a few outstanding tetrahedra

A tetrahedron ("four faces") is the most simple of the polyhedra: its four vertices define four triangular faces assembled with six edges.
The regular tetrahedron and the regular right-angled tetrahedron can easily be obtained from a cube.

a regular tetrahedron a regular tri-right-angled tetrahedron a diamond (compound
of these two regular pyramids)
Schläflis' birectangular tetrahedron
(four rectangular faces)

We can do so using a rectangular parallelepiped to obtain other types of tetrahedra.

an equifacial tetrahedron
a tri-right-angled tetrahedron an equifacial tetrahedron
with isosceles faces
(tetragonal disphenoid)
an other tetrahedron with
right-angled faces (non equifacial)

Here are two other interesting tetrahedra defined in a cube (1/24 of the cube's volume):
  • if we cut out the first (blue) along each of the twelve edges of a cube it remains a Kepler star (see puzzles),
  • 6 of the second (magenta) can build an oblate rhombohedron, 8 an octahedron (longest edge in common), 24 a cube (two along each edge), 6×8=48 a rhombic dodecahedron (six octahedra with a common vertex); 12 pairs (bipyramids) with a common vertex build a stellated rhombic dodecahedron.

See also the Hill's tetrahedra (paving of the space, decomposition).

a few properties of tetrahedra

the equifacial tetrahedra (isometric faces)

⇔  the opposite edges have same length
⇔  the circumsphere and the insphere (tangent to the four faces) are concentric
⇔  at each vertex the sum of the face angles is 180°
 •   the four altitudes are equal
 •   the common perpendiculars to opposite edges come through the midpoints, are convergent and mutually perpendicular
 •   the isobarycentre G of the vertices is center of the circumscribed and inscribed spheres; the contact point of the insphere with a face is the circumcenter of that face
 •   the feet of the altitudes, the orthocentres of the faces and the midpoints of the altitudes lay on a sphere with center G
 •   the volume V is given by V²=(b²+c²-a²)(c²+a²-b²)(a²+b²-c²)/72  where a, b and c are the lengths of the pairs of opposite edges
examples: the regular tetrahedra, two of the tetrahedra above (see net below)

the orthocentric tetrahedra

⇔  two pairs of opposite edges are orthogonal (characteristic property)
⇔  three heights are concurrent (existence of an orthocenter H)
In fact we have a sufficient condition much more weak (Georges Lion).
If one of the altitude is secant with two others, not necessarily a priori at the same point, then the tetrahedron is orthocentric.
proof: For line AB to be orthogonal to line CD it is necessary and sufficient that h(A), altitude going through A, is secant with h(B).
Let us notice that generally h(A) and h(B) are orthogonal to CD without being parallels. We have then:
   h(A) and h(B) are secant  ⇔  h(A) and h(B) are coplanar  ⇔  there is a plane going through A and B and perpendicular to CD.
If this condition is carried out the edges AB and CD are orthogonal and so are the edges AC and BD, thus the tetrahedron is orthocentric.
Remark: if h(A) and h(B) are secant then so are h(C) and h(D) but the tetrahedron ABCD is not necessarily orthocentric.
⇔  the foot of one altitude is orthocentre of the opposite face
⇔  a²+a'²=b²+b'²=c²+c'² where l and l' are the lengths of two opposite edges
 •   the three pairs of opposite edges are orthogonal
 •   the feet of the altitudes are orthocentres of the opposite faces
 •   the three common perpendicular to opposite edges are convergent in H
 •   the three segments joining the midpoints of opposite edges have same length
 •  the midpoints of the edges and the feet of the common perpendiculars to opposite edges lay on a sphere with center the isobarycentre G of the vertices (first Euler's sphere); the Euler's circles of the faces belong to this sphere
 •   G is midpoint of [OH] (Euler's line of the tetrahedron) where O is the center of the circumscribed sphere
 •   the perpendiculars to the faces on their centers of gravity are convergent in I on the Euler's line
 •   in a tetrahedron ABCD, the feet of the altitudes, the centers of gravity of the faces, and the points laying on the thirds of [HA], [HB], [HC] and [HD] lay on a sphere with center the midpoint of [HI] (second Euler's sphere)
 •   volume:  V=(1/6)×l×l'×h  where h is the length of the common perpendicular to two opposite edges of lengths l and l'
For any tetrahedron ABCD  V=(1/6)×AB×CD×h×sin(x)  where x is then angle (AB,CD)
The angle of two non coplanar lines is the one of their parallels going through some point (on the construction H is such a point).

