The five regular solids
There is a regular polygon with any number of sides you like. How many solids can be built from copies of one regular polygon, with every corner exactly alike?
Where you are 7 in Three-dimensional shapes
Why this exists
In a flat plane, regular polygons never run out. There is an equilateral triangle, a square, a regular pentagon, a regular hexagon, and one for every number of sides after that, each with its own interior angle.
Go up one dimension and ask the same question of solids. Take copies of a single regular polygon and fit them together into a closed solid, so that every face is like every other face and every corner is like every other corner. A fair die is made this way: no face and no corner is special, so no side is favored. You might expect one such solid for each polygon, an endless list again.
There are exactly five. Not five that people happened to find: five, and a proof that a sixth cannot exist. The proof needs only the angle of a regular polygon and one fact about corners, and it is one of the oldest complete classifications in mathematics. The last book of Euclid’s Elements builds all five and ends by showing that no other can be built. They are often called the Platonic solids, after Plato, who wrote about them in his dialogue the Timaeus.
Definition
A polyhedron is convex when the segment joining any two of its points stays inside it: no dents, no holes, no faces folded inward.
A regular polyhedron, or Platonic solid, is a convex polyhedron in which
- every face is a regular polygon, and all the faces are congruent; and
- the same number of faces meet at every vertex.
Both conditions are needed. The Boundaries section shows a solid that meets the first and not the second.
A regular polyhedron is described by two whole numbers: p, the number of sides of each face, and q, the number of faces that meet at each vertex. It is written \{p, q\}. A cube has square faces with three at every corner, so the cube is \{4, 3\}.
Throughout, V, E and F count the vertices, edges and faces of a solid. Figure 1 shows the five.
Intuition
Look at a single corner. Take a cardboard corner, cut it along one edge and press it flat on the table. The faces round the vertex lie down as a fan of angles about one point, and because the corner used to stand up out of the table, the fan does not reach all the way round. A gap is left where the cut was. Fold the gap shut and the corner stands up again.
So the face angles at a corner must add to less than 360^\circ. If they add to exactly 360^\circ, the faces lie flat and there is no corner to fold. If they add to more, they cannot be laid round a point without overlapping, and they cannot make an outward-pointing corner.
Now build corners from regular polygons, using the interior angle of each. An equilateral triangle has angles of 60^\circ: three, four or five of them leave a gap, and six fill the turn exactly. A square has 90^\circ angles: three leave a gap and four lie flat. A regular pentagon has 108^\circ angles: three make 324^\circ and leave a gap, while four make 432^\circ and overlap. A regular hexagon has 120^\circ angles, so even three of them fill the turn, and every polygon with more sides has a larger angle still.
That leaves exactly five corners, the five in Figure 2, and each one turns out to close up into one of the solids of Figure 1.
The fans that close up with no gap at all are worth seeing too. They cannot make a corner, but they cover the flat floor round the point (Figure 3).
Results
Corner condition. At least three faces meet at every vertex of a convex polyhedron, and the face angles at the vertex add to less than 360^\circ.
There are exactly five regular solids. The symbol \{p, q\} of a regular solid must satisfy
which is the same as
and only five pairs of whole numbers p \ge 3, q \ge 3 do. Each pair is the pattern of exactly one solid, apart from its size.
| Solid | \{p, q\} | F | E | V |
|---|---|---|---|---|
| tetrahedron | \{3, 3\} | 4 | 6 | 4 |
| cube | \{4, 3\} | 6 | 12 | 8 |
| octahedron | \{3, 4\} | 8 | 12 | 6 |
| dodecahedron | \{5, 3\} | 12 | 30 | 20 |
| icosahedron | \{3, 5\} | 20 | 30 | 12 |
The face angles at one vertex add to 3 \cdot 60^\circ = 180^\circ for the tetrahedron, 3 \cdot 90^\circ = 270^\circ for the cube, 4 \cdot 60^\circ = 240^\circ for the octahedron, 3 \cdot 108^\circ = 324^\circ for the dodecahedron and 5 \cdot 60^\circ = 300^\circ for the icosahedron, leaving gaps of 180^\circ, 90^\circ, 120^\circ, 36^\circ and 60^\circ.
