Solution To Mensa Tetrahedron Challenge

 

Solution:

The edges connecting the base vertices are face diagonals of the cube. Therefore, the base of the tetrahedron is an equilateral triangle  of edge length  Ö2. 

 

The height h of the equilateral triangle can be found from the Pythagorean theorem:

 Ö2 / 2) ²    +   h ²    =   Ö(2) ²   Þ  (½)   +   h ²    =   2

h ²    =   3/2  Þ   h     =   Ö6/ (2)

The distance from a base vertex to the center of the base is: Ö6/ (2)  -   x

Then:

( Ö2 / 2) ²    +   x ²    =  (Ö6/ (2)  -   x )   x ²

And:

(½)   +   x ²    =   3/2   -   Ö6 x )  -  x ²

^{Ö6 x )  =   1}

^Þ 

  ^{x  =    1  /} Ö6  =   ( Ö6/ 6)

The distance from a base vertex to the center of the base is now:

Ö6/ (2)  -   ( Ö6/ 6  )  =   2 ( Ö6/ 6  ) =  ( Ö6/ 3)

The height H of the tetrahedron can then be found from  Pythagorean theorem, i.e.:

( Ö6/ 3  ) ²  +    H  ²  =  1 ²

^Þ 

 

2/3 +    H  ²  =   1    

H  ²  =   1    -  2/3   = 1/3

H  =   Ö3/ 3  )

The volume  V of the tetrahedron can then be found from  

 V  = 1/3 Area of the base x height H

Then the volume:

V = 1/3    [½] (Ö2) (Ö6/ 2)   (Ö3/ 3  )

V =  1/6  cubic units