Congratulations to Ang Zhi Ping, age 15, River Valley High School, Singapore for this solution.

Take the vertex of the cube at the origin $(0, 0, 0)$ as shown in the diagram. The tetrahedron has vertices $O, X, Y, Z$ , where the three centres $X$ , $Y$ and $Z$ are as follows:

$X = (1, \frac{1}{2}, \frac{1}{2}), Y = (\frac{1}{2}, 1, \frac{1}{2}), Z = (\frac{1}{2}, \frac{1}{2}, 1) .$

Solution.

The area required is the sum of the areas of the triangles $XYZ$ (which is equilateral), and $OXY$ , $OYZ$ , $OZX$ (which are congruent to each other and isosceles).

Now $XY = 1 / \sqrt{2)}$ so that

$area (XYZ) = \frac{1}{2} {(\frac{1}{\sqrt{2}})}^{2} \sin 60^{\circ} = \frac{\sqrt{3}}{8}$

Next,

$OX = OY = OZ = \sqrt{\frac{3}{2}} .$

Thus

$OX = OY = \sqrt{\frac{3}{2}}, XY = \frac{1}{\sqrt{2}} .$

This gives the area of $OXY$ as $\sqrt{11} / 8$ . Thus the surface area of the tetrahedron is

$\frac{\sqrt{3}}{8} + \frac{3 \sqrt{11}}{8} .$