{2}y + \sqrt{3}z = 8$ .

The equation of the plane with $A$ , $E$ , and $F$ is $z = 0$ .

The normal to $x + 2\sqrt{2}y + \sqrt{3}z = 8$ is $(1, 2\sqrt{2}, \sqrt{3})$ and the normal to $z = 0$ is $(0, 0, 1)$ . The angle $\theta$ between these normals (and also the dihedral angle between the two planes) is $\theta = \cos^{-1} ( \frac{(1, 2\sqrt{2}, \sqrt{3}) \cdot (0, 0, 1)}{\sqrt{1^2 + (2\sqrt{2})^2 + (\sqrt{3})^2} \cdot \sqrt{0^2 + 0^2 + 1^2}}) = \cos^{-1} (\frac{1}{2}) = \boxed{60°}$ .