Count the corners, edges and faces of almost any solid without holes and the three numbers always satisfy one short equation. A cube has 8 vertices, 12 edges and 6 faces. A pentagonal pyramid has 6, 10 and 6. Take the vertices, subtract the edges, add the faces, and both give 2.
This is Euler's formula for polyhedra. It is one of the first places where geometry turns into topology, because it does not care about lengths or angles at all, only about how the pieces connect. Our article on the Platonic solids uses it briefly. Here it gets its own page: how to use it to find missing counts, why it is true, and the one situation where it breaks. Turn the solids below and let the program count for you.
The formula
A polyhedron is a solid whose surface is made of flat polygons called faces. Two faces meet along an edge, and edges meet at corners called vertices. Write $latex V$ for the number of vertices, $latex E$ for edges and $latex F$ for faces. For any polyhedron with no holes:
$$V-E+F=2$$
You may also see it written $latex F+V=E+2$ or $latex F+V-E=2$. All three say the same thing, and each one lets you solve for a missing count.
$$E=V+F-2\qquad\qquad V=E-F+2\qquad\qquad F=E-V+2$$
Checking the formula on familiar solids
Here are the five Platonic solids. Every face of a Platonic solid is the same regular polygon, and the same number of faces meet at each vertex.
| Solid | Faces | V | E | F | V − E + F |
|---|---|---|---|---|---|
| Tetrahedron | 4 triangles | 4 | 6 | 4 | 2 |
| Cube | 6 squares | 8 | 12 | 6 | 2 |
| Octahedron | 8 triangles | 6 | 12 | 8 | 2 |
| Dodecahedron | 12 pentagons | 20 | 30 | 12 | 2 |
| Icosahedron | 20 triangles | 12 | 30 | 20 | 2 |
Prisms and pyramids follow patterns you can write as formulas. A prism with an $latex n$-sided base has $latex 2n$ vertices, $latex 3n$ edges and $latex n+2$ faces. A pyramid with an $latex n$-sided base has $latex n+1$ vertices, $latex 2n$ edges and $latex n+1$ faces.
| Solid | V | E | F | V − E + F |
|---|---|---|---|---|
| Triangular prism | 6 | 9 | 5 | 2 |
| Pentagonal prism | 10 | 15 | 7 | 2 |
| Hexagonal prism | 12 | 18 | 8 | 2 |
| Square pyramid | 5 | 8 | 5 | 2 |
| Hexagonal pyramid | 7 | 12 | 7 | 2 |
For the prism, $latex 2n-3n+(n+2)=2$, and for the pyramid, $latex (n+1)-2n+(n+1)=2$. The formula holds for every $latex n$, so you can check it by moving the slider in the tool above.
Two counting rules that pair with it
Euler's formula has three unknowns, so usually you need one more relation. Two counting rules provide it.
- Every edge is shared by exactly two faces. So adding up the number of sides of all faces counts each edge twice, so $latex 2E$ equals that total.
- Every edge has two ends. So adding up the number of edges that meet at each vertex also counts each edge twice, so $latex 2E$ equals that sum.
For a polyhedron with 8 triangular faces, the total number of sides is $latex 8\cdot 3=24$, so $latex E=12$. Euler's formula then gives $latex V=2+12-8=6$.
Why the formula is true
Imagine the polyhedron made of rubber. Cut out one face, then stretch the rest flat on a table. You get a flat map made of vertices, edges and the faces, with the missing face becoming the outside region, so the count of faces stays the same.
Now build that map from nothing. Start with a single vertex: $latex V=1$, $latex E=0$ and one face (the outside), so $latex V-E+F=1-0+1=2$. Every step you take preserves the value:
- Add a new vertex joined by an edge to an existing one. Then $latex V$ and $latex E$ each grow by 1, and the difference does not change.
- Add an edge between two vertices that already exist. This closes a loop and splits a face in two, so $latex E$ and $latex F$ each grow by 1, and again the total stays at 2.
Every connected map can be built this way, so $latex V-E+F$ is always 2. The proof breaks when the surface has a hole, because then a loop around the hole does not split a face.
When the formula fails
A picture frame is a solid with a rectangular hole through it. Build it from flat faces and you get $latex V=16$, $latex E=32$ and $latex F=16$, so $latex V-E+F=0$. Choose "Picture frame" in the tool above to see the count. In general, a solid with $latex g$ holes gives:
$$V-E+F=2-2g$$
The formula also needs every face to be a flat polygon with no hole of its own, and the solid to be one connected piece. A solid can be non-convex, with dents, and still give 2, as long as it has no holes.
