ITSPHUN geometry

Each ITSPHUN construct is a model, or approximation, of some abstract, mathematical object.
The relation between the octahedron and an ITSPHUN model (built with our spiral-shaped wood pieces) is illustrated below.
ITSPHUN octahedron
ITSPHUN piece
ITSPHUN "octahedron"
Spiral piece and its underlying polygon

ITSPHUN activities

Here are some fun activities, games and puzzles you can do with ITSPHUN pieces. Contact us if you have any suggestions,

Build objects from a model or picture

ITSPHUN models
This is a great way to develop spatial visualization and pattern recognition skills. Check out our  photo gallery or download the file at right to get you started.

For an extra challenge, try counting how many pieces are required before starting to build your model.

Perfection

Icosahedron
Build the five  Platonic solids  whose perfect shapes have fascinated humans for thousands of years.
Icosahedron, drawing by Leonardo da Vinci

Have a ball!

Rhombicosidodecahedron

Hint: If by "ball" we mean the highly symmetrical  Platonic  and  Archimedean  solids whose vertices are on a sphere, then the answer is no, there are only 18 such objects. If by "ball" we mean anyconvex polyhedron with regular faces, the answer is still no - there are only 89 convex polyhedra with faces that are regular triangles, squares, pentagons, and hexagons.
Rhombicosidodecahedron

Billions and billions of stars

Stella octangula

Hint: Assuming you have enough pieces, there is no limit for the number and size of the non-convex polyhedra that you can build.
Stella octangula

Build by connecting

Connected objects
Two objects that contain a same-shaped piece can be joined by removing that piece from each and then connecting the freed notches. The picture shows two dodecahedra (red and yellow) and a pentagonal prism (blue). They are connected by common pentagonal faces.

Hint: Two objects can be connected only when they have the same orientation (the notches of each face point in the same direction, either clockwise or counter-clockwise). If the two objects have opposite orientations, you will have to reconstruct one of the objects (turn it "inside out").

What cannot be built with ITSPHUN?

  • Try to make a Möbius strip with ITSPHUN pieces.

You will not be able to close the  Möbius strip since the orientation of the notches is reversed!
In general, you can build only  orientable surfaces with ITSPHUN.

Build by connecting, part 2

Augmented dodecahedron
The model at right can be build as follows:
  1.  Build a dodecahedron
  2.  Connect, as described above, 12 pentagonal pyramids to the 12 faces of the dodecahedron
  3.  Connect a tetrahedron to each triangular face of each pentagonal pyramid

Hint: Note that the faces of the original dodecahedron and of all the pentagonal pyramids have been removed in the final object. It is easier to build this model directly using 180 triangles, but the best way to describe it is in terms of simpler components.

Build from description

Truncated octahedron
  • Build a polyhedron that has only square faces and  hexagonal faces and has the following properties: (a) each square has only hexagonal faces as neighbors, and (b) each hexagonal face has three square and three hexagonal neighbors that alternate.

Hint: The polyhedron has 6 squares and 8 hexagons. 
Answer: 
truncated octahedron
 
  • Build the object obtained by replacing every face of a cube with a square pyramid  without the base.

Build from description game for two-players teams: one team member looks at an object (or a picture of an object) and describes it to the other team member who must build it. The team get points if the built object matches the original one.
Augmented cube

Build with colors

  • For a given object, what is the minimum number of colors needed so that no two same-colored faces touch each other along an edge?

Answer: tetrahedron: 4, cube: 3, octahedron: 2, dodecahedron: 4, icosahedron: 3, cuboctahedron: 2, truncated tetrahedron: 4, truncated octahedron: 3, etc. At most 4 colors are needed to build any convex object in this manner (the four color theorem).

  • Which objects can be colored in this manner with just two colors?

Answer: A simple theorem (prove it!) states that an object can be colored with two colors iff it has an even number of faces meeting at each vertex. For example: octahedron, cuboctahedron, icosidodecahedron, rhombicuboctahedron, rhombicosidodecahedron, antiprisms, many Johnson solids (elongated square bipyramid, ortho/gyro bicupolae, birotunda, cupola-rotunda, etc.)
 
  • What if two faces of the same color are not allowed to touch even at a vertex?
     