The names count faces in Greek: tetra four, hexa six (the cube is also called a hexahedron), octa eight, dodeca twelve, icosa twenty.
Counting edges. Every edge borders exactly two faces and has exactly two ends, so for a regular solid
Euler’s formula. Every convex polyhedron has
which is proved through planar graphs. For a regular solid \{p, q\} it fixes the number of edges:
Total gap. Call 360^\circ minus the sum of the face angles at a vertex the gap at that vertex. For every convex polyhedron, regular or not, the gaps at all its vertices add to exactly 720^\circ.
Duals. Swapping p and q turns one regular solid into another: the cube \{4, 3\} and the octahedron \{3, 4\} are a pair, so are the dodecahedron and the icosahedron, and the tetrahedron \{3, 3\} is paired with itself. Joining the centers of neighboring faces of one solid of a pair builds the other. The pair have the same E, and their V and F are swapped.
Why it is true
At least three faces meet at a vertex. The faces round a vertex form a ring, each sharing an edge with the next. If only two faces met there, both would contain the same two edges leaving the vertex. Two rays from one point lie in a single plane, so the two faces would lie in that plane on top of each other near the vertex, which is a folded sheet and not a corner.
The angles at a convex corner add to less than 360^\circ. Euclid proves this in Book XI, Proposition 21 of the Elements. The flattening picture in Intuition shows why it should be so: the gap left when the corner is pressed flat is the amount by which the corner stands up.
Only five pairs pass the corner test. The interior angle of a regular p-gon is \displaystyle \frac{(p - 2) \cdot 180^\circ}{p}, which equals \displaystyle 180^\circ - \frac{360^\circ}{p} and so grows as p grows. With q \ge 3 faces at each vertex, test each p in turn.
| p | Angle | Test | Allowed q |
|---|---|---|---|
| 3 | 60^\circ | q < 6 | 3, 4, 5 |
| 4 | 90^\circ | q < 4 | 3 |
| 5 | 108^\circ | \displaystyle q < \frac{10}{3} | 3 |
| \ge 6 | \ge 120^\circ | 3 \cdot 120 \ge 360 | none |
Each test is q times the angle less than 360: 60q < 360, 90q < 360, 108q < 360. For p \ge 6 even the smallest q, which is 3, fails.
The same count comes out of the second form of the condition. Multiply \displaystyle q \cdot \frac{(p - 2) \cdot 180}{p} < 360 by \displaystyle \frac{p}{180}, which is positive, to get q(p - 2) < 2p, that is pq - 2p - 2q < 0. Adding 4 to both sides gives pq - 2p - 2q + 4 < 4, and the left side factorizes as (p - 2)(q - 2). Both factors are whole numbers at least 1, and the only ways to make a product less than 4 are 1 \cdot 1, 1 \cdot 2, 2 \cdot 1, 1 \cdot 3 and 3 \cdot 1. These are \{3, 3\}, \{3, 4\}, \{4, 3\}, \{3, 5\} and \{5, 3\}.
Each of the five exists. Passing the corner test only says one corner can be folded. Whether the corners can be fitted together into a closed solid is a separate question, answered by building each one. Coordinates in space do it. The eight points (\pm 1, \pm 1, \pm 1) are the corners of a cube. Example 5 builds the octahedron from the centers of its faces and Example 6 the tetrahedron from four of its corners. Example 7 checks that the twelve points (0, \pm 1, \pm\varphi), (\pm 1, \pm\varphi, 0) and (\pm\varphi, 0, \pm 1), where \displaystyle \varphi = \frac{1 + \sqrt{5}}{2}, are the corners of an icosahedron. The centers of the icosahedron’s twenty faces are then the corners of a dodecahedron, its dual. That each pattern builds only one solid, apart from size, is also true, but its proof is longer and is not given here.
