Polyhedron Angles Add Up to 720°

Every vertex in a 3D shape has an angle, more technically called its angle defect, that basically tells you how flat or sharp it is. Why do the angle defects of every convex polyhedron add up to exactly 720°? This article is the written version of our video, with illustrations you can turn and fold yourself.

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A lopsided tetrahedron, with the angle defect at each of its vertices.

If it's flat, the angle defect is 0°. In a cube, the angle defect is 90°. In a really sharp corner, like the last one below, the angle defect is 270°. But in any convex polyhedron, these angle defects will add up to 720°.

Flat

Cube

Sharp corner

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A flat vertex, a corner of a cube, and a really sharp corner, with their angle defects.

What the angle defect is

Let's clarify what we mean by the angle defect. When you put polygons together to make a tessellation, the internal angles must all add up to 360°. But if we want the polygons to form part of a polyhedron, then we have to leave a gap, so that there's room for the vertex to fold up. The amount those internal angles add up to must be less than 360°. How much they're less than 360° is the angle defect.

Top: the angles round a vertex of a tiling add up to 360°. Bottom: take one polygon away, and the gap that's left is the angle defect.

You can see this for yourself by folding up some vertices below. However many polygons there are, the gap closes up just as the vertex folds into a corner.

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Fold up the vertex

Pick a vertex and fold it up. The orange gap is the angle defect.

But why do a polyhedron's angle defects add up to 720°?

Splitting faces into triangles

Let's simplify the question by assuming all the faces are triangles. We can reasonably assume that, because any polygon can be broken up into triangles. And it won't change the angle defect, because there will still be the same sum of angles at every vertex. Some angles might have been split up, but without changing the total.

A hexagon and two squares at a vertex, split into triangles two different ways. On the left, the vertex keeps its angles; on the right, they are split up. Either way, they still add up to 300°.

Euler's formula: V - E + F = 2

One thing we need to know is that for any convex shape, the number of vertices, V, minus the number of edges, E, plus the number of faces, F, equals 2. If that fact isn't familiar to you, we'll just give a quick demonstration by drawing a shape on the surface of a sphere, as we can do with any convex shape. The amount we're interested in is V - E + F.

Let's take away an edge between two faces, turning two faces into one, and so reducing the number of edges and the number of faces by one. V - E + F doesn't change. We can keep doing that until there's only one face left: the face of the whole sphere.

Since the sphere is the only face, any edges remaining are just joining vertices that are dotted about on the sphere. Those vertices can be removed, also removing the edge that joins them to the rest. So you remove one vertex and one edge. Again, V - E + F stays the same. If you keep doing that, you'll end up with just the one vertex left: one vertex, one face, and no edges. So V - E + F = 2.

A tetrahedron drawn on a sphere. Each step takes away an edge between two faces, or a vertex with the edge joining it to the rest, and V − E + F never changes.

And since it never changed, V - E + F was 2 all along. That's true for any shape that can be drawn on a sphere, which any convex polyhedron can.

	V	E	F	V - E + F
Tetrahedron	4	6	4	2
Cube	8	12	6	2
Dodecahedron	20	30	12	2

A shape made of triangles

So let's imagine a convex shape, completely made of triangles. We already know that V - E + F = 2. The number of triangles, or faces, is F. Since every triangle has three edges, but each edge belongs to two triangles, not just one, the number of edges is the number of faces multiplied by 3, then divided by 2:

\displaystyle E = \frac{3}{2}F

Substituting the number of edges into V - E + F = 2 gives:

\displaystyle \begin{aligned} V - \frac{3}{2}F + F &= 2 \\ V - \frac{F}{2} &= 2 \\ V &= 2 + \frac{F}{2} \end{aligned}

That means we know that the number of vertices is 2 plus half the number of faces. Remember that this is only guaranteed when the shape is made of triangles, and notice that it also means that the number of triangles can't be odd, otherwise the number of edges and vertices wouldn't be an integer.

	V	E	F
Tetrahedron	4	6	4
Triangular bipyramid	5	9	6
Octahedron	6	12	8
Pentagonal bipyramid	7	15	10
Icosahedron	12	30	20
Any shape made of triangles	2 + \frac{F}{2}	\frac{3}{2}F	F

The total angle defect is 720°

What's the total angle defect of our shape made of triangles? If every vertex's angle defect is 360° minus the internal angles there, then the total must be 360° for each vertex, minus the internal angles at all vertices.

The sum of all internal angles in a triangle is 180°, so the sum of all internal angles is 180° multiplied by the number of triangles. So our total angle defect is 360° for each vertex, minus 180° for each triangle.

Fold up the vertices

The lopsided tetrahedron (top), and its four vertices opened out flat; slide to fold them back up into its corners. Each colour's three marked angles are the corners of one triangle, so they add up to 180°. The four vertices have 4 × 360° round them, less 4 × 180° for the four triangles.

Since we had a formula for our number of vertices in terms of the number of triangles, V = 2 + \frac{F}{2}, we can substitute in that formula there, and find that our terms all cancel out except for 2 multiplied by 360°:

\displaystyle \begin{aligned} &\text{total angle defect} \\ &\quad = 360^\circ \times V - 180^\circ \times F \\ &\quad = 360^\circ \times \left(2 + \frac{F}{2}\right) - 180^\circ \times F \\ &\quad = 2 \times 360^\circ + 180^\circ \times F - 180^\circ \times F \\ &\quad = 2 \times 360^\circ \\ &\quad = 720^\circ \end{aligned}

So the total angle defect is 720°.

Any polyhedron, and Archimedean solids

Since we already showed that triangles can make up any shape, we can extend this proof to shapes with any faces. And if every vertex is the same, like in a Platonic solid or Archimedean solid, then we could find the angle defect just by dividing by the number of vertices. Or vice versa.

For example, if we know an Archimedean solid has an octagon, hexagon and square at a vertex, the internal angles of those shapes add up to 345°, so the angle defect is 15°. The number of vertices must be 720 divided by 15, or 48 vertices.

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Every vertex of a hexagonal prism has two squares and a hexagon, leaving 60°, so it has 720 ÷ 60 = 12 vertices. The truncated cuboctahedron has a square, a hexagon and an octagon at every vertex, leaving 15°, so it has 48.

Try it yourself

If you'd like to experiment with shapes like these yourself, there's now a demo of 4D Builder on itch.io. While it's mainly meant for playing with four-dimensional shapes, the sandbox also lets you build three-dimensional polyhedra as well, so you can experiment with the ideas we've seen here.

Wishlist 4D Builder

Let us know what you thought of this, and what else you'd like to see in the future, in the comments on the video.

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