Test problems

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

Here are some objects one might explore and problems one might solve with dyna3. We'll use this list to guide and evaluate our progress in developing the dyna3 engine and interface. We're striving to build an engine that can represent and solve as many of these problems as possible, and an interface that makes them as convenient as possible to set up.

Basic elements and constraints

Points on a sphere

Statement

Choose a whole number n \ge 1. Make a sphere and constrain n points to lie on it.

Notes

In a generic solution, the points are all different, but there are degenerate solutions where some or all of the points coincide. These are valid solutions to the problem as stated here, but it's also interesting to test whether one can gravitate toward generic solutions or deform a degenerate solution into a generic one.

Mutually tangent spheres

Statement

Choose a graph G. Make a sphere for each vertex of G, and constrain the spheres to be tangent whenever the corresponding vertices are connected by an edge.

Notes

Here are some interesting tangency graphs:

Triangle
Tetrahedron
5-cell
Triangular bipyramid

Points at the intersection of two spheres

Statement

Choose a whole number n \ge 1. Make two spheres and constrain n points to each lie on both spheres.

Notes

In a generic solution, both spheres and all the points are different, but there are degenerate solutions where the two spheres or some or all of the points coincide. These are valid solutions to the problem as stated here, but it's also interesting to test whether one can gravitate toward generic solutions or deform a degenerate solution into a generic one.

Rigid hexagon

Source

Author: Noam Elkies
Published: P12 from the Playground in Math Horizons 2 (1)

Statement

“(Proposed by Noam Elkies, Harvard University.) Suppose A, B, C, D, E, F are points in 3-space such that

\begin{alignat*}{6}
AB &\;=\;& CD &\;=\;& EF &\;=\;& r, \\
BC &\;=\;& DE &\;=\;& FA &\;=\;& s, \\
AD &\;=\;& BE &\;=\;& CF &\;=\;& r + s
\end{alignat*}

for some positive r, s. Show that A, B, C, D, E, F must be the vertices of an equiangular planar hexagon.”

Coverage

As of pull request #84, you can place six points in a rough planar hexagon, impose the distance constraints for some chosen r and s, and see that the assembly becomes an equiangular planar hexagon, which seems to be rigid when you nudge the points.

Tetrahedron radius ratio

Statement

Find the ratio between the inradius and the circumradius of a regular tetrahedron.

Irisawa's hexlet

Source

Author: Irisawa Shintarō Hiroatsu (入澤新太郎博篤)
Published: Donated to Samukawa Shrine in Sagami Province as a sangaku in 1822. Recorded by Irisawa’s teacher Uchida Itsumi in the book Kokon sankan (Mathematics, past and present) in 1832.
Access: Abe Haruki. “Japan’s ‘Wasan’ Mathematical Tradition.” Nippon.com (April 30, 2024).

Statement

Abe gives the following translation of the problem as stated in Kokon sankan. The length unit sun is unrelated to the “sun” sphere.

As shown in the figure, place two spheres (“sun” and “moon”) inside a larger external sphere, such that both touch the external sphere’s inner surface, and then create a chain of connected spheres in the remaining space within the external sphere. If the diameters of the external, sun, and moon spheres are 30, 10, and 6 sun respectively, and the diameter of the sphere marked kō is 5 sun, find the diameters of the other spheres.

Here's an alternative statement.

Six spheres form a cyclic chain, each one externally tangent the next. Two spheres, the “sun” and the “moon,” are each externally tangent to all the chain spheres. An outer sphere encloses and is tangent to all the other spheres. Fix the diameters of these spheres:

Sphere	Diameter
Outer	30
Sun	10
Moon	6
One chain sphere	5

What are the diameters of the other chain spheres?

Coverage

As of pull request #84, the function engine::examples::realize_irisawa_hexlet sets up and solves this problem for use in the unit test irisawa_hexlet_test and the example irisawa-hexlet.rs.

You can also set up and solve the problem by hand. A good way to do this is to first impose all the tangencies and then start fixing diameters. Right now, the engine isn't very robust, so you'll have to adjust the spheres as you set up the problem to help the engine satisfy each additional constraint.

Frugal firepower (combinatorial)

Source

A version of the “Frugal firepower” problem that doesn't require hierarchical constraints.

Statement

Place five unit spheres tangent to each other so that their centers form either a square pyramid or a triangular bipyramid. Build a rectangular box whose faces are parallel to the coordinate axes. By exploring different patterns of tangency between the spheres and the walls of the box, try to make the box enclose the spheres while minimizing the sum of its length, width, and height.