Euler's formula: worked examples
Work each problem before opening the solution.
EXAMPLE 1
A polyhedron has 12 faces and 20 vertices. How many edges does it have?
Solution
$$E=V+F-2=20+12-2=30$$
It has 30 edges. The regular dodecahedron has exactly these counts.
EXAMPLE 2
A prism has a hexagonal base. Find its vertices, edges and faces, and check Euler's formula.
Solution
With $latex n=6$: $latex V=2n=12$, $latex E=3n=18$ and $latex F=n+2=8$ (two hexagons and six rectangles).
$$V-E+F=12-18+8=2$$
EXAMPLE 3
A polyhedron has 8 faces, all triangles. Find its edges and vertices.
Solution
The faces have $latex 8\cdot 3=24$ sides in total, and each edge is shared by two faces, so $latex E=24\div 2=12$.
$$V=2+E-F=2+12-8=6$$
With 6 vertices, 12 edges and 8 triangular faces, this is an octahedron (or any solid with the same counts).
EXAMPLE 4
A soccer ball pattern has 12 pentagons and 20 hexagons. How many edges and vertices does the solid have?
Solution
The faces number 32. The sides add up to $latex 12\cdot 5+20\cdot 6=180$, so $latex E=180\div 2=90$.
$$V=2+E-F=2+90-32=60$$
The solid has 60 vertices, 90 edges and 32 faces. This shape is the truncated icosahedron, and you can build it in the tool above.
EXAMPLE 5
A student claims a convex polyhedron has 7 faces, 12 vertices and 15 edges. Is this possible?
Solution
$$V-E+F=12-15+7=4$$
The result is 4, not 2, so no such polyhedron exists. At least one of the three numbers is wrong.
EXAMPLE 6
A picture frame solid has 16 vertices, 32 edges and 16 faces. Why does Euler's formula not give 2?
Solution
$$V-E+F=16-32+16=0$$
A solid with one hole has $latex g=1$, so the value is $latex 2-2g=0$. Euler's formula of 2 applies only to solids with no holes.
Common mistakes
- Missing the hidden edges and faces. Count the back of the solid too, or rotate it until you have seen every part.
- Counting the base as a face of a pyramid or prism. It is a face, so include it.
- Mixing up the order, writing $latex V+E-F$. Vertices and faces are added, edges are subtracted.
- Using the formula on a solid with a hole, such as a frame or a doughnut shape.
- Counting edges twice when adding the sides of the faces. Divide that total by 2.
Euler's formula: practice problems
Use the formula and the two counting rules. Open each answer to check.
1. A cube has 8 vertices and 12 edges. How many faces does it have?
Answer
$latex F=2-8+12=6$.
2. A polyhedron has 14 vertices and 21 edges. How many faces does it have? Name a solid with these counts.
Answer
$latex F=2-14+21=9$. A heptagonal prism has $latex V=14$, $latex E=21$ and $latex F=9$.
3. Find the vertices, edges and faces of a pyramid with a 7-sided base.
Answer
$latex V=8$, $latex E=14$, $latex F=8$. Check: $latex 8-14+8=2$.
4. A polyhedron has 6 faces, all quadrilaterals. How many edges and vertices?
Answer
$latex E=6\cdot 4\div 2=12$ and $latex V=2+12-6=8$. For example, a cube.
5. An icosahedron has 20 triangular faces. Find its edges and vertices.
Answer
$latex E=20\cdot 3\div 2=30$ and $latex V=2+30-20=12$.
6. A convex solid is said to have 10 vertices, 18 edges and 9 faces. Is that possible?
Answer
$latex 10-18+9=1$, which is not 2. It is not possible.
Frequently asked questions
Does Euler's formula work for non-convex polyhedra?
Yes, as long as the solid has no holes and is one connected piece. A solid with dents or a star shape still gives $latex V-E+F=2$.
Why is it 2?
Because a sphere has an Euler number of 2. Any solid you could inflate into a ball gives 2. A solid with a hole through it is like a doughnut, and a doughnut gives 0.
Is there a version for flat shapes?
Yes. A connected planar map with $latex V$ vertices, $latex E$ edges and $latex F$ regions (counting the outside) also satisfies $latex V-E+F=2$. That is how the proof above works.
Who discovered it?
Leonhard Euler described it in the 1750s. René Descartes had found a related result about a century earlier.
Related calculators
Once you know a solid's faces and edges, find its volume and surface area with these tools:
See also
More on solids, faces, edges and vertices: