Assertion: icosahedron's angle | /r/HasProperty | equal
Facets:
value: to the value in the table
type: OTHER
value: at this size
type: OTHER
Source sentences:
text: This can be verified by computing V V T {\displaystyle VV^{T}} ; the resulting matrix has 0.5 on the diagonal (so all points are equidistant from the origin) and plus-or-minus 0.2236 on the off-diagonal (so all dot products and hence all angles are equal).
text: The regular dodecahedron represents a special intermediate case where all edges and angles are equal.
text: Continuing from there in that direction, we pass through a degenerate case where twelve vertices coincide in the centre, and on to the regular great stellated dodecahedron where all edges and angles are equal again, and the faces have been distorted into regular pentagrams.
text: The regular dodecahedron represents a special intermediate case where all edges and angles are equal.
text: Continuing from there in that direction, we pass through a degenerate case where twelve vertices coincide in the centre, and on to the regular great stellated dodecahedron where all edges and angles are equal again, and the faces have been distorted into regular pentagrams.
text: All the dihedral angles of the polyhedron are equal
text: The duals of the isohedral Catalan polyhedra are the isogonal semiregular Archimedean polyhedra, in turn characterized by faces made up by two or three different regular polygons (whose sides and angles are therefore all equal) and by equal but non-regular angoloids.
text: At this size, the dihedral angle between all neighbouring triangular faces is equal to the value in the table above.
text: Along the bases edges of length $\alpha\sqrt{2}$, the diheral angle is equal the dihedral angle of an icosahedron plus two times the face angle of a right triangular pyramid: \[Height_{right triangular pyramid} = \frac{\alpha\sqrt{2}}{2}\sqrt{1-Tan^2(\pi/6)}\] \[Midradius_{equal lateral triangle} = \frac{\alpha\sqrt{2}}{2}Tan(\pi/6)}\] \[FaceAngle_{equal lateral triangle} = Atan(\frac{\sqrt{1-Tan^2(\pi/6)}}{Tan(\pi/6)})\] \[FaceAngle_{equal lateral triangle} = 0.955316618 or 54.73561032^{\circ}\] \[dihedral_{icosahedron}
text: all angles are equal so that