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 .

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 .