Answer: tetrahedron: 4, cube: 3, octahedron: 4, dodecahedron: 4, icosahedron: 5, cuboctahedron: 5, truncated tetrahedron: 4, truncated octahedron: 3, etc.
 
  • For a given object, how many faces of the same color can you place so that they do not touch each other at an edge/corner?

Answer: tetrahedron: 1/1, cube: 2/2, octahedron: 4/2, dodecahedron: 3/3, icosahedron: 8/4, cuboctahedron: 8/4, truncated tetrahedron: 4/4, truncated: octahedron: 6/6, etc.
 
Cuboctahedron
Cuboctahedron:
same color faces do not touch - two colors are sufficient

Build from net

Tridiminished icosahedron net
Tridiminished icosahedron
 
  • Build an object given its net (see picture at right).
 
Answer:  tridiminished icosahedron .

Net puzzle

Triangular prism
Which figure below is a net of the triangular prism ?
 
Triangular prism net
Triangular prism net?
Triangular prism net?
A
B
C
Triangular prism
Answer: B.

Build as a tag game

  • Build an object as follows. Choose the first piece (the it piece). You can only make connections to the it piece. After each connection, the piece that is connected to itbecomes the new it (this can be a new piece or one that is already part of the object). Continue until you cannot make any new connections ("it" is already connected on all sides). Can you build the entire object (make all connections) in this manner?

For example, in the picture at right, assume Red is it. After connecting it to Green,
Green becomes it. After the connection Green-Yellow, Yellow becomes it, and then Blue.
If we connect Blue to Green, Green is it again and we need to add a new piece to Green, etc.

Hint: always keep it in your hand and make the connections with your other hand.
Tetrahedron: no, cube: yes, octahedron: no, triangular prism: yes,
dodecahedron: no, truncated octahedron: yes, etc.

Does it matter:
  1. which piece you choose as your starting it, and
  2. in which order you make the connections?

Answer: (1) cube: no, triangular prism: yes, truncated octahedron: no, etc. (2) yes.
 
  • Can you predict beforehand which objects can be completed this way?

Hint: this is the famous "Seven Bridges of Königsberg" problem in graph theory; an object can be built in this way if the graph that has as nodes the object faces has an  Eulerian path .
 
  • Change the game so that a new it can only be a new piece that you're adding (make all connections to the old it before moving to the new one). Which objects can you build now?

Hint: you must now find a  Hamiltonian path , a much harder problem.
Platonic solids - yes, cuboctahedron, icosidodecahedron - no (prove it!), etc.
ITSPHUN tag game

Fun with words

Increase your vocabulary! Impress your friends! Improve your diction!
  • Build the  Johnson solid  with the most impressive name. Casually mention the name to anybody who asks what you are doing (make sure to practice saying the name aloud).
 
Suggestions: gyrobifastigium, parabiaugmented hexagonal prism, metagyrate diminished rhombicosidodecahedron, hebesphenomegacorona, disphenocingulum, etc.
 
  • Build all Platonic/Archimedean/Johnson solids with fewer faces than letters in their names. Can you find any that have exactly as many faces as there are letters in their name?

Geometry word game. Setup: list of Johnson solids and a set of 92 numbered cards.
In turn, a player picks a card and reads aloud the name of the corresponding Johnson solid. If (s)he stumbles then (s)he must keep the card, otherwise (s)he discards it. After 5 rounds each player must build the objects on the cards (s)he has. First one to finish wins, last one must clean the room, or read aloud the list of all 92 Johnson solids (while standing on one leg), or just buy the next round of drinks.
Gyroelongated pentagonal cupolarotunda
Gyroelongated pentagonal cupolarotunda

Geometric art

Owl
Owl profile
  1. Build a  bilunabirotunda
  2. Build a  pentagonal orthobicupola
  3. Connect them to make the head and body of the owl; add the beak, legs, and tail. (Does an owl even have a tail?)

Hint: add additional pieces as needed. The plastic pieces are flexible and can be bended as needed. And yes, owls do have a short tail.
 
  • Build your own creations

Geometric art game

ITSPHUN creature
Each team builds a sculpture; the most creative/decorative/symmetric/funny/colorful/tall/etc. sculpture wins. Taking turns, each team member must add 3 pieces to the growing sculpture.
 