Counting edges. List the sides of every face: there are pF of them. Each edge is a side of exactly two faces, so it is listed twice, and pF = 2E. In the same way, list the edges at each vertex: there are qV, and each edge is listed once at each of its two ends, so qV = 2E.
Euler’s formula gives the same five. Put \displaystyle F = \frac{2E}{p} and \displaystyle V = \frac{2E}{q} into V - E + F = 2:
so
Since E and 2 are positive, the bracket must be positive. Multiplying \displaystyle \frac{2}{p} + \frac{2}{q} > 1 by pq gives 2q + 2p > pq, which is pq - 2p - 2q < 0: the same inequality the corners gave, reached by counting instead of by angles. Solving for E gives the formula in Results.
Why the gaps add to 720^\circ. A face with n sides has angles adding to (n - 2) \cdot 180^\circ, by the polygon angle sum. Add this over every face. The numbers n add to 2E, since each edge is a side of two faces, and there are F faces, so all the face angles of the solid add to (2E - 2F) \cdot 180^\circ. Those same angles, grouped by vertex instead of by face, are the angle sums at the V vertices. So the gaps add to
Duals. Put a point at the center of each face of \{p, q\} and join two points when their faces share an edge. Round each old vertex sit q faces, and their q centers become a new face with q sides. Each old face has p corners, so its center lies on p new faces. The new solid has faces with q sides, p at each vertex: it is \{q, p\}. Its vertices are the old faces and its faces are the old vertices, so V and F swap, and each new edge crosses one old edge, so E stays.
Worked examples
Counting the icosahedron
An icosahedron has 20 faces, all equilateral triangles, with five meeting at every vertex. Find its number of edges and of vertices, and check them with Euler’s formula.
SolutionThe 20 triangles have 20 \cdot 3 = 60 sides between them. Each edge of the solid is a side of exactly two triangles, so the 60 sides count every edge twice:
The triangles also have 60 corners between them. Each vertex of the solid is a corner of exactly five triangles, so the 60 corners count every vertex five times:
Check. V - E + F = 12 - 30 + 20 = 2, as Euler’s formula requires.
Which corners can be built
Decide whether each of these can be the corner of a regular solid: (a) four regular pentagons; (b) four equilateral triangles; (c) three regular octagons; (d) four squares.
SolutionA corner can be folded only when the face angles add to less than 360^\circ, so add them.
(a) A regular pentagon’s angle is 108^\circ, and 4 \cdot 108^\circ = 432^\circ > 360^\circ. The four pentagons overlap when laid round a point, so no corner.
(b) 4 \cdot 60^\circ = 240^\circ < 360^\circ, leaving a gap of 120^\circ. This corner works; it is the corner of the octahedron \{3, 4\}.
(c) A regular octagon’s angle is \displaystyle \frac{(8 - 2) \cdot 180^\circ}{8} = 135^\circ, and 3 \cdot 135^\circ = 405^\circ > 360^\circ. Even the fewest faces a corner can have already overlap, so no regular solid has octagonal faces.
(d) 4 \cdot 90^\circ = 360^\circ exactly. The squares lie flat round the point, like floor tiles, and there is no gap to fold shut. No corner.
Euler’s formula finds the dodecahedron
A regular solid has faces that are regular pentagons, with three at each vertex. Using only pF = 2E = qV and Euler’s formula, find F, E and V.
SolutionHere p = 5 and q = 3, so 5F = 2E and 3V = 2E, which give \displaystyle F = \frac{2E}{5} and \displaystyle V = \frac{2E}{3}. Substitute both into V - E + F = 2:
Multiply both sides by 15 to clear the fractions:
so E = 30. Then \displaystyle F = \frac{2 \cdot 30}{5} = 12 and \displaystyle V = \frac{2 \cdot 30}{3} = 20.
