The cosine of the angle between normals to the faces of the dodecahedron is equal to the angle between the vectors to the vertices of the icosahedron, $\frac{\sqrt{5}}{5}$ . Similarly, the cosine of the angle between normals to the faces of the icosahedron is equal to the angle between the vectors to the vertices of the dodecahedron, $\frac{\sqrt{5}}{3}$ .

The exact formulas for the pentagonal and fivefold symmetrical platonic solids is rendered with MathML. The expressions are useful in constructing geodesic domes and planetariums.

component\object	dodecahedron	dodecahedron	icosahedron	icosahedron
inscribed radius	1	$1 ∙ \frac{\sqrt{5 + 2 ∙ \sqrt{5}}}{\sqrt{15}}$ = 0.794654	1	$1 ∙ \frac{\sqrt{5 + 2 ∙ \sqrt{5}}}{\sqrt{15}}$ = 0.794654
circumscribed radius	$\sqrt{3} ∙ \frac{\sqrt{5 - 2 ∙ \sqrt{5}}}{1}$ = 1.258409	1	$\sqrt{3} ∙ \frac{\sqrt{5 - 2 ∙ \sqrt{5}}}{1}$ = 1.258409	1
edge length	$\sqrt{2} ∙ \frac{\sqrt{25 - 11 ∙ \sqrt{5}}}{1}$ = 0.898056	$\sqrt{\frac{2}{3}} ∙ \sqrt{3 - \sqrt{5}}$ = 0.713644	$\sqrt{6} ∙ \frac{\sqrt{7 - 3 ∙ \sqrt{5}}}{1}$ = 1.323169	$\sqrt{\frac{2}{5}} ∙ \sqrt{5 - 5 ∙ \sqrt{5}}$ = 1.051462
edge from center, radius	$\sqrt{1} ∙ \frac{\sqrt{5 - \sqrt{5}}}{\sqrt{2}}$ = 1/0.850650	$\sqrt{1} ∙ \frac{\sqrt{3 + 5 ∙ \sqrt{5}}}{\sqrt{6}}$ = 0.934172	$\sqrt{\frac{3}{2}} ∙ \sqrt{3 - \sqrt{5}}$ = 1/0.934172	$\sqrt{1} ∙ \frac{\sqrt{5 + 5 ∙ \sqrt{5}}}{\sqrt{10}}$ = 0.850650

The derivation of the exact formulas, assuming some symmetries, with enough detail to follow is at proof_new.txt; more detail with proof of the symmetry without any assumptions on symmetry is at proof.txt. At proof_new.txt enough progress is made to check the expressions; further blazing into new territory is at plato.txt where additional expressions are derived. They are rendered in MathML at pentagon.xml. Calculators using the exact formulas for the pentagon, dodecahedron, icosahedron, and an other example available, as well as an illustration. are available. Another amazing expression for π is

π = 3 + \frac{1^{2}}{6 +} \frac{3^{2}}{6 +} \frac{5^{2}}{6 +} \frac{7^{2}}{6 +} \frac{9^{2}}{6 +}

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The pentagon can be constructed using the ruler and staight edge but the angle pi/7 cannot be. For pi/7 angles the solution is transcendental but for pi/60 the solution is still irrational