File:Octaedro Luis Abinader.gif

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Octaedro_Luis_Abinader.gif ‎(566 × 409 pixels, file size: 3.53 MB, MIME type: image/gif, looped, 167 frames, 12 s)

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It is a three-dimensional solid that has 24 uniform Jose isosceles triangles, 14 vertices and 36 edges

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Description
Español: Octaedro estrellado Luis Abinader: es un sólido tridimensional cóncavo estrellado, que está compuesto por 24 caras uniformes, representadas por triángulos isósceles de Jose. Además, posee 6 vértices intermedios uniformes entre sí, ochos vértices exteriores uniforme, para un total de 14 vértices y posee 36 aristas deformes. Este politopo tridimensional fue descubierto por el profesor Jose Joel Leonardo el 4 de diciembre del 2021 y fue nombrado por el, en honor a la memoria del presidente de la republica dominicana, Lic. Luis Abinader Corona.

El dual del octaedro estrellado Luis Abinader es el sólido de Arquímedes, conocido con el nombre de hexaedro truncado. Estos seis vértices forman un octaedro regular: A1 = (0, 0, 10.3), B1 = (0, 0,- 10.3), C1 = (0, 10.3, 0), D1 = (0, -10.3, 0), E1 = (10.3, 0, 0), F1 = (-10.3, 0, 0). Estos ochos vértices forman un Hexaedro regular: G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3), K1 = (8.3, 8.3, -8.3), L1 = (-8.3, 8.3, -8.3), M1 = (8.3, -8.3, -8.3), N1 = (-8.3, -8.3, -8.3). Con la combinación de estos 14 vértices: A1 = (0, 0, 10.3), B1 = (0, 0,- 10.3), C1 = (0, 10.3, 0), D1 = (0, -10.3, 0), E1 = (10.3, 0, 0), F1 = (-10.3, 0, 0), G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3), K1 = (8.3, 8.3, -8.3), L1 = (-8.3, 8.3, -8.3), M1 = (8.3, -8.3, -8.3), N1 = (-8.3, -8.3, -8.3), formamos el octaedro estrellado Luis Abinader. Estos vértices representan el punto central de las 14 caras que componen el hexaedro truncado.

Estos 38 puntos constituyen la representación dual del octaedro estrellado Luis Abinader: A = (4.3, 10.3, 10.3), B = (-4.3, 10.3, 10.3), C = (4.3, -10.3, 10.3), D = ( -4.3, -10.3, 10.3), E = (10.3, 4.3, 10.3), F = (-10.3, 4.3, 10.3), G = (10.3, -4.3, 10.3), H = (-10.3, -4.3, 10.3), I = (10.3, 10.3, 4.3), J = (-10.3, 10.3, 4.3), K = (10.3, -10.3, 4.3), L = (-10.3, -10.3, 4.3), M = (10.3, 10.3, -4.3), N = (-10.3, 10.3, -4.3), O = (10.3, -10.3,- 4.3), P = (-10.3, -10.3,- 4.3), Q = (4.3, 10.3, -10.3), R = (-4.3, 10.3, -10.3), S = (4.3, -10.3, -10.3), T = ( -4.3, -10.3, -10.3), U = (10.3, 4.3, -10.3), V = (-10.3, 4.3, -10.3), W = (10.3, -4.3, -10.3), Z = (-10.3, -4.3, -10.3), A1 = (0, 0, 10.3), B1 = (0, 0,- 10.3), C1 = (0, 10.3, 0), D1 = (0, -10.3, 0), E1 = (10.3, 0, 0), F1 = (-10.3, 0, 0), G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3), K1 = (8.3, 8.3, -8.3), L1 = (-8.3, 8.3, -8.3), M1 = (8.3, -8.3, -8.3), N1 = (-8.3, -8.3, -8.3).
English: Stellated octahedron Luis Abinader: it is a stellate concave three-dimensional solid, which is composed of 24 uniform faces, represented by Jose's isosceles triangles. In addition, it has 6 intermediate vertices that are uniform to each other, eight uniform outer vertices, for a total of 14 vertices, and it has 36 deformed edges. This three-dimensional polytope was discovered by Professor Jose Joel Leonardo on December 4, 2021 and was named by him, in honor of the memory of the President of the Dominican Republic, Lic. Luis Rodolfo Abinader Corona. The dual of the stellated octahedron Luis Abinader is the Archimedean solid, known as the truncated hexahedron.

These six vertices form a regular octahedron: A1 = (0, 0, 10.3), B1 = (0, 0, - 10.3), C1 = (0, 10.3, 0), D1 = (0, -10.3, 0), E1 = (10.3, 0, 0), F1 = (-10.3, 0, 0). These eight vertices form a regular Hexahedron: G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3 ), K1 = (8.3, 8.3, -8.3), L1 = (-8.3, 8.3, -8.3), M1 = (8.3, -8.3, -8.3), N1 = (-8.3, -8.3, -8.3). With the combination of these 14 vertices: A1 = (0, 0, 10.3), B1 = (0, 0, - 10.3), C1 = (0, 10.3, 0), D1 = (0, -10.3, 0), E1 = (10.3, 0, 0), F1 = (-10.3, 0, 0), G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3), K1 = (8.3, 8.3, -8.3), L1 = (-8.3, 8.3, -8.3), M1 = (8.3, -8.3, -8.3), N1 = (-8.3, -8.3, -8.3), we form the stellated octahedron Luis Abinader. These vertices represent the central point of the 14 faces that make up the truncated hexahedron.