Check. The formula in Results gives \displaystyle E = \frac{2 \cdot 5 \cdot 3}{10 + 6 - 15} = 30 as well, and 20 - 30 + 12 = 2.
Euler’s formula rules out hexagons
Show that no convex polyhedron has every face a hexagon, regular or not.
SolutionSuppose a convex polyhedron had only hexagons for faces. Each face has 6 sides and each edge borders two faces, so 6F = 2E and \displaystyle F = \frac{E}{3}.
At every vertex at least three faces meet, and so at least three edges meet. Counting edge ends vertex by vertex gives 2E in total, at least 3 at each vertex, so 2E \ge 3V, which is \displaystyle V \le \frac{2E}{3}. Then
But Euler’s formula says V - E + F = 2, and 2 is not at most 0. So there is no such polyhedron.
The argument never used that the hexagons were regular, or even equal. Every convex polyhedron needs some faces with fewer than six sides, which is why a soccer, Example 8, cannot be made of hexagons alone.
The octahedron inside a cube
A cube has its corners at the eight points (\pm 1, \pm 1, \pm 1). The centers of its six faces are (\pm 1, 0, 0), (0, \pm 1, 0) and (0, 0, \pm 1) (Figure 4). Show that these six points are the corners of a regular octahedron, and find its edge length.
Take two centers on faces that are not opposite, such as (1, 0, 0) and (0, 1, 0). Both have third coordinate 0, so both lie in one flat plane, and the distance formula in that plane gives
Any other two centers on faces that are not opposite also have a single \pm 1 each, in different places, and a 0 in their shared third place, so the same calculation gives \displaystyle \sqrt{2} every time. A center and the one opposite it, such as (1, 0, 0) and (-1, 0, 0), are 2 apart and are not joined. So each point is \displaystyle \sqrt{2} from the four points that are not itself or its opposite.
For the faces, choose one point from each opposite pair: one of (\pm 1, 0, 0), one of (0, \pm 1, 0) and one of (0, 0, \pm 1). That is 2 \cdot 2 \cdot 2 = 8 choices, and each gives a triangle whose three sides are all \displaystyle \sqrt{2}: an equilateral triangle. A point such as (1, 0, 0) lies in the 2 \cdot 2 = 4 triangles that choose it.
So the solid has 8 congruent equilateral faces with 4 at every vertex. It is the regular octahedron \{3, 4\}, with edge length \displaystyle \sqrt{2}.
The tetrahedron inside a cube
A cube of side 1 has its corners at the eight points whose coordinates are each 0 or 1. Show that (0, 0, 0), (1, 1, 0), (1, 0, 1) and (0, 1, 1) are the corners of a regular tetrahedron (Figure 5), and find its edge length.
Compare any two of the four points: (0, 0, 0) and (1, 1, 0) differ in the first two coordinates; (1, 1, 0) and (1, 0, 1) in the last two; (1, 0, 1) and (0, 1, 1) in the first two; and so on for all six pairs. Each pair differs in exactly two coordinates, so the two points are opposite corners of one face of the cube, a square of side 1. By Pythagoras that diagonal is
So all six distances between the four points are \displaystyle \sqrt{2}. Any three of the points form an equilateral triangle of side \displaystyle \sqrt{2}, there are four such triangles, and three of them meet at each point. The solid is the regular tetrahedron \{3, 3\} with edge length \displaystyle \sqrt{2}.
The four corners left over, (1, 0, 0), (0, 1, 0), (0, 0, 1) and (1, 1, 1), make a second tetrahedron the same way.
The icosahedron by coordinates
Let \displaystyle \varphi = \frac{1 + \sqrt{5}}{2}, and take the twelve points (0, \pm 1, \pm\varphi), (\pm 1, \pm\varphi, 0) and (\pm\varphi, 0, \pm 1). Show that the point A = (0, 1, \varphi) is exactly 2 from five of the others, and that those five sit round A in a ring, each 2 from the next.
