Twenty triangles, one edge
A regular icosahedron is assembled from 20 congruent equilateral triangles. Five faces meet at every vertex. That packing produces 12 vertices and 30 equal edges. Because the solid is regular, a single length — the edge a — fixes volume, surface, and the three radii.
Why the golden ratio appears
The twelve vertices of a regular icosahedron can be written as the corners of three mutually perpendicular golden rectangles — rectangles whose side ratio is φ = (1 + √5)/2. That construction is why φ walks into the midradius and inradius instead of sitting in a footnote.
vertices ⊆ { (0, ±1, ±φ), (±1, ±φ, 0), (±φ, 0, ±1) }
Volume from twenty pyramids
Each face is an equilateral triangle of area (√3/4)a², so the surface is twenty faces:
For volume, join every face to the center. You get 20 congruent pyramids whose base is one face and whose height is the inradius r. Adding them produces the compact formula below — same number the calculator uses.
Three radii, three endpoints
| Quantity | Ends at | Formula | ≈ × a |
|---|---|---|---|
| r inradius | center of a face | a φ² / (2√3) | 0.7558 |
| ρ midradius | midpoint of an edge | a φ / 2 | 0.8090 |
| R circumradius | a vertex | a √(10 + 2√5) / 4 | 0.9511 |
Always r < ρ < R. That order is a checksum: if a result ever violates it, the wrong radius formula was used.
Worked example 1 — a = 2
Thirty edges of length 2. Small enough to compute by hand; the same steps scale to any a.
| Measure | Calculation | Result |
|---|---|---|
| Volume | (5/12)(3 + √5) · 8 | 17.4536 |
| Surface | 5√3 · 4 | 34.6410 |
| Circumradius R | 2 · √(10 + 2√5) / 4 | 1.9021 |
| Midradius ρ | 2φ / 2 = φ | 1.6180 |
| Inradius r | 2 φ² / (2√3) | 1.5115 |
Worked example 2 — start from volume
Suppose the volume is given as 100 and you need the edge. Invert the volume formula:
Then S = 5√3 a² ≈ 67.61 and R ≈ 0.9511 a ≈ 2.656. That is what the calculator does when you type in the volume field and leave edge blank as the last edited input.
Dual: icosahedron ↔ dodecahedron
The regular dodecahedron is the dual. Faces and vertices swap roles. If an icosahedron has circumradius R, its dual dodecahedron can be scaled so that its inradius matches that R — a face of one points at a vertex of the other.
| Icosahedron | Dodecahedron | |
|---|---|---|
| Faces | 20 triangles | 12 pentagons |
| Vertices | 12 | 20 |
| Edges | 30 | 30 |
| Faces at a vertex | 5 | 3 |
Where it shows up
Twenty-sided dice, geodesic scraps, and any model that wants a nearly spherical Platonic solid. Icosahedral symmetry also appears in some viral capsids: twenty triangular facets is a compact way to enclose volume with identical pieces.
Common mistakes
- Swapping this solid with its dual, the dodecahedron — face and vertex counts reverse.
- Borrowing the tetrahedron or octahedron surface formula because those faces are also triangles. Those solids do not have five triangles at a vertex.
- Mixing r, ρ, and R. Face center, edge midpoint, vertex — three different endpoints.
- Scaling volume with a². Surface is a²; volume is a³.
FAQ
20 equilateral triangles, 30 edges, 12 vertices. Five faces meet at each vertex.
V = 5(3 + √5)a³ / 12. It is exactly twenty pyramids of height r on triangular bases of area (√3/4)a².
R = a√(10 + 2√5)/4 — center to any vertex. For a quick check, R ≈ 0.9511 a.
Because the vertices sit on golden rectangles. The distance from the origin to an edge midpoint collapses to aφ/2.