Notes

One might reasonably hope that solving this problem will provide a solution of the original Frugal Firepower problem. The idea is that a box that solves the original problem should have “maximal contact” with the spheres, and should therefore be determined by its tangencies with the spheres.

Johnson solid

Statement

Choose a Johnson solid. Realize it and confirm that it's rigid.

Notes

There are various ways to represent the solid and to constrain the faces to be regular. Here's a way that has the advantage of requiring only point–point distance, point–sphere incidence, and curvature constraints.

For each vertex, create a point.
For each edge, add a unit distance constraint between vertices.
Constrain the faces to be planar. For each face with more than three vertices:
- Create a plane—a sphere whose curvature is constrained to be zero.
- Constrain each vertex of the face to lie on the plane.
Constrain the faces to be regular. For each face with more than three vertices:
- Triangulate the face.
- For each triangulation diagonal, add a distance constraint to enforce the desired length.

Nets for the Johnson solids might help with assembly. Here are some interesting Johnson solids with relatively low vertex, edge, and face counts:

Triangular cupola (J3)
Triangular dipyramid (J12)
Gyrobifastigium (J26)
Tridiminished icosahedron (J63)
Snub disphenoid (J84)

Tridiminished icosahedron rigidity

Statement

Take the 1-skeleton (vertices and edges) of a tridiminished icosahedron, keeping the edges fixed at unit length but allowing the angles between them to vary. This assembly is flexible. Now, constrain each pentagonal face to be planar. Is this enough to make the assembly rigid?

Notes

The answer yes is suggested by:

Glen's physical experiments with Geometiles, using chains of squares to represent the planar pentagonal faces.
Aaron's numerical experiments with dyna3, using the version of the engine from pull request #92.

5-5-4 acrohedron

Source

Author: Jim McNeill
Published: “Acrohedra” and “5-5-4 Near Misses.” Polyhedra.

Statement

Try to realize a 5-5-4 acrohedron, which is a polyhedron with the following properties:

Every face is a regular polygon.
There’s some vertex whose neighboring faces are a 5-gon, a 5-gon, and a 4-gon, in that cyclic order.

If the candidate is a known near miss, confirm that realization fails.

Notes

The problem of finding a 5-5-4 acrohedron is open as of June 2025.

Ring of polyhedra

Source

Rhombic ring of icosahedra

Author: RobertLovesPi
Published: “A Rhombic Ring of Icosahedra, Leading to a Rhombic Dodecahedron Made of Icosahedra.” RobertLovesPi.net (September 15, 2015).

Almost-ring of 48 tetrahedra

Author: Ian Stewart
Published: “Steinhaus’s Problem on Chains of Regular Tetrahedra.”

Quadrahelix and octahelix

Author: Michael Elgersma and Stan Wagon
Published: “The Quadrahelix: A Nearly Perfect Loop of Tetrahedra.” arXiv:1610.00280 (November 7, 2016).

Almost-ring of thirteen square anti-prisms

Author: Jim McNeill
Published: “A ring of anti-prisms”. Polyhedra.

Statement

Choose a rigid ring or almost-ring of polyhedra. Realize it by constraining distances between vertices or angles between faces. If it's a ring, verify that it closes and is rigid. If it's an almost-ring, verify that it doesn't close.

Notes

Here are some examples of interesting rings:

A ring of eight dodecahedra
A rhombic ring of eight icosahedra

Here are some examples of interesting almost-rings

An almost-ring of 48 tetrahedra (Stewart 2023, Figure 2)
A quadrahelix almost-ring of tetrahedra (Wagon and Elgersma, 2016)
An octahelix almost-ring of tetrahedra (Wagon and Elgersma, 2016)
An almost-ring of thirteen square anti-prisms

Exploring configuration spaces

The configuration space of equiangular, equilateral hexagons

Source

Author: Jun O'Hara
Published: "The configuration space of equilateral and equiangular hexagons." Osaka Journal of Mathematics 50 (2)
Access: Also available from Project Euclid

Statement

O’Hara has described the configuration space of hexagons in 3-space which are equiangular, equilateral, and not necessarily planar. For most realizable choices of angle, the configuration space has both one-dimensional and zero-dimensional components. It becomes zero-dimensional at each end of the realizable range of angles.