SolutionDistances come from the distance formula in space, which adds a third square under the root: \displaystyle \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}.
One fact about \varphi does all the work:
From A = (0, 1, \varphi):
- to (0, -1, \varphi) only the middle coordinate changes, by 2, so the distance is 2;
- to (1, \varphi, 0) the changes are 1, \varphi - 1 and \varphi, and 1 + (\varphi - 1)^2 + \varphi^2 = 2\varphi^2 - 2\varphi + 2, which is 2(\varphi + 1) - 2\varphi + 2 = 4, so the distance is 2, and the same holds for (-1, \varphi, 0);
- to (\varphi, 0, 1) the changes are \varphi, 1 and \varphi - 1, the same three numbers in another order, so the distance is again 2, and the same holds for (-\varphi, 0, 1).
That is five points at distance 2. The other six are further: (0, 1, -\varphi) is 2\varphi away, so are (\pm 1, -\varphi, 0) and (\pm\varphi, 0, -1), since 1 + (\varphi + 1)^2 + \varphi^2 = 2\varphi^2 + 2\varphi + 2, which is 4\varphi + 4 = 4\varphi^2, and the point (0, -1, -\varphi) opposite A is further still (Exercise 20).
Now go round the five neighbors in the order (0, -1, \varphi), (\varphi, 0, 1), (1, \varphi, 0), (-1, \varphi, 0), (-\varphi, 0, 1) and back. Each step is 2 long: for instance (1, \varphi, 0) to (-1, \varphi, 0) changes only the first coordinate, by 2, and (\varphi, 0, 1) to (1, \varphi, 0) has changes \varphi - 1, \varphi and 1, which give 2 as before. So five equilateral triangles of side 2 meet at A.
Finally, every point of the list looks like A. Changing signs, or moving each coordinate one place along as (x, y, z) \to (y, z, x), carries the list of twelve to itself and keeps every distance, and these moves can carry A to any of the twelve. So every point has five triangles round it, and there are \displaystyle \frac{12 \cdot 5}{3} = 20 triangles in all: the icosahedron \{3, 5\}.
Counting a soccer by its gaps
The panels of a classic soccer (a soccer ball) follow a convex polyhedron with every vertex alike: at each one, a regular pentagon and two regular hexagons meet. How many vertices does it have, and how many pentagons and hexagons?
SolutionThis solid is not regular, but the total gap of 720^\circ holds for every convex polyhedron. The angles at one vertex add to 108^\circ + 120^\circ + 120^\circ = 348^\circ, so each vertex has a gap of 360^\circ - 348^\circ = 12^\circ. All the vertices are alike, so
Each vertex lies on exactly one pentagon, and each pentagon has 5 corners, so there are \displaystyle \frac{60}{5} = 12 pentagons. Each vertex lies on two hexagons, and each hexagon has 6 corners, so there are \displaystyle \frac{60 \cdot 2}{6} = 20 hexagons.
Check. F = 12 + 20 = 32, and the sides number 12 \cdot 5 + 20 \cdot 6 = 180, so E = 90. Then V - E + F = 60 - 90 + 32 = 2.
Boundaries
Convex is part of the definition. If faces may pass through one another, or may be star-shaped polygons, there are four more solids with all faces alike and all corners alike, the Kepler-Poinsot solids. Two of them have five-pointed stars for faces, and in the other two the faces round each corner wrap twice round it, so the corner test of this lesson does not reach them. “Exactly five” is a statement about convex solids.
Both conditions are needed. Glue two regular tetrahedra face to face (Figure 6). The result is convex and its six faces are congruent equilateral triangles, but three faces meet at the top and bottom points and four at each of the three points round the middle. It is not regular. The opposite failure happens too: a prism whose ends are equilateral triangles and whose sides are squares has every corner alike (one triangle and two squares) but two kinds of face.
The corner test is necessary, not enough. It shows that a single corner can be folded. It does not show that copies of that corner fit together into a closed solid; that has to be built, as in Example 7.
