Some formulas for disphenoids

We need some properties concerning angles and distances of a disphenoid. Leech [4] for example gives an overview. But it is not too difficult to find the relations we need directly. The computations are done with Mathematica.

With A(0,0,0), B(c,0,0) and C(q,0,h) we first reflect the triangle ABC across the perpendicular bisector of c. Then we rotate the triangle ABC' along axis AB into the y- dimension by an angle γ until the segment CC'' has the same length as c. The edge C'' then is the 4th corner of the disphenoid and we rename it with D. All skew edges have the same length now.

All skew edges have the same length and the angle γ is the dihedral angle at edge c.

We define the coordinates of the corners first

PA={0,0,0}; PB={c,0,0}; PC={q,0,h}; PD = {c-q,h Sin[gamma],h Cos[gamma]};

and solve for h and q with the Pythagorean theorem

shq=Solve[q^2+h^2==b^2 &&(c-q)^2+h^2==a^2,{h,q}]

Then we plug in h an q into PC and PD

PC = PC /. shq[[2]] ; PD = PD /. shq[[2]]

Now we can solve for the dihedral angle γ at edge c with the equation |AB|=|CD|

sg=Solve[(PB-PA).(PB-PA)== (PC-PD).(PC-PD),gamma]//FullSimplify

Leech [4] has provided an equivalent formula.

We plug this into PD do get the coordinate of PD finally

PD = PD /. sg[[2]]

We do not need PD explicitely here, but now we can compute the squared distance between c and c''. In a disphenoid it is just the squared length of the midpoints of the corresponding edges

((PA + PB)/2 - (PC + PD)/2).((PA + PB)/2 - (PC + PD)/2) // Simplify

which also is a well known result.

For the angle between c an c'' we do

angle[v1_,v2_]:=ArcCos[v1.v2/(Norm[v1]Norm[v2])]
Simplify[angle[PB-PA,PD-PC],Assumptions->{a^2+b^2>c^2,b^2+c^2>a^2,c^2+a^2>b^2}]

Without assuming the triangle to be acute (which is the case in a disphenoid) the expression would not simplify.

These formulas suffice to tackle the kaleidocycle problem.

References

4. J. Leech, Some properties of the isosceles tetrahedron,Math. Gazette34(1950)269–27

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