reminder: construction of the common perpendicular Δ to two non coplanar lines D and D'
Let D" be the orthogonal projection of D' on the plane P going through D and parallel to D', and H its intersection point with D; Δ is the line perpendicular to P going through H.
This drawing is interactive: you may modify the tetrahedron by moving the red points B, C and D, and thus you can verify that the common perpendicular can be outside of the tetrahedron.

examples: the regular tetrahedra, the trirectangle tetrahedra (the orthocentre is one vertex)
a construction of an orthocentric tetrahedron: let ABC be a triangle with orthocentre D' and let D be a point distinct from D' on the perpendicular to the plane of ABC going through D'.

the isodynamic tetrahedra

aa'=bb'=cc'  where l and l' are the length of two opposite edges
 ⇔  the segments joining one vertex to the center of the circle inscribed in the opposite face are concurrent

Crelle's tetrahedra

a+a'=b+b'=c+c'  where l and l' are the length of two opposite edges
 ⇔  there exists a sphere tangent to the six edges (mid-sphere)

example: the "4-ball tetrahedra"
a+a'=b+b'=c+c'=r1+r2+r3+r where the ri are the radii of four spheres with centers at the vertices and mutually tangent.
The tangent points of these spheres are also the tangent points of the edges with the mid-sphere; the intersections of the mid-sphere with the tetrahedron's faces are the incircles of the faces.
The three segments joining the tangent points on two opposite edges converge.

See also Soddy's "kissing spheres".

nets of tetrahedra

The net of the regular tetrahedron is well known (an equilateral triangle with its midpoint triangle); the net of the regular trirectangle tetrahedron can be easily deduced. regular tetrahedron (net 1) tetrahedron (net 3)
Here is the second net of the regular tetrahedron: regular tetrahedron (net 2)
A triangle folded along the sides of its midpoints triangle is the net of an equifacial tetrahedron (the triangle must have three acute angles).
Three folds a square into the net of a tri-right-angled tetrahedron.
A square may also be the net of an equifacial tetrahedron.
tetrahedra (nets 4)
If we pull one lateral face of a pyramid down on the plane of its base, we notice that the altitude of this face swings around its foot. We deduce the construction of a net of a tetrahedron:
We choose the base triangle and a point S (projection of the summit) which may be chosen on a side or outside of the base triangle. Each lateral triangle has an altitude coming through S, on which lays its third vertex. We choose one of them (S' for example, with HS'>HS) and deduce the two others by transferring the lengths of the sides.
We may choose the height SS'=h of the tetrahedron by constructing the right-angled triangle HSS'.
tetrahedron (net 5)

cuttings through a tetrahedron

If we cut a tetrahedron SABC using a plane parallel to the face ABC, for example at mid-high, we get a pentahedron (truncated pyramid with three trapezoid lateral faces), whose volume is 7/8 of the tetrahedron's volume.

The altitudes of the three trapeziums are obviously not equal, but there is a direction in the space according to which they seem to be. We may search it experimentally by moving the solid.

Let's switch from the space to the plane by projecting the summit S to S' on the plane (ABC). The section at mid-high is then a triangle A'B'C', and the midpoint theorem in a triangle proves that A'B'C' is a reduction of ABC (for a reduction ratio different from 1/2 we use Thales theorem, or an homothecy with center S'). In order to have three trapeziums with same altitude, it is necessary and sufficient for each vertex of A'B'C' to be equidistant from two sides of ABC, thus laying on the bisector of the sector. Finally, S' has to be the incenter I of ABC. The searched direction is (SI). tetrahedron (projection)

Here are three exercises (computations and constructions) to test your mastery of the space :
•   Compute the volume of a regular tetrahedron with edge a, and the measure of the angle of two faces.
How to verify these results?
•   Draw in perspective a tetrahedron SABC, and chose on its surface three points P, Q an R not on the same face. Construct the section of SABC by the plane (PQR).
    The construction is easy if one point lays on an edge or if two points lay on a same face.
If not you have to tire yourself a little more to get the solution.
    •   Construct a tetrahedron with given volume knowing the directions and the lengths of two opposite edges.

reference: Géométrie de l'espace et du plan  (in French) by Yvonne and René Sortais (published by Hermann - 1988) pages 305-336

See also the pages devoted to the decompositions of the parallelepiped, to the tetrahedral symmetries and to the kaleidocycles.

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convex polyhedra - non convex polyhedra - interesting polyhedra - related subjects April 1999
updated 03-01-2014