Euler’s formula needs a solid without holes. A polyhedron shaped like a picture frame, with a hole through the middle, has V - E + F = 0, not 2. The count of five, the edge formula and the 720^\circ total all rest on the formula, so they are statements about convex solids.
At exactly 360^\circ the solid becomes a floor. The three pairs with (p - 2)(q - 2) = 4, namely \{3, 6\}, \{4, 4\} and \{6, 3\}, give no corner, but they tile the flat plane: these are the three regular Tessellations.
Five shapes, any size. Every regular solid can be scaled up or down. “Exactly five” counts shapes, not sizes.
Where this goes next
The counting in this lesson is graph theory in disguise. Flatten a solid’s vertices and edges onto the plane and it becomes a drawing with no crossings, and Euler’s formula for planar graphs is the general fact that made Examples 3, 4 and 8 work.
The flat corners of Figure 3 lead to Tessellations: the same angle test, with “less than” turned into “equal to”, finds every way to tile a floor with one regular polygon.
The regular solids are also the most symmetric solids there are. Every face can be turned onto every other, and the turns that carry a solid onto itself number 12 for the tetrahedron, 24 for the cube and octahedron and 60 for the dodecahedron and icosahedron (Exercise 25). That symmetry is why all five are used as fair dice.
The same question can be asked in more dimensions, with the solids as the faces. In four dimensions there are six regular shapes of this kind, and in every dimension above four there are only three.
Their surface areas follow from the area of one face, since all the faces are the same.
Exercises
Reading the definition
- A cube has square faces with three at every vertex. Write its symbol \{p, q\}, then use pF = 2E = qV with F = 6 to find E and V.
- A regular octahedron has 8 equilateral triangle faces, with four at every vertex. Find E and V, and check them with Euler’s formula.
- A square pyramid has all eight of its edges the same length, so its four sloping faces are equilateral triangles. Is it a regular polyhedron? Explain.
- A prism has equilateral triangles for its two ends and squares for its three sides. Every vertex has one triangle and two squares meeting at it. Is it a regular polyhedron? Explain.
- Which regular solid has faces that are regular pentagons? How many faces meet at each of its vertices, and why could it not be four?
Angles at a corner
- Find the sum of the face angles at one vertex of a regular octahedron, and the gap.
- Find the sum of the face angles at one vertex of a regular dodecahedron, and the gap.
- Find the interior angle of a regular heptagon (seven sides). Can a regular solid have heptagonal faces?
- Using the gap at each vertex, list the five regular solids from the sharpest corner (largest gap) to the bluntest (smallest gap).
- Work out (p - 2)(q - 2) for \{3, 6\}, \{4, 4\} and \{6, 3\}. What does the answer say about these corners?
Counting with Euler’s formula
- A convex polyhedron has 14 faces and 24 edges. How many vertices does it have?
- A convex polyhedron has 8 equilateral triangles and 6 squares for faces, with four faces meeting at every vertex. Find E and V, and check Euler’s formula.
- At every vertex of a convex polyhedron, one equilateral triangle and two regular octagons meet. Use the gaps to find V, then find the number of triangles and of octagons, and check Euler’s formula.
- At every vertex of a convex polyhedron, two equilateral triangles and two regular pentagons meet. Find V, the number of triangles and the number of pentagons.
- Check that the gaps of the glued pair of tetrahedra in Figure 6 add to 720^\circ, and check Euler’s formula for it.
Duals and solids inside solids
- Use \displaystyle E = \frac{2pq}{2p + 2q - pq} to find E for the cube \{4, 3\} and the octahedron \{3, 4\}, then find V and F for each. What do you notice?
- Put a point at the center of each face of a regular octahedron and join the points on neighboring faces. Which solid do you get? Give its number of vertices and faces.
- A regular octahedron has its corners at (\pm 2, 0, 0), (0, \pm 2, 0) and (0, 0, \pm 2). Find its edge length, and the distance between two opposite corners.
