Cuboctahedron atop small rhombicuboctahedron

{{Infobox polytope Cuboctahedron atop small rhombicuboctahedron, or coasirco, is a CRF segmentochoron (designated K-4.61 on Richard Klitzing's list). As the name suggests, it consists of a cuboctahedron and a small rhombicuboctahedron as bases, connected by 8 octahedra, 12 square pyramids, and 6 square antiprisms.
 * img=sirco=co.png
 * off=auto
 * type=Segmentotope
 * dim = 4
 * obsa = Coasirco
 * cells = 12 square pyramids, 8 octahedra, 6 square antiprisms, 1 cuboctahedron, 1 small rhombicuboctahedron
 * faces = 8+8+24+24+24 triangles, 6+6+12 squares
 * edges = 24+24+24+48
 * vertices = 12+24
 * verf = 12 rectangular-square trapezoprisms, one base a rectangle of edge lengths 1 and $\sqrt{2}$, other base a square of edge length 1, side lengths 1
 * verf2 = 24 skewed wedges, edge lengths 1 (6) and $\sqrt{2}$ (3)
 * coxeter = ox4xo3ox&#x
 * army=Coasirco
 * reg=Coasirco
 * symmetry = BC3×I, order 48
 * circum = $$\frac{\sqrt7+2\sqrt{14}}{7} ≈ 1.44701$$
 * height = $$\frac{\sqrt{2\sqrt2-1}}{2} ≈ 0.67610$$
 * hypervolume = $$\frac{\sqrt{103+74\sqrt2}}{4} ≈ 3.60253$$
 * dich = Oct–3–squippy: $$\arccos\left(\frac{1-3\sqrt2}{4}\right) ≈ 144.16048°$$
 * dich2 = Oct–3–squap: $$\arccos\left(-\frac{\sqrt{4+3\sqrt2}}{4}\right) ≈ 135.86903°$$
 * dich2 = Squippy–3–squap: $$\arccos\left(-\frac{\sqrt{4+3\sqrt2}}{4}\right) ≈ 135.86903°$$
 * dich4 = Co–4–squap: $$\arccos\left(-\frac{\sqrt[4]{2}}{2}\right) ≈ 126.48438°$$
 * dich5 = Co–3–oct: \arccos\left(\frac{2-3\sqrt2}{4}\right) ≈ 124.10147°
 * dich6 = Sirco–4–squippy: $$\arccos\left(\frac{2-\sqrt2}{2}\right) ≈ 72.96875°$$
 * dich7 = Sirco–3–oct: $$\arccos\left(\frac{3\sqrt2-2}{4}\right) ≈ 55.89854°$$
 * dich8 = Sirco–4–squap: $$\arccos\left{\frac{\sqrt[4]{2}}{2}\right) ≈ 53.51562°$$
 * dual=Rhombic dodecahedral-deltoidal icositetrahedral tegmoid
 * conjugate=Cuboctahedron atop small rhombicuboctahedron
 * conv = Yes
 * orientable=Yes
 * nat=Tame}}

Vertex coordinates
The vertices of a cuboctahedron atop small rhombicuboctahedron segmentochoron of edge length 1 are given by:
 * $$\left(±\frac{\sqrt2}{2},\,±\frac{\sqrt2}{2},\,0)\,\frac{\sqrt{2\sqrt2-1}}{2}\right)$$ and all permutations of first three coordinates
 * $$\left(±\frac{1+\sqrt2}{2},\,±\frac12,\,±\frac12,\,0\right)$$ and all permutations of first three coordinates