These 38 points constitute the dual representation of the Luis Abinader stellated octahedron: A = (4.3, 10.3, 10.3), B = (-4.3, 10.3, 10.3), C = (4.3, -10.3, 10.3), D = (-4.3 , -10.3, 10.3), E = (10.3, 4.3, 10.3), F = (-10.3, 4.3, 10.3), G = (10.3, -4.3, 10.3), H = (-10.3, -4.3, 10.3) , I = (10.3, 10.3, 4.3), J = (-10.3, 10.3, 4.3), K = (10.3, -10.3, 4.3), L = (-10.3, -10.3, 4.3), M = (10.3, 10.3, -4.3), N = (-10.3, 10.3, -4.3), O = (10.3, -10.3, - 4.3), P = (-10.3, -10.3, - 4.3), Q = (4.3, 10.3, -10.3), R = (-4.3, 10.3, -10.3), S = (4.3, -10.3, -10.3), T = (-4.3, -10.3, -10.3), U = (10.3, 4.3, -10.3 ), V = (-10.3, 4.3, -10.3), W = (10.3, -4.3, -10.3), Z = (-10.3, -4.3, -10.3), A1 = (0, 0, 10.3), B1 = (0, 0, - 10.3), C1 = (0, 10.3, 0), D1 = (0, -10.3, 0), E1 = (10.3, 0, 0), F1 = (-10.3, 0, 0 ), G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3), K1 = (8.3 , 8.3, -8.3), L1 = (-8.3, 8.3, -8.3), M1 = (8.3, -8.3, -8.3), N1 = (-8.3, -8.3, -8.3).
Français : Octaèdre étoilé Luis Abinader : c'est un solide tridimensionnel concave étoilé, composé de 24 faces uniformes, représentées par les triangles isocèles de José. De plus, il a 6 sommets intermédiaires qui sont uniformes les uns par rapport aux autres, huit sommets externes uniformes, pour un total de 14 sommets, et il a 36 arêtes déformées. Ce polytope tridimensionnel a été découvert par le professeur José Joel Leonardo le 4 décembre 2021 et a été nommé par lui, en l'honneur de la mémoire du président de la République dominicaine, le lieutenant-colonel Luis Rodolfo Abinader Corona. Le dual de l'octaèdre étoilé Luis Abinader est le solide d'Archimède, connu sous le nom d'hexaèdre tronqué.

Ces six sommets forment un octaèdre régulier : A1 = (0, 0, 10.3), B1 = (0, 0, - 10.3), C1 = (0, 10.3, 0), D1 = (0, -10,3, 0), E1 = (10,3, 0, 0), F1 = (-10,3, 0, 0). Ces huit sommets forment un Hexaèdre régulier : G1 = (8.3, 8.3, 8.3), H1 = (-8.3, 8.3, 8.3), I1 = (8.3, -8.3, 8.3), J1 = (-8.3, -8.3, 8.3 ), K1 = (8,3, 8,3, -8,3), L1 = (-8,3, 8,3, -8,3), M1 = (8,3, -8,3, -8,3), N1 = (-8,3, -8,3, -8,3). Avec la combinaison de ces 14 sommets : A1 = (0, 0, 10.3), B1 = (0, 0, - 10,3), C1 = (0, 10,3, 0), D1 = (0, -10,3, 0), E1 = (10,3, 0, 0), F1 = (-10,3, 0, 0), G1 = (8,3, 8,3, 8,3), H1 = (-8,3, 8,3, 8,3), I1 = (8,3, -8,3, 8.3), J1 = (-8,3, -8,3, 8,3), K1 = (8,3, 8,3, -8,3), L1 = (-8,3, 8,3, -8,3), M1 = (8,3, -8,3, -8,3), N1 = (-8,3, -8,3, -8,3), nous formons l'octaèdre étoilé Luis Abinader. Ces sommets représentent le point central des 14 faces qui composent l'hexaèdre tronqué. Ces 38 points constituent la double représentation de l'octaèdre étoilé de Luis Abinader : A = (4.3, 10.3, 10.3), B = (-4,3, 10,3, 10,3), C = (4,3, -10,3, 10,3), D = (- 4,3, -10,3, 10,3), E = (10,3, 4,3, 10,3), F = (-10,3, 4,3, 10,3), G = (10,3, -4,3, 10,3), H = (-10,3, -4,3, 10,3 ) , I = (10,3, 10,3, 4,3), J = (-10,3, 10,3, 4,3), K = (10,3, -10,3, 4,3), L = (-10,3, -10,3, 4,3), M = (10,3 , 10,3, -4,3, N = (-10,3, 10,3, -4,3), O = (10,3, -10,3, - 4,3), P = (-10,3, -10,3, - 4,3), Q = (4,3, 10,3 , -10,3), R = (-4,3, 10,3, -10,3), S = (4,3, -10,3, -10,3), T = (-4,3, -10,3, -10,3), U = (10,3, 4,3, - 10,3, V = (-10,3, 4,3, -10,3), W = (10,3, -4,3, -10,3), Z = (-10,3, -4,3, -10,3), A1 = (0, 0, 10,3), B1 = (0, 0, - 10,3), C1 = (0, 10,3, 0), D1 = (0, -10,3, 0), E1 = (10,3, 0, 0), F1 = (-10,3, 0, 0 ), G1 = (8,3, 8,3, 8,3), H1 = (-8,3, 8,3, 8,3), I1 = (8,3, -8,3, 8,3), J1 = (-8,3, -8,3, 8,3), K1 = ( 8,3, 8,3, -8,3), L1 = (-8,3, 8,3, -8,3), M1 = (8,3, -8,3, -8,3), N1 = (-8,3, -8,3, -8,3). Más información sobre este texto de origenPara obtener más información sobre la traducción, se necesita el texto de origen Enviar comentarios

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