Exploring 4-ball tetrahedra with Cabri 3D
Adrian Oldknow
August 2005
The first step: to construct the `3-disc problem’ in the plane
Three points A,B,C have been chosen in the `ground-plane’, and joined by segments to
form the sides of a triangle. The bisector of an angle such as A has been found by
constructing the ray AC, and the perpendicular to the plane through A. Let the circle
around this perpendicular through B cut the ray in C’. Then the midpoint M of BC’ lies
on the angle bisector, which is the ray AM. The intersections of any two such bisectors
are the incentre I of ABC. The planes through I perpendicular to the edges BC, CA, AB
cut the edges in P,Q,R. The circle around the perpendicular to the plane through I
passing through P,Q,R is the incircle of ABC. The circle around the perpendicular to the
plane through A passing through Q,R is the vertex circle for A etc. Thus we have
modelled the plane construction for the `3-coin problem’.
VW is a vector which we shall use as a `slider’ for the radius d of the 4-th vertex sphere.
We will construct the 4-th vertex D as the intersections of the three extended vertex
spheres, such as the sphere centre A and radius a+d. The ray through A parallel to VW
meets the A vertex circle in A1, and this point is translated by vector VW to the point A2.
The sphere centre A through A2 has the desired radius a+d.
The intersection of the extended A and B vertex spheres is the dotted circle, and that of
the extended A and C vertex spheres is the solid circle. The intersections of the two
circles are D and D’. To ensure we systematically select the `upper’ choice, the point E
is the mid-point of DD’, which is
translated by the `unit z-vector’ to point
F. Then D,D’ are hidden, and the
intersection is found between the ray EF
and either of the circle – this point is
now named D, and is the required 4-th
vertex of the 4-ball tetrahedron ABCD.
As W slides so the vertex D describes a
path in space. We can now construct
the sphere centre D with radius VW and
find the points where it intersects the
sides DA, DB, DC. The resulting 4-ball
configuration is shown alongside.
The mid-sphere touches each edge, such as AD, at the point of tangency where the
corresponding vertex spheres touch. So its centre M is the point of intersection of the
planes perpendicular to each edge through the corresponding tangent point. The
perpendiculars from M to the faces meet the faces at the face incentres, which are the
intersections of the faces with the midsphere.
The three segments joining tangency points of opposite edges of ABCD are coincident
in the Zeeman point Z. The four rays, such AZ, meet the opposite face in its Gergonne
point. The line ZM is the Soddy line for the 4-ball tetrahedron ABCD. In order to find
the Soddy points on this line we will need to construct some vertex-locus hyperbolae.
In the `ground plane’ we have the face
Soddy line as the line joining the face
incentre and Gergonne point. The vertex locus
of B, say, in the plane ABC is the hyperbola
with A and C as foci through the point of
tangency on AC. Using the perpendicular
bisector of AC as mirror we can reflect both B
and the tangent point. Reflecting B in the plane
through AC perpendicular to the ground-plane
gives a 5th point lying on the locus and hence
we can construct the 5-point conic through B.
This intersects the face Soddy line of ABC in
the face Soddy centres, which are also points
lying in the space vertex locus of D. In order to
find two more points of the locus we reflect D
in the ground plane, and also in the
perpendicular bisector of the ABC Soddy points.
By sliding W we can see that the vertex locus of D is indeed the constructed hyperbola.
By introducing a second slider we can create another 4-ball tetrahedron ABCK in the
same way as above, and see dynamically the way the K-ball tracks along the hyperbola.
We can locate the Soddy points of ABCD as the points of intersection of the D-vertex
hyperbola with the Soddy line MZ. We denote that between M and Z as S, the first
Soddy point, and the other as S’, the second Soddy point. The intersections of each
vertex sphere, such as A, with the corresponding segment, such as SA, give the contact
points for the first Soddy sphere with the vertex spheres. So we can construct the first
Soddy sphere and its contacts with the vertex spheres.
These four contact points form the vertices of the first Soddy tetrahedron. For each
vertex, such as A, we can construct the plane perpendicular to SA through the contact
point on SA as the common tangent plane to the A-vertex sphere and the first Soddy
sphere. The four such tangent planes form the first Soddy tangent tetrahedron.
Repeating using the second Soddy point S’ we can also construct the second Soddy
tetrahedron through the contact points, and also the corresponding second Soddy
tangent tetrahedron. We just have to take care that we produce S’A to find the correct
point of intersection with the A-vertex sphere as the contact point.
The figures above show the second Soddy
sphere surrounding the 4-balls, and also the
way that the second Soddy tetrahedron is
formed.
The figure on the right shows the
corresponding second Soddy tangent
tetrahedron.
To complete the `cast list’ we have the
tetrahedron formed by the Gergonne points of
the faces, and also the original 4-ball
tetrahedron ABCD.
The Gergonne
tetrahedron is in
point perspective
with ABCD from the
Zeeman point Z, so
the corresponding
sides of these two
tetrahedra meet in 6
points lying in 4
(Fox) lines in the
perspective (Soddy)
plane which is
orthogonal to the
Soddy line ZM
meeting it in the
Fletcher point F.