- A cube has side 3. Four of its corners, no two joined by an edge of the cube, are the corners of a regular tetrahedron. Find the tetrahedron’s edge length.
- Using the coordinates of Example 7, find the distance from (0, 1, \varphi) to the opposite point (0, -1, -\varphi). Give it in the form \displaystyle \sqrt{a + b\sqrt{5}}.
Reasoning
- Explain why at least three faces must meet at every vertex of a polyhedron.
- Five equilateral triangles fit round a point with a 60^\circ gap. Explain why this alone does not prove that an icosahedron exists.
- Explain why no regular polyhedron has faces with six or more sides.
- Let \displaystyle L = q \cdot \frac{(p - 2) \cdot 180}{p} - 360. Show that \displaystyle L = \frac{180}{p}\big((p - 2)(q - 2) - 4\big), and explain why this shows the two forms of the condition in Results say the same thing.
- A turn of a regular solid can carry any face onto any chosen face, and then the face can be turned in p ways onto itself. Explain why the number of turns that carry the solid onto itself is pF, find it for each of the five, and explain why the two solids in a dual pair give the same number.
Answers
- \{4, 3\}, with E = 12 and V = 8. From 4 \cdot 6 = 2E, E = 12; from 3V = 24, V = 8.
- E = 12 and V = 6. The triangles have 8 \cdot 3 = 24 sides, each edge counted twice, so E = 12; their 24 corners count each vertex four times, so V = 6. Then 6 - 12 + 8 = 2.
- No. Its faces are four triangles and a square, so they are not all congruent. The corners differ too: four faces meet at the top and three at each corner of the base.
- No. Its corners are all alike, but its faces are of two kinds, triangles and squares, so they are not all congruent.
- The dodecahedron, with three faces at each vertex. Three pentagon angles make 3 \cdot 108^\circ = 324^\circ < 360^\circ, but four make 4 \cdot 108^\circ = 432^\circ, which is more than a full turn.
- Sum 240^\circ, gap 120^\circ. Four triangles: 4 \cdot 60^\circ = 240^\circ, and 360^\circ - 240^\circ = 120^\circ.
- Sum 324^\circ, gap 36^\circ. Three pentagons: 3 \cdot 108^\circ = 324^\circ, and 360^\circ - 324^\circ = 36^\circ.
- \displaystyle \left(\frac{900}{7}\right)^\circ \approx 128.6^\circ, and no. The angle is \displaystyle \frac{5 \cdot 180^\circ}{7}. Even three of them make \displaystyle \left(\frac{2700}{7}\right)^\circ \approx 385.7^\circ, more than 360^\circ.
- Tetrahedron (180^\circ), octahedron (120^\circ), cube (90^\circ), icosahedron (60^\circ), dodecahedron (36^\circ). Each gap is 360^\circ minus the angle sum at a vertex, as listed under the table in Results.
- All three give 4: 1 \cdot 4, 2 \cdot 2 and 4 \cdot 1. A product of exactly 4 means the angles add to exactly 360^\circ, so the polygons lie flat and make no corner. They tile the plane instead.
- V = 12. From V - E + F = 2, V = 2 + 24 - 14 = 12.
- E = 24 and V = 12. The faces have 8 \cdot 3 + 6 \cdot 4 = 48 sides, so E = 24. Four edges meet at each vertex, so 4V = 2E = 48 and V = 12. Then 12 - 24 + 14 = 2.
- V = 24, with 8 triangles and 6 octagons. The angles at a vertex add to 60^\circ + 135^\circ + 135^\circ = 330^\circ, a gap of 30^\circ, so \displaystyle V = \frac{720}{30} = 24. Each vertex is on one triangle: \displaystyle \frac{24}{3} = 8. Each is on two octagons: \displaystyle \frac{24 \cdot 2}{8} = 6. Sides 8 \cdot 3 + 6 \cdot 8 = 72, so E = 36, and 24 - 36 + 14 = 2.
