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Platonic Solids Platonic Solids

Plato (427 BC – 347 BC) • Greek philosepher • 387 BC based a Plato (427 BC – 347 BC) • Greek philosepher • 387 BC based a school in Athina • Died 80 years old

Archeological discovery from Scotland 2000 BC Archeological discovery from Scotland 2000 BC

Platonic Solids in the antic Platonic Solids in the antic

Johannes Kepler (27. 12. 1571 Weil der Stadt – 15. 11. 1630 Regensburg) • Johannes Kepler (27. 12. 1571 Weil der Stadt – 15. 11. 1630 Regensburg) • German mathematican and astronomer • Several years lived in Prague as a guest of Rudolf II

Karlova ulice 4, Prague, house of Johannes Kepler Karlova ulice 4, Prague, house of Johannes Kepler

Solar System Model Solar System Model

Platonic Solid • Convex polyhedron • All faces are equal regular polygon (K vertecies) Platonic Solid • Convex polyhedron • All faces are equal regular polygon (K vertecies) • In each vertex of the polyhedron the same number of edges intersects (L edges)

Which Platonic Solids exist? • • V … number of vertecies F … number Which Platonic Solids exist? • • V … number of vertecies F … number of faces E … number of edges Euler formula: For each convex polyhedron V + F = E +2

Which Platonic Solids exist? • The conditions of Platonic Solids – K. F = Which Platonic Solids exist? • The conditions of Platonic Solids – K. F = 2. E – L. V = 2. E F = 2 E/K V = 2 E/L • Substitution into Euler formula 2 E/L + 2 E/K = E +2 1/L + 1/K = 1/E + 1/2

For which numbers holds 1/K+1/L = 1/E +1/2 • K and L are integers For which numbers holds 1/K+1/L = 1/E +1/2 • K and L are integers bigger or equal to 3 • For K = 3 – – L=3, E=6: 1/3+1/3 = 1/6+1/2 L=4, E=12: 1/3+1/4 = 1/12+1/2 L=5, E=30: 1/3+1/5 = 1/30+1/2 For L=>6 not possible • For K = 4 – L=3, E=12: 1/4+1/3 = 1/12+1/2 – For L=>4 not possible • For K = 5 – L=3, E=30: 1/5+1/3 = 1/30+1/2 – Pro L=>4 not possible • For K => 6 not possible even for L=3

Důkaz neexistence více než 5 ti pravidelných mnohostěnů • From previous equatation we know Důkaz neexistence více než 5 ti pravidelných mnohostěnů • From previous equatation we know that only following polyhedrons can exist as a Platonic Solids K L Edges Verticies Faces 3 3 6 4 4 3 4 12 6 8 3 5 30 12 20 4 3 12 8 6 5 3 30 20 12 • Ti proof their existence we must construct them

Regular Tetrahedron • 4 3 -angle faces, 4 verticies, in each 3 edges, total Regular Tetrahedron • 4 3 -angle faces, 4 verticies, in each 3 edges, total 6 edges

Regular Tetrahedron • 4 3 -angle faces, 4 verticies, in each 3 edges, total Regular Tetrahedron • 4 3 -angle faces, 4 verticies, in each 3 edges, total 6 edges

Regular Octahedron • 8 3 -angle faces, 6 verticies, in each 4 edges, total Regular Octahedron • 8 3 -angle faces, 6 verticies, in each 4 edges, total 12 edges

Regular Octahedron • 8 3 -angle faces, 6 verticies, in each 4 edges, total Regular Octahedron • 8 3 -angle faces, 6 verticies, in each 4 edges, total 12 edges

Regular Isosahedron • 20 3 -angle faces, 12 vertecies, in each 5 edges, total Regular Isosahedron • 20 3 -angle faces, 12 vertecies, in each 5 edges, total 30 edges

Regular Isosahedron • 20 3 -angle faces, 12 vertecies, in each 5 edges, total Regular Isosahedron • 20 3 -angle faces, 12 vertecies, in each 5 edges, total 30 edges

Regular 6 -hedron • 6 4 -angle faces, 8 vertecies, in each 3 edges, Regular 6 -hedron • 6 4 -angle faces, 8 vertecies, in each 3 edges, total 12 edges

Cube • 6 4 -angle faces, 8 vertecies, in each 3 edges, total 12 Cube • 6 4 -angle faces, 8 vertecies, in each 3 edges, total 12 edges

Regular Dodecahedron • 12 5 -angle faces, 20 vertecies, in each 3 edges, total Regular Dodecahedron • 12 5 -angle faces, 20 vertecies, in each 3 edges, total 30 edges

Regular Dodecahedron • 12 5 -angle faces, 20 vertecies, in each 3 edges, total Regular Dodecahedron • 12 5 -angle faces, 20 vertecies, in each 3 edges, total 30 edges