Variants:
  1. No communication is allowed between team members.
  2. Limited, verbal only, communication is allowed (no pointing).
  3. Removing pieces is also allowed - each team member is allowed 3 "moves", a move being either adding or removing a piece from the sculpture.

Near misses and large balls

Goldeberg polyhedron G(2,1)
Build some  Goldberg polyhedra  using hexagons and exactly 12 pentagons. We said above that you cannot build larger and larger convex polyhedra - what's the catch?

Hint: The hexagonal faces of the Goldberg polyhedra are not regular, but the difference is very small. Try building other near misses, e.g., a  tetrated dodecahedron .
Goldeberg polyhedron G(2,1)

Build with no rules

ITSPHUN object
Bend and twist the ITSPHUN pieces to connect them in ways they were not meant to be connected!
 
  • Build the model at right

Hint: reverse the weave of the bottom (hidden) hemisphere.
 
 

Build to infinity

Connected truncated octahedra
Certain polyhedra can be stacked together, in all directions, without leaving any holes or gaps. A familiar example is the cube; a less familiar one is the truncated octahedron. We can join (see "build by connection" above) such space filling objects forever and ever, creating larger and larger structures, and using more and more ITSPHUN pieces.
 
  • Connect a few truncated octahedra to get a feel on how they stack together to fill the space.
  • If we continue adding new such space-filling polyhedra, in all possible ways, ad infinitum we'll end up with an infinitely large object containing infinitely many pieces, right?
Answer: Wrong. We'll end up with nothing and we'll use 0 pieces. Indeed, assume the infinite construct has a face F. Then it is possible to add another polyhedron to the construct at F thus removing F, which is a contradiction. This shows that the final "object" has no faces.
Another way to think about it: every time we join two objects we remove the "interior" common face(s) and leave only the exterior faces. The partial constructs contain only these exterior faces that form the border of an increasingly large object. But, at infinity, the final, space-filling "object" has no border at all, and thus has no faces.

Build to infinity, part 2

Muoctahedron
Do not get discouraged by the previous activity - infinite polyhedra do exist! As a first example, start with the same space-filling with truncated octahedra.
Using 8 hexagons build a (partial) truncated octahedron (TO) without square faces.
Build a second TO with the opposite notch orientation and attach it to the first at a hexagonal face. Note that, unlike our usual connection method (see "build by connecting" above), we need to remove only one of the common faces when connecting - the internal face which is the intersection of the two TOs remains in the model.
Continue adding TOs until you fill the space (or until you run out of pieces).
 
 

Hint: a good coloring scheme will help. For example: 4 colors corresponding to the 4 families of parallel planes (pictured), or 2 colors so that only different colored hexagons are connected.

The result is a partial model of the 
muoctahedron , an infinite polyhedron discovered by H.S.M. Coxeter. This partial model is not a polyhedron since it always has some unfilled notches, i.e., edges with only one face (only at the limit, in the infinite muoctahedron, two faces meet at each edge). Unlike the polyhedron built in the previous activity, this model has all the hexagonal faces, including the internal ones, but none of the square faces.
 

Build to infinity, part 3

Mucube
There are many infinite polyhedra.
 
  • Build a model of the  mucube , another infinite regular polyhedron discovered by Coxeter.

Hint: Choose again either a coloring scheme with 3 colors corresponding to the 3 families of parallel planes (pictured), or 2 colors such that only different colored squares are connected.
 
  • Find another infinite polyhedron related to the space filling with truncated octahedra (see part 2).

Hint: Put back the square faces but remove half of the hexagonal ones. 
 

Counting

Dodecahedron
  • Use 12 pentagons to build a dodecahedron. How many connections did you make? (or, how many edges are there in a dodecahedron?)

Answer: Count the dodecahedron edges by counting the edges of its pentagonal faces. Each face has 5 edges for a total of 5x12=60 edges, but each dodecahedron edge is counted twice (from the 2 faces meeting at that edge), so the answer is 60/2=30.
 
  • How many vertices?

Answer: Each face has 5 vertices and there are 3 faces meeting at each dodecahedron vertex, so, using the same argument as above, we get 5x12/3=20 vertices.
 
  • Can we use the same method of counting edges for all (closed) polyhedra?

