A closer look at the solutions

Using our result

we define a function

valid[a_,b_,c_]:=-8 (a^2-b^2)^2 (a^2+b^2)-5 c^6+11 (a^2-b^2)^2 c^2+2 (a^2+b^2) c^4

which is >= 0 if the kaleidocycle works with the hinges at edge c and <0 else. Without loss of generality we can set c=1 in the following region plot of valid(a,b,c) (hinges at edge c), valid(b,c,a) (hinges at edge a) and valid(c,a,b) (hinges at edge b). In red regions we have values >=0.


  • We see that the function valid[a,b,c ]>=0 (valid[b,c,a]>=0, valid[c,a,b]>=0) implies that a, b and c not only define a triangle but in fact an acute triangle.
  • In each of the three red hinge regions the hinge edge is the shortest edge
  • The three regions are separated. There is no triangle/disphenoid, where two different edges can be used as hinges.

Proposition, stronger version: Let a, b, c > 0 and without loss of generality c<=a and c<=b. Then a physically realizable kaleidocycle made of 3 disphenoids with edge lengths a, b and c and 3 corresponding chiral partners exists if and only if

In case of equality the kaleidocycles neigbouring faces touch 4 times during a cycle and the kaleidocycle has no hole in these 4 positions.

We now search for a few extremal points, which all lie on the boundary of the red region, and there we have valid[a,b,c]=0.

  1. For a disphenoid with hinge edge c=1 we are looking for the smallest solution with a=b

    Solve[valid[a, b, c] == 0 /. b -> a, a]

    This gives a=b=Sqrt[5]/2*c ≈1.12 c. This is the sort of disphenoids Schattschneider and Walker [1] use in their kaleidocycles. The skew angle is 90°.

    Face of disphenoid with a=b and shortest possible legs

  2. The most "irregular" faces can be defined as those where for given hinge c the difference delta between a and b is maximal. This is the case where the line b=a+delta is a tangent to the valid region. Without loss of generality we choose delta>0. We solve valid[a,b,1]=0 for b, then solve b = a + delta for a and look for which delta there only is one solution. The expressions get complicated but the final result is simple. We get

    delta = 1/2 (2-Sqrt[3]) c ≈0.13 c,
    a= 1/4 (-2+Sqrt[3]+Sqrt[2 (8+5 Sqrt[3])]) c≈1.38 c and
    b= a+ delta = 1/4 (2-Sqrt[3]+Sqrt[2 (8+5 Sqrt[3])]) ≈1.51 c

    For the skew angle we get

    ArcCos[|(a^2 - b^2)/c^2|]=ArcSec[2 Sqrt[2/(-4+3 Sqrt[3])]] ≈ 67.3°

    Face of disphenoid where the non-hinge edges have maximal difference

  3. To find the smallest possible value for b (for given hinge length c) we solve value[a,b,1]=0 for a. There are 6 solutions but only two of them have positive values for a. By equating these solutions and solving for b we find for the minimal b

    ≈ 1.106 c
    The corresponding value for a is
    ≈ 1.151 c

    Using Mathematica Root expressions these complicated radicals can be written simpler as
    1/4 Root[-6591+991 #1^2-53 #1^4+#1^6&,2] c and
    1/4 Root[-1359+399 #1^2-37 #1^4+#1^6&,2] c

    The skew angle ist ArcCos[Root[-1 + 8 #1 + 16 #1^2 + 16 #1^3 &, 1]] ≈84.1°

There is a class of tetrahedra which is called Heronian tetrahedra. Here all edge lengths, face areas and the volume have integer values. The subclass of Heronian disphenoids plays an important role in the theory of Heronian tetrahedra. This motivates a similar question: which integer lengths disphenoids can serve as the basis for a kaleidocycle?

The following code computes the number a(n) of disphenoids which can serve as kaleidocycle which have integer edge lengths <= n.

value[a_,b_,c_]:=-8 (a^2-b^2)^2 (a^2+b^2)-5 c^6+11 (a^2-b^2)^2 c^2+2 (a^2+b^2) c^4

{0,1,3,6,10,15,21,28,36,44,54,66,80,96,113,132,153,176,200,225} (OEIS A338334 )

There are for example three possible disphenoids with edge lenghts <=3: {2,2,1}, {3,3,1} and {3,3,2}. The other triangles do not work: {1,1,1}, {2,2,2} and {3,3,3} are acute but blocking occurs with the corresponding kaleidocycles. {3,2,2} is not acute and hence does not allow a disphenoid. And {2,1,1} , (3,1,1) and {3,2,1 }are no triangles.

If we request scalene triangles for the faces of the disphenoid which have integer edge lengths <= n we get


{0,0,0,0,0,0,0,0,0,0,1,3,6,10,14,19,25,32,40,48,57,68,82,97,113,131,151,173,196,220} (OEIS A338335 )

The only scalene possible triangle with edge lenghts <= 11 is {11,10,8}.

If we are looking for integer triangles for the faces of the disphenoid which have perimeter n we get


(OEIS A338336 )

The smalles possible perimeter is 5 with the triangle {2,2,1}.

Are there integer length triangles such that with the corresponding kaleidocycle neighbouring faces touch 4 times during a cycle? This would mean equality in the relation between a, b and c:

I did not find an answer to this question.


1. Doris Schattschneider and Wallace Walker, M.C. Escher Kaleidocycles, 1977. ISBN 0-906212-28-6
2. Baker, J.E., 1980. An analysis of Bricard linkages. Mechanism and Machine Theory 15, 267–286.
3. Chen, Y., You, Z., and Tarnai, T., 2005, “Threefold-Symmetric Bricard Linkages for Deployable Structures,”Int. J. Solids Struct.,42(8), pp. 2287–2300.
4. J. Leech, Some properties of the isosceles tetrahedron,Math. Gazette34(1950)269–27
5.Safsten C, Fillmore T, Logan A, Halverson D, Howell L (2016) Analyzing the stability properties of kaleidocycles. J Appl Mech 83:051001.

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<2. Some formulas for disphenoids>
<3. The kinematics of the linkage>
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<7. The 8-kaleidocyce polyhedron>

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