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= æ
ç
è 		1, 1
2
 , 1
2
 ö
÷
ø 		, Y= æ
ç
è 		1
2
 ,1, 1
2
 ö
÷
ø 		, Z= æ
ç
è 		1
2
 , 1
2
 ,1 ö
÷
ø 		.

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/   _
Ö2)
 

so that
area(XYZ) = 1
2
 æ
ç
è 		1
Ö2
 ö
÷
ø 		2

 
 sin60° = Ö3
8
Next,
OX = OY = OZ =   æ
 ú
Ö
3
2
 
 .
Thus
OX = OY =   æ
 ú
Ö
3
2
 
 , XY = 1
Ö2
 .
This gives the area of OXY as
  __
Ö11
 
 /8

. Thus the surface area of the tetrahedron is
Ö3
8
 +
3   __
Ö11
 
8
 .