Answer: yes, since there are exactly 2 faces meeting at each edge. In case of a polyhedron with more than one kind of face, we have to count faces of each kind separately; for example, a cuboctahedron has (8x3 (from its triangular faces) + 6x4 (from its square faces)) / 2 = 24 edges.
 
  • Can we apply the same method of counting vertices to all polyhedra?

Answer: no, this method works only for polyhedra where the same number of faces meet at each vertex, for example isogonal (vertex-transitive) polyhedra. The Platonic and Archimedean solids are isogonal.

Beyond counting: interesting number patterns

  • Build several objects, then count their number of faces (F), edges (E) and vertices (V). Verify  Euler's formula  F - E + V = 2.
  • Build the Platonic solids and count their numbers of faces, edges and vertices. Can you find any patterns?

Hint: The cube and the octahedron are duals, so E(cube) = E(octahedron), F(cube) = V(octahedron), and V(cube) = F(octahedron).
Similar relations hold for the dodecahedron and icosahedron which are also duals.
The tetrahedron is its own dual, and indeed F(tetrahedron) = V(tetrahedron).

Symmetry and color

  • Build a cube using 6 squares of 3 colors, 2 squares of each color. Build several more cubes using squares of the same colors.
    Compare the cubes: are they the same or different? What does "different" mean?

Answer: two cubes are different if we cannot rotate one to obtain the other.
 
  • How many different cubes have this color scheme?​

Answer: 6. Try to build them all!
 
  • Does the answer change if we allow the cubes to be "turned inside out"?
    (This can be done by swapping opposite faces.)

Answer: yes, there are now only 5 different cubes.
With the new transformation, 2 of the 6 previously different cubes are no longer different. Which ones?
 
  • Is one cube "more symmetric" than the others? In what way?

Answer: the cube where the opposite faces have the same color is symmetric in the following sense: choose two faces which have the same color, for example top and bottom are both red. Rotate the cube in any way (even turn it inside out) and look at the top and bottom faces - they will still have the same color (not necessarily red).
If you start with two faces of different colors then, after any rotation of the cube, the same faces will still have different colors.
 
  • Repeat this activity with other coloring schemes (e.g. 2 colors, 3+3); note that symmetric cubes do not exist for all schemes.
  • Repeat with other simple objects, e.g., octahedron.
Cube

Symmetry and color, part 2

Cube
  • Build the 2 different cubes with the 3+3 coloring scheme (2 colors, 3 faces of each color).
    Which cube rotations are compatible (as defined above) with each coloring scheme?

Answer: For the first cube at right (F=front face, B=back, U=up, D=down, L=left, R=right): 
  1. the 120 and 240 degrees rotations around the cube diagonal from the blue vertex FUL to the orange vertex BDR (they preserve the colors exactly), and
  2. 180 rotations around the 3 semi-diagonal axes from one edge midpoint to the opposite edge midpoint: FR-BL, FD-BU, UR-DL (they swap the colors). Total: 6 rotations (including the "identity", i.e., the rotation that leaves the cube unchanged)
For the second cube:
  1. 180 degree rotation around the face axis UD preserves the colors,
  2. 180 degrees rotations around the semi-diagonal axes FL-BR and FR-BL swap the colors. Total 4 rotations, including the identity.
 
  • Repeat the activity for other objects and coloring schemes.
 
Advanced topic: the set of rotations compatible with the coloring form a subgroup G of the symmetry group  of the object. The set of rotations that preserve the colors exactly form a subgroup of G.
Cube

Triangular and tetrahedral numbers

Mutetrahedron
  • Build the object at right (a tetrahedron-shaped partial model of the  mutetrahedron , another infinite polyhedron) using 40 hexagons of 4 colors. Note that the parallel planes have the same color (two hexagons have the same color if and only if they are parallel).
  • How many green hexagons are needed to add another layer to this model?
 
Answer: Counting from the green face, through parallel planes, 4+3+2+1=10.
 
  • How many green hexagons are needed for the n-th layer?
Answer: 1+2+...+n = n(n+1)/2 ( triangular number )
 
  • What is the total number of green hexagons in a tetrahedron with n layers?

Answer: the sum of the first n triangular numbers = (1^2+2^2+...n^2 + 1+2+...+n)/2 = n(n+1)(n+2)/6 ( tetrahedral number )
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