“The configuration space \mathcal{M}_6(\theta) of $\theta$-equiangular unit equilateral hexagons (\theta \neq \pi/3) is homeomorphic to a point if \theta = 0, 2\pi/3, the union of two circles and four points if 0 < \theta < \pi/3, and the empty set if \theta < 0 or $\theta > 2\pi/3$” (O’Hara, Corollary 3.2).
“The configuration space \mathcal{M}_6(\pi/3) of equilateral and $\pi/3$-equiangular hexagons is homeomorphic to the union of a pair of points and the space X illustrated in Fig. 11 [see source] which is a 1-skeleton of a tetrahedron with edges being doubled” (O’Hara, Corollary 3.5).

Notes

According to Dan Piker, the configuration space of right-angled equilateral heptagons also has at least one one-dimensional component.

Kaleidocycle

Statement

A kaleidocycle has seven degrees of freedom: six from the Euclidean motions and one from twisting.

Notes

In the notes “Kaleidocycles with 6 Disphenoids,” Herbert Kociemba describes the edge lengths of all kaleidocycles that can be realized and rotated through a full twist without self-intersecting.

Solution

The unit test tangent_test_kaleidocycle uses a kaleidocycle and initial configuration for which the infinitesimal twist motion can be written algebraically and verified by hand. To verify it, take the derivatives of the distances between the connected vertices of the kaleidocycle and confirm that they vanish at the configuration space tangent vector that we claim generates the twist motion.

To find the full configuration space of a kaleidocycle, rather than just its tangent space at a particular point, one could use the procedure that Shizuo Kaji, Kenji Kajiwara, and Shota Shigetomi describe in the preprint “An explicit construction of Kaleidocycles by elliptic theta functions” (2024). The paper comes with an implementation of the procedure.

Coverage

As of pull request #84, the function engine::examples::realize_kaleidocycle sets up a kaleidocycle using distance constraints between points and finds its tangent space. The unit test tangent_test_kaleidocycle confirms that the tangent space includes the twist motion. The example kaleidocycle.jl finds the twist motion.

Generalized conic: multifocal ellipsoid

Source

Two-dimensional version

Author: James Clerk Maxwell
Published: "Paper on the Description of Oval Curves" (February 1846), in The Scientific Letters and Papers of James Clerk Maxwell: 1846-1862

Statement

Fix two or more “pin” points in 2-space or 3-space. Constrain a movable “pencil” point by running a fixed-length string through all the points in some sequence. Find the locus of possible positions of the pencil.

Notes

To avoid topological issues, it would be simplest to model the length of the string as a whole-number linear combination of the distances from the points to the pencil. To get an ellipse, use two pins and fix the unweighted sum of the distances.

Generalized conic: triangle of directrices

Source

No source found yet, but it seems to be in the spirit of generalized conics as described on Wikipedia.

Statement

Fix three “pin” points in 3-space. Constrain a movable “pencil” point by fixing the surface area of the tetrahedron whose vertices are the pins and the pencil.

Equivalently, fix three “directrix” lines in 3-space that intersect pairwise, forming a triangle. Constrain a movable “pencil” point by fixing the sum of the distances from the directrices to the pencil.

Notes

To get an ellipse or an ellipsoid, use two pins in 2- or 3-space and fix the area of the triangle whose vertices are the pins and the pencil.

Bubble flowers in 2d

Statement

In a plane, draw two concentric $n$-gons, and choose a congruence between them. Connect each vertex of the inner $n$-gon to the corresponding vertex of the outer $n$-gon using a line segment “spoke.” Rotate the $n$-gons so that the “spokes” don't intersect them or each other. We now have a planar graph whose faces are an inner $n$-gon surrounded by n quadrilaterals.

Relax the spokes and the edges of the $n$-gons from line segments to constant-curvature arcs. Constrain the arcs to meet at 120° angles at the vertices of the graph. This turns the graph into a 2d bubble cluster obeying Plateau’s laws.

The configuration space should be an algebraic variety with two components. On one component, the assembly should be n + 5 degrees of freedom: 4 from the Euclidean motions and n + 1 from motions that vary the pressures in the bubbles. The other component intersects this one at the configuration where the inner and outer $n$-gons are regular, and their edges have the same curvature. Physically, this means that the inner $n$-gon is at the same pressure as the exterior. From this configuration, the assembly can move like a necklace of n beads while the pressures in all the beads are held constant. On the “necklace component,” the assembly should have 2n + 1 degrees of freedom: 4 from the Euclidean motions, n from motions that vary the pressures in the necklace bubbles, and n - 3 from the constant-pressure necklace motions. Note that on most of the necklace component, the pressure in the inner $n$-gon can’t vary.

Pappas' Theorem

(suggested by Aaron Abrams)

Statement

In a plane, draw 3 points on each of two lines and connect them in a hexagon that alternates lines. That produces three new intersection points of pairs of sides of the hexagon. Those three intersection points are collinear.

