Editing Regular icosahedron (section)

=== Mensuration ===
The [[insphere]] of a convex polyhedron is a sphere inside the polyhedron, touching every face. The [[circumsphere]] of a convex polyhedron is a sphere that contains the polyhedron and touches every vertex. The [[midsphere]] of a convex polyhedron is a sphere tangent to every edge. Therefore, given that the edge length <math> a </math> of a regular icosahedron, the radius of insphere (inradius) <math> r_I </math>, the radius of circumsphere (circumradius) <math> r_C </math>, and the radius of midsphere (midradius) <math> r_M </math> are, respectively:<ref>{{multiref
 |{{harvnb|MacLean|2007|p=[https://books.google.com/books?id=vINuAwAAQBAJ&pg=PA44 43&ndash;44]}}
 |{{harvnb|Coxeter|1973}}, Table I(i), pp. 292–293. See column "<math>{}_1\!\mathrm{R}/\ell</math>", where <math>{}_1\!\mathrm{R}</math> is Coxeter's notation for the midradius, also noting that Coxeter uses <math>2\ell</math> as the edge length (see p. 2).
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<math display="block">
 r_I = \frac{\varphi^2 a}{2 \sqrt{3}} \approx 0.756a, \qquad
 r_C = \frac{\sqrt{\varphi^2 + 1}}{2}a \approx 0.951a, \qquad
 r_M = \frac{\varphi}{2}a \approx 0.809a. 
</math>

[[File:Regular icosahedron.stl|thumb|3D model of a regular icosahedron]]
The [[surface area]] of a polyhedron is the sum of the areas of its faces. Therefore, the surface area <math>(A)</math> of a regular icosahedron is twenty times that of each of its equilateral triangle faces. The volume <math>(V)</math> of a regular icosahedron can be obtained as twenty times that of a pyramid whose base is one of its faces and whose apex is the icosahedron's center; or as the sum of two uniform [[pentagonal pyramid]]s and a [[pentagonal antiprism]]. The expressions of both are:<ref>{{multiref
|{{harvnb|MacLean|2007|p=[https://books.google.com/books?id=vINuAwAAQBAJ&pg=PA44 43&ndash;44]}}
|{{harvnb|Berman|1971}}
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<math display="block">
 A = 5\sqrt{3}a^2 \approx 8.660a^2, \qquad
 V = \frac{5 \varphi^2}{6}a^3 \approx 2.182a^3. 
</math>
A problem dating back to the ancient Greeks is determining which of two shapes has a larger volume, an icosahedron inscribed in a sphere, or a dodecahedron inscribed in the same sphere. The problem was solved by [[Hero of Alexandria|Hero]], [[Pappus of Alexandria|Pappus]], and [[Fibonacci]], among others.{{sfn|Herz-Fischler|2013|p=[https://books.google.com/books?id=aYjXZJwLARQC&pg=PA138 138–140]}} [[Apollonius of Perga]] discovered the curious result that the ratio of volumes of these two shapes is the same as the ratio of their surface areas.{{sfn|Simmons|2007|p=[https://books.google.com/books?id=3KOst4Mon90C&pg=PA50 50]}} Both volumes have formulas involving the [[golden ratio]], but taken to different powers.{{sfn|Sutton|2002|p=[https://books.google.com/books?id=vgo7bTxDmIsC&pg=PA55 55]}} As it turns out, the icosahedron occupies less of the sphere's volume (60.54%) than the dodecahedron (66.49%).{{efn|1=Numerical values for the volumes of the inscribed Platonic solids may be found in {{harvnb|Buker|Eggleton|1969}}.}}

The [[dihedral angle]] of a regular icosahedron can be calculated by adding the angle of pentagonal pyramids with regular faces and a pentagonal antiprism. The dihedral angle of a pentagonal antiprism and pentagonal pyramid between two adjacent triangular faces is approximately 38.2°. The dihedral angle of a pentagonal antiprism between pentagon-to-triangle is 100.8°, and the dihedral angle of a pentagonal pyramid between the same faces is 37.4°. Therefore, for the regular icosahedron, the dihedral angle between two adjacent triangles, on the edge where the pentagonal pyramid and pentagonal antiprism are attached is 37.4° + 100.8° = 138.2°.<ref>{{multiref
 |{{harvnb|Johnson|1966|loc=See table II, line 4.}}
 |{{harvnb|MacLean|2007|p=[https://books.google.com/books?id=vINuAwAAQBAJ&pg=PA44 43&ndash;44]}}
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