File:Cuboedro de Joel.gif
Cuboedro_de_Joel.gif (639 × 483 pixels, file size: 1.35 MB, MIME type: image/gif, looped, 146 frames, 11 s)
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[edit]DescriptionCuboedro de Joel.gif |
Español: La séptima estelación del hexaedro está representada por el cuboedro de Joel. Este poliedro posee 24 caras interiores triangulares isósceles. Ostenta 6 vértices interiores y 8 vértices intermedios, para un total de 14 vértices. Además tiene 24 aristas interiores y 12 aristas intermedias, para un total de 36 aristas.
Siendo m = 4, n =3 sustituyendo en el segundo miembro del bispar poliédrico (m, n) = (4, 3), aplicando fórmulas y resolviendo la ecuación, AI = -2mn / 2m + 2n – mn = -2 (4) (3) / 2 (3) + 2 (3) -4 (3) = -24/2 = -12 entonces AI = - 12, satisface la ecuación. CI = 2AI / n = 2 (-12) / 3 = -8, C = -8, V = 2AI / m = 2 (-12) / 4 = -6, V = -6. Sustituyendo, m = 3, n =3, en el primer miembro del bispar poliédrico- (2m, n) = (2(3), 3) = -(6, 3), aplicando fórmulas y resolviendo Ai = 2AI = 2 (-12) = -24, Ai = -24, Ci= Ai = -24, Ci = -24, Vi = 2 (Ai/2m) = 2 (-24 /6) = 2 (-4) = -8, V = -8. Probando Dc: Experimentemos con el hexaedro regular (3, 4). Dc ≈ w(0.866), siendo w ˃ 0, w∈R entonces w =4Cm entonces Dc= 4Cm(0.866) =3.464 Cm Dc = 3.464Cm. Como 4 ˃ 3.46, entonces el hexaedro regular no genera ningún poliedro regular hueco porque Dc < A. El hexaedro regular genera interiormente un poliedro semirregular llamado el cuboedro de Joel – V (6, 3s) – V (4, 3s), que está compuesto por un conjunto de triángulos isósceles. En todas las aristas de este poliedro, comprobamos que “k” sí varía su resultado debido a que existen dos clases de aristas diferentes, por lo tanto, el cuboedro de Joel no es válido como poliedro regular. File:Cuboedro_de_Joel_Vértices_Poliedricos.jpg https://www.geogebra.org/classic/fcEtPrJF File:Séptima_Estelación_del_Cubo_o_Hexaedro.jpg |
Date | |
Source | Own work |
Author | J.Joel Leonardo |
https://www.geogebra.org/classic/fcEtPrJF
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