Therefore, ideally if we set this up (currently with a "drawing plane" and modeling lines as perpendicular planes and all points incident to the drawing plane) we can add all the hypotheses of the theorem. It should produce a configuration in which the intersection points are collinear of course. And now if we add the constraint that these three points are collinear, ideally Dyna3 should realize that this new constraint does not remove any degrees of freedom or even better that the configuration space is identical to what it was before.

Note that in Geometry Expressions, if you set up the hypotheses of Pappas' Theorem, and then attempt to add the constraint representing the conclusion (i.e., that one of the three new intersection points of the sides of the hexagon is incident to the line through the other three), the option for adding that constraint is greyed out. Apparently it realizes that the relative position of the third intersection point and the line between the other two is determined by all of the givens. However, it does not seem to realize that this determined situation is one of incidence. On the other hand, since it is able to compute algebraic expressions for the positions and (e.g.) cosines of angles at determined points in terms of variables representing the coordinates of given points, it should actually have enough information to generate and analytic-geometry-style proof of Pappas' Theorem. As long as we are sticking with numerical methods in Dyna3, we won't be able to generate actual proofs but it would of course be nice nevertheless to notice that the relationship between that point and line is determined, and even better, that it always seems to be incidence so that if we have a sufficiently generic occurrence of incidence, we can be sure that it will always happen. (Jurgen R-G did mention in one Cinderella presentation that they could show in some generality that if an incidence held in a sufficient neighborhood of a configuration, then in fact it was a necessary occurrence in all configurations with the construction sequence of that configuration.)

Origami with rigid sheets

{Stub from Savannah Cofer's Illustrating Mathematics Reunion/Expansion talk}

Statement

Represent a folding pattern for a rigid sheet by fixing the lengths of the creases and forcing faces to be planar. Explore the sheet's range of motion and count its degrees of freedom.

Waterbomb octahedron

{Stub from Savannah Cofer's Illustrating Mathematics Reunion/Expansion talk}

Statement

Show how the configuration space of the waterbomb octahedron is described by a sphere passing through a plane.

Hierarchical constraints

These problems impose various kinds of soft constraints on top of the hard constraints that an assembly must satisfy to qualify as a solution. Here are some possible kinds of soft constraints.

Optimizing constraints: Loss function terms whose sum should be minimized.
Optional constraints: Constraints that may be relaxed if they can't be satisfied.

Frugal firepower (optimizing)

Source

Author: David Seppala-Holtzman
Published: P370 from the Playground in Math Horizons 25 (4)
Access: Openly available from Sheridan Digital Editions

Statement

“David Seppala-Holtzman of St. Joseph’s College New York gives us the following problem. A customer orders five identical perfectly spherical cannonballs from Adderley’s Cannonball Emporium, and it’s your job to pack them for shipping. The Emporium ships only in rectangular boxes but can construct such boxes with any desired dimensions. You have a choice of packing the cannonballs so that their centers form a square pyramid or two triangular pyramids, as in figure 1 [see source], but you can orient the arrangements however you like inside the box. The shipping cost will be proportional to the sum of the length, width, and height of the box. In Frugal Firepower, determine which arrangement allows you to minimize the shipping cost.”

Near-miss Johnson solid

Statement

Choose a near-miss Johnson solid and a measure of distortion. Look for a minimal-distortion realization of the solid. You could include some hard constraints by restricting distortion to faces below a certain number of sides, like Jim McNeill does for near-miss Johnson solids and acrohedra.

Notes

Some distortion measures include:

An \ell^p norm on the vector of lengths of edges and face diagonals.
- The \ell^1 norm might be equivalent to the E + P measure used by Jim McNeill.

Origami with face bending

{Stub from Savannah Cofer's Illustrating Mathematics Reunion/Expansion talk}

Statement

Represent a folding pattern for a flexible sheet by fixing the lengths of the creases and adding a bending energy for faces. Find the energy barriers between different folding states.

Algebraic relations

Circular string art

Statement

Draw a pair of perpendicular lines, which we'll treat as coordinate axes. Draw a unit circle in the positive quadrant which is tangent to both axes. Draw a line which is tangent to the circle, letting s and t be the coordinates where it crosses the axes. Find the relationship between s and t.

Solution

s + t = 1 + \tfrac{1}{2} st

Note

Geometry Expressions can solve this problem handily!

Other ambient geometries

Crystal structures

Source

Peter Fenyes

Statement

Assemble triply periodic sphere packings like the ones sometimes used to illustrate crystal structures.