Exploring 4-Ball Tetrahedra with Cabri 3D

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.