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An Archimedean solid is a convex uniform polyhedron, excluding the Platonic solids and the infinite families of uniform prisms and antiprisms. Here uniform means that every face is a regular polygon and that the symmetry group acts transitively on the vertices of a polyhedron: a symmetry of the entire solid carries any chosen vertex of a polyhedron to any other. Merely having equal numbers of faces of each type at every vertex of a polyhedron is a weaker local condition.^[ref]

There are thirteen types under this convention, counting a mirror-image pair as one type. The infinite prism and antiprism families are excluded by convention even though their uniform members satisfy the same regular-face and global vertex-equivalence conditions.^[ref]

For example, the truncated octahedron has six square faces and eight regular hexagon faces. Counting face-edge incidences gives $(6\cdot 4+8\cdot 6)/2=36$ edge of a polyhedron; Euler characteristic then gives $36-14+2=24$ vertices of a polyhedron. At each vertex of a polyhedron, one square and two regular hexagons meet. The icosidodecahedron, with twenty equilateral triangles and twelve regular pentagons, gives a different regular-face combination.

Let $\Phi$ be a nonempty set of unit vectors in a real Euclidean plane, closed under the reflections $s_v(u)=u-2⟨u,v⟩v$ and satisfying $2⟨u,v⟩\in \mathbb{Z}$ for every pair. Then it consists of an antipodal pair, a square, or a regular hexagon. Relative to any fixed member, the inner products can only be $0,\pm 1/2,\pm 1$, so only eight angles are initially possible. Reflection gives every negative vector. If a perpendicular pair occurs, it and its negatives exhaust the set, since any extra sixty-degree direction would have forbidden inner product $\pm \sqrt{3}/2$ with one of them. Otherwise any noncollinear pair gives, after negation, a sixty-degree angle; reflection supplies the third direction and negation supplies the other three hexagon vertices. If no such pair occurs, only the antipodal pair remains. Each configuration directly satisfies the conditions.

These are the equal-length cases of the classification of rank-two root systems, with the antipodal case having rank one. The square is the orthogonal sum of two A1 root systems, and the hexagon is an A2 root system. The root system is considered in the span of its vectors when the ambient plane is not spanned.

For equal-length crystallographic root configurations in a plane, taking $u=v$ in the closure condition gives $S_v(v)=-v$, so every vector occurs with its negative. For fixed $v\in \Phi$, a vector $u\in \Phi$ has $⟨u,v⟩\in \{-1,-1/2,0,1/2,1\}$, because both vectors are unit vectors. Thus its angle relative to $v$ belongs to the finite list $0,\pm \pi /3,\pm \pi /2,\pm 2\pi /3,\pi$.

If all vectors are collinear, the set is exactly $\{v,-v\}$, giving two opposite points of the unit circle.

If an orthogonal vector $w$ occurs, closure supplies $\pm v,\pm w$. A unit vector making an angle of $\pi /3$ or $2\pi /3$ with $v$ would have inner product $\pm \sqrt{3}/2$ with $w$, contradicting integrality of twice that inner product. Thus these four points are the whole set: the vertices of a square.

In the remaining case a noncollinear pair makes an angle of $\pi /3$ or $2\pi /3$. Changing a sign, take their angles as $0,\pi /3$. Reflecting the second in the line perpendicular to the first supplies the angle $2\pi /3$, and negation supplies the other three hexagon vertices. No perpendicular vector can occur, by the preceding inner-product obstruction. The initial finite list now leaves only these six points. Conversely the opposite pair, square and regular hexagon each satisfy both reflection closure and the integrality condition. Hence

For equal-length crystallographic root configurations in a plane, taking $u=v$ in the closure condition gives $S_v(v)=-v$, so every vector occurs with its negative. For fixed $v\in \Phi$, a vector $u\in \Phi$ has $⟨u,v⟩\in \{-1,-1/2,0,1/2,1\}$, because both vectors are unit vectors. Thus its angle relative to $v$ belongs to the finite list $0,\pm \pi /3,\pm \pi /2,\pm 2\pi /3,\pi$.

If all vectors are collinear, the set is exactly $\{v,-v\}$, giving two opposite points of the unit circle.

If an orthogonal vector $w$ occurs, closure supplies $\pm v,\pm w$. A unit vector making an angle of $\pi /3$ or $2\pi /3$ with $v$ would have inner product $\pm \sqrt{3}/2$ with $w$, contradicting integrality of twice that inner product. Thus these four points are the whole set: the vertices of a square.

In the remaining case a noncollinear pair makes an angle of $\pi /3$ or $2\pi /3$. Changing a sign, take their angles as $0,\pi /3$. Reflecting the second in the line perpendicular to the first supplies the angle $2\pi /3$, and negation supplies the other three hexagon vertices. No perpendicular vector can occur, by the preceding inner-product obstruction. The initial finite list now leaves only these six points. Conversely the opposite pair, square and regular hexagon each satisfy both reflection closure and the integrality condition. Hence

The phase step between adjacent slits isFor equal amplitudes $a_0$, the six unmodified contributions form a closed regular hexagon in the complex plane:Reversing one slit at an array minimum replaces slit $j$'s contribution by its negative, so $S=S_0-2a_0{e}^{ij\pi /3}$. ConsequentlyThis is independent of which slit is reversed. The result uses the far-field, narrow-slit and small-angle approximations; retaining $\sin \theta$ changes the phase step only by about $7\times 10^{-7}$ radians.

A regular hexagon is a hexagon with all side lengths equal and all interior angles equal to $2\pi /3$. Joining its centre to the six vertices divides it into six equilateral triangles. Consequently its circumradius equals its side length $a$, its inradius is $\sqrt{3}a/2$, and its area is $3\sqrt{3}{a}^2/2$. Three regular hexagons meet around a vertex because $3(2\pi /3)=2\pi$, producing the familiar edge-to-edge tessellation of the Euclidean plane.