- V = 30, with 20 triangles and 12 pentagons. The angles add to 60^\circ + 60^\circ + 108^\circ + 108^\circ = 336^\circ, a gap of 24^\circ, so \displaystyle V = \frac{720}{24} = 30. Triangles: \displaystyle \frac{30 \cdot 2}{3} = 20. Pentagons: \displaystyle \frac{30 \cdot 2}{5} = 12.
- Both hold. The two points with three triangles have angle sum 180^\circ and gap 180^\circ; the three points with four triangles have angle sum 240^\circ and gap 120^\circ. In all, 2 \cdot 180^\circ + 3 \cdot 120^\circ = 720^\circ. The solid has V = 5, E = 9 and F = 6, and 5 - 9 + 6 = 2.
- Both have E = 12. For the cube, \displaystyle E = \frac{24}{8 + 6 - 12} = 12, \displaystyle F = \frac{2 \cdot 12}{4} = 6 and \displaystyle V = \frac{2 \cdot 12}{3} = 8. For the octahedron, \displaystyle E = \frac{24}{6 + 8 - 12} = 12, F = 8 and V = 6. The formula for E does not change when p and q swap, while \displaystyle F = \frac{2E}{p} and \displaystyle V = \frac{2E}{q} trade places: the two are duals.
- A cube, with 8 vertices and 6 faces. There is one new point for each of the 8 faces. Round each of the 6 old vertices sit four faces, whose centers make a square face. Each old face is a triangle, so each new point lies on three new faces: \{4, 3\}.
- Edge \displaystyle 2\sqrt{2}; opposite corners 4 apart. For (2, 0, 0) and (0, 2, 0), in the plane of the first two coordinates, the distance is \displaystyle \sqrt{4 + 4} = 2\sqrt{2}. From (2, 0, 0) to (-2, 0, 0) is 4.
- \displaystyle 3\sqrt{2}. Each edge is a diagonal of a square face of side 3: \displaystyle \sqrt{9 + 9} = 3\sqrt{2}.
- \displaystyle \sqrt{10 + 2\sqrt{5}} \approx 3.80. The changes are 0, 2 and 2\varphi, so the square of the distance is 4 + 4\varphi^2 = 4 + 4(\varphi + 1), which is \displaystyle 8 + 4\varphi = 10 + 2\sqrt{5}.
- The faces round a vertex form a ring, each sharing an edge with the next. With only two faces, both would contain the same two edges from the vertex. Two rays from a point lie in one plane, so the two faces would lie flat on each other there, and there would be no corner.
- The test shows that one corner of five triangles can be folded. It does not show that twenty triangles can be fitted together so that every vertex is such a corner and the surface closes up. That needs a construction, such as the coordinates of Example 7.
- A regular polygon with p \ge 6 sides has interior angle \displaystyle 180^\circ - \frac{360^\circ}{p} \ge 120^\circ. At least three faces meet at a vertex, so the angles there add to at least 3 \cdot 120^\circ = 360^\circ, which a convex corner cannot have.
- Expanding the right side gives \displaystyle \frac{180}{p}(pq - 2p - 2q), which is \displaystyle 180q - 360 - \frac{360q}{p}, and L itself is \displaystyle 180q - \frac{360q}{p} - 360, the same. Since \displaystyle \frac{180}{p} > 0, L is negative exactly when (p - 2)(q - 2) - 4 is. So the angle sum is less than 360 exactly when (p - 2)(q - 2) < 4.
- 12, 24, 24, 60 and 60. A turn is fixed by which face lands on the chosen face, F choices, and how that face is turned, p choices, so there are pF. Tetrahedron 3 \cdot 4 = 12; cube 4 \cdot 6 = 24; octahedron 3 \cdot 8 = 24; dodecahedron 5 \cdot 12 = 60; icosahedron 3 \cdot 20 = 60. Since pF = 2E and a dual pair share the same E, they share the same number of turns.