File:Tetraedro Estrellado Ultra Hueco de Leonardo.gif
Tetraedro_Estrellado_Ultra_Hueco_de_Leonardo.gif (412 × 398 pixels, file size: 2.12 MB, MIME type: image/gif, looped, 226 frames, 18 s)
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English: Leonardo's ultra-hollow star tetrahedron is composed internally of a regular convex tetrahedron, to which 4 ultra-hollow triangular pyramids of Leonardo are attached externally. This polyhedron has 36 triangular interior faces and 72 triangular inner faces, for a total of 108 triangular faces.
In addition, it has 36 interior vertices, 4 intermediate vertices, 12 ultra-outer vertices and 4 exterior vertices for a total of 56 vertices. Likewise, it has 12 external edges, 72 internal edges, 6 intermediate edges, 36 ultra-outer edges and 36 ultra-inner edges for a total of 162 edges. This polyhedron belongs to the set of triangular polyhedrons that occupy the place (L = 53) of the three polyhedral sequences discovered in the month of May 2010, by the Dominican José Joel Leonardo. A = 3L + 3, V = L + 3, C = 2l + 2 If we join the set of outer vertices, 6 external imaginary edges are formed, with which an imaginary regular tetrahedron is perfectly constructed. The hollow Leonardian tetrahedron belongs to the set of hollow starred polyhedra. The hollow stelae of the tetrahedron are represented by an infinite set of hollow polyhedra, this is because the hollow spaces of a selected polyhedron are infinitely divisible. Area and volume of Leonardo's ultra-hollow starred tetrahedron. Area of the Leonardian Hollow Tetrahedron is equal to the area of 36 isosceles triangles, 36 (0.177 L2), plus the area of 72 Scalene triangles, 72 (0.07038L2). Area of the Leonardiano Hueco tetrahedron = Atelh Area of the Leonardian tetrahedron Hollow = 36 (0.177 L raised to power 2) +72 (0.07038 L raised to power 2) = 6,372 L raised to power 2 + 5.06736 L raised to power 2 = 11.43936 L raised to power 2. Atelh = 11.43936 L raised to power 2. Volume of the Leonardiano Hueco tetrahedron = Vtelh Vtelh = (Atelh) (E) + 0.117776 L raised to power 3 = 11.43936 L raised to power 2 (E) + 0.117776 L raised to power 3. Vtelh = 11.43936 L raised to power 2 (E) + 0.117776 L elevada to power 3.Español: El tetraedro estrellado ultra hueco de Leonardo está compuesto internamente por un tetraedro regular convexo, al que se le unen de forma externa 4 pirámides triangulares ultra hueca de Leonardo.Este poliedro posee 36 caras interiores triangulares y 72 cara ultra interiores triangulares, para un total de 108 caras triangulares.
Además, tiene 36 vértices interiores, 4 vértices intermedios, 12 vértices ultra exteriores y 4 vértices exteriores para un total de 56 vértices. asimismo, ostenta 12 aristas exteriores, 72 aristas interiores, 6 aristas intermedias, 36 arista ultra exteriores y 36 arista ultra interiores para un total de 162 aristas. Este poliedro pertenece al conjunto de poliedros triangulares que ocupan el lugar (L=53) de las tres secuencias poliédricas descubiertas en el mes de mayo del año 2010, por el Dominicano José Joel Leonardo. A = 3L + 3, V = L + 3, C = 2l + 2 Si unimos el conjunto de vértices exteriores se forman 6 aristas imaginarias exteriores, con la que se construye perfectamente un tetraedro regular imaginario. El tetraedro Leonardiano hueco pertenece al conjunto de los poliedros estrellados hueco. Las estelaciones huecas del tetraedro esta representadas por un conjunto infinito de poliedros huecos, este es debido a que los espacios huecos de un poliedro seleccionado son infinitamente divisibles. Área y volumen del tetraedro estrellado ultra hueco de Leonardo. Área del tetraedro Leonardiano Hueco es igual al área de 36 triángulos isósceles, 36 (0.177 L2), más el área de 72 triángulos escaleno, 72 (0.07038L2). Área del tetraedro Leonardiano Hueco = Atelh Área del tetraedro Leonardiano Hueco =36 (0.177 L elevada a potencia 2) +72 (0.07038 L elevada a potencia 2) = 6.372 L elevada a potencia 2 + 5.06736 L elevada a potencia 2 = 11.43936 L elevada a potencia 2. Atelh = 11.43936 L elevada a potencia 2. Volumen del tetraedro Leonardiano Hueco = Vtelh Vtelh = (Atelh) (E) + 0.117776 L elevada a potencia 3 =11.43936 L elevada a potencia 2 (E) + 0.117776 L elevada a potencia 3. Vtelh =11.43936 L elevada a potencia 2 (E) + 0.117776 Lelevada a potencia 3.Français : Le tétraèdre étoiles ultra-creuses de Leonardo est composé intérieurement d'un tétraèdre convexe régulier auquel sont fixées 4 pyramides triangulaires ultra-creuses de Léonard. Ce polyèdre à 36 faces intérieures triangulaires et 72 faces internes triangulaires 108 faces triangulaires.
De plus, il possède 36 sommets intérieurs, 4 sommets intermédiaires, 12 sommets ultra-extérieurs et 4 sommets extérieurs pour un total de 56 sommets. De même, il possède 12 bords externes, 72 bords internes, 6 bords intermédiaires, 36 bords ultra-extérieurs et 36 bords ultra-intérieurs pour un total de 162 bords. Ce polyèdre appartient à l'ensemble des polyèdres triangulaires qui occupent la place (L = 53) des trois séquences polyédriques découvertes au mois de mai 2010 par le dominicain José Joel Leonardo. A = 3L + 3, V = L + 3, C = 2l + 2 Si nous joignons l'ensemble des sommets externes, 6 arêtes imaginaires externes sont formées, avec lesquelles un tétraèdre imaginaire régulier est parfaitement construit. Le tétraèdre Leonardian creux appartient à l'ensemble des polyèdres creux étoilés. Les stèles creuses du tétraèdre sont représentées par un ensemble infini de polyèdres creux, car les espaces creux d’un polyèdre sélectionné sont divisibles à l’infini. Zone et volume du tétraèdre étoilé ultra-creux de Leonardo. La surface du tétraèdre creux de Leonardian est égale à la surface de 36 triangles isocèles, 36 (0,177 L2), plus la surface de 72 triangles scalènes, 72 (0,07038L2). Région du tétraèdre Leonardiano Hueco = Atelh zone Tetrahedron Leonardiano brèche = 36 (0177 L-2 haute puissance) 72 (0,07038 L soulevée pour alimenter 2) = 6372 L-haute puissance 5,06736 2 + L 2 = haute puissance à haute puissance 11,43936 L 2. Atelh = 11,43936 L-haute puissance deux. Volume du tétraèdre Leonardiano Hueco = Vtelh Vtelh = (Atelh) (E) + 0,117776 L soulevé pour alimenter 3 = 11,43936 L élevés pour alimenter 2 (E) + 0,117776 L soulevé pour alimenter 3. Vtelh = 11,43936 L élevé à la puissance 2 (E) + 0,117776 Lelevada à la puissance 3. |
| Date | |
| Source | Own work |
| Author | Jose J. Leonard |
File:Tetraedro_Estrellado_Ultra_Hueco_de_Leonardo.jpg
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