File:Tetraedro Davinciano Dual Del Tetraedro Truncado.gif

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Tetraedro_Davinciano_Dual_Del_Tetraedro_Truncado.gif ‎(434 × 385 pixels, file size: 2.48 MB, MIME type: image/gif, looped, 151 frames, 11 s)

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In the truncated Archimedean tetrahedron, determine a centralized point on each of its faces, with these centralized vertices we define the Davincian stellated tetrahedron.

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Español: Con ochos puntos centralizados, en las ochos caras poliédricas del tetraedro truncado de Arquímedes, definimos perfectamente el tetraedro estrellado Davinciano. Para la demostración grafica definimos estos 20 vértices: A=(1.6666666666666667,-2.886751345948129,-2.041241452319315), B=(1.6666666666666667,2.886751345948129,-2.041241452319315), C=(-1.6666666666666667,-2.886751345948129,-2.041241452319315), D=(-1.797762114336972,2.886751345948129,-2.041241452319316), E=(3.3333333333333335,0,-2.041241452319315), F=(-3.3333333333333335,0,-2.041241452319315), G=(3.3333333333333335,1.924500897298752,0.680413817439772), H=(-3.3333333333333335,1.924500897298752,0.680413817439772), I=(0,-3.849001794597505,0.680413817439772), J=(-1.6666666666666667,0.962250448649376,3.402069087198858), K=(1.6666666666666667,0.962250448649376,3.402069087198858), L=(0,-1.924500897298753,3.402069087198858), M=(0,0,3.402069087198858), N=(-2.777777777777778,1.603750747748961,-1.134023029066286), O=(2.777777777777778,1.603750747748961,-1.134023029066286), P=(0,-3.207501495497921,-1.134023029066286), Q=(-1.666666666666666,-0.962250448649377,0.680413817439772), R=(1.666666666666667,-0.962250448649377,0.680413817439772), S=(0,1.924500897298753,0.680413817439772), T=(0,0,-2.041241452319315). Utilizamos los primeros doce tríos cartesianos (A, B, C, D, E, F, G, H, I, J, K, L), desde el punto A, hasta el punto L, los cuales definen el tetraedro truncado de Arquímedes. Luego desde el punto M, hasta el punto T, definen el centro de cada una de las caras del politopo de Arquímedes. Combinando los puntos (M, N, O, P, Q, R, S, T), determinamos el tetraedro estrellado Davinciano, demostrando la dualidad existente entre ambos solidos geométricos.
English: With eight centralized points, on the eight polyhedral faces of the truncated Archimedean tetrahedron, we perfectly define the Davincian stellated tetrahedron. To define the graph showing the 20 vertices: A = (1.6666666666666667, -2.886751345948129, -2.041241452319315), B = (1.6666666666666667,2.886751345948129, -2.041241452319315), C = (- 1.6666666666666667, -2.886751345948129, -2.041241452319315), D = (- 1.797762114336972 , 2.886751345948129, -2.041241452319316), E = (3.3333333333333335,0, -2.041241452319315), F = (- 3.3333333333333335,0, -2.041241452319315), G = (3.3333333333333335,1.924500897298752,0.680413817439772), H = (- 3.3333333333333335,1.924500897298752,0.680413817439772) , I = (0, -3.849001794597505,0.680413817439772), J = (- 1.6666666666666667,0.962250448649376,3.402069087198858), K = (1.6666666666666667,0.962250448649376,3.402069087198858), L = (0, -1.924500897298753,3.402069087198858), M = (0,0 , 3.402069087198858), N = (- 2.777777777777778,1.603750747748961, -1.134023029066286), O = (2.777777777777778,1.603750747748961, -1.134023029066286), P = (0, -3.207501495497921, -1.134023029066286), Q = (- 1.666666666666666, -0.962250448649377,0.680413817439772) , R=(1.666666666666667 ,-0.962250448649377,0.680413817439772), S=(0,1.924500897298753,0.680413817439772), T=(0.0,-2.041241452319315). We use the first twelve Cartesian triplets (A, B, C, D, E, F, G, H, I, J, K, L), from point A to point L, which define the truncated Archimedean tetrahedron . Then from point M, to point T, define the center of each of the faces of the Archimedean polytope. Combining the points (M, N, O, P, Q, R, S, T), we determine the Davincian star tetrahedron, demonstrating the duality between both geometric solids.
Français : Avec huit points centralisés, sur les huit faces polyédriques du tétraèdre archimédien tronqué, on définit parfaitement le tétraèdre étoilé Davincien. Pour définir le graphique représentant les 20 sommets: A = (1,6666666666666667, -2,886751345948129, -2,041241452319315), B = (1.6666666666666667,2.886751345948129, -2,041241452319315), C = (- 1,6666666666666667, -2,886751345948129, -2,041241452319315), D = (- 1,797762114336972 , 2,886751345948129, -2,041241452319316), E = (3.3333333333333335,0, -2,041241452319315), F = (- 3.3333333333333335,0, -2,041241452319315), G = (3.3333333333333335,1.924500897298752,0.680413817439772), H = (- 3.3333333333333335,1.924500897298752,0.680413817439772) , i = (0, -3.849001794597505,0.680413817439772), J = (- 1.6666666666666667,0.962250448649376,3.402069087198858), K = (1.6666666666666667,0.962250448649376,3.402069087198858), L = (0, -1.924500897298753,3.402069087198858), M = (0,0 , 3,402069087198858), N = (- 2.777777777777778,1.603750747748961, -1,134023029066286), O = (2.777777777777778,1.603750747748961, -1,134023029066286), P = (0, -3,207501495497921, -1,134023029066286), Q = (- 1,666666666666666, -0.962250448649377,0.680413817439772) , R=(1.666666666666667 ,-0,962250448649377,0,680413817439772), S=(0,1,924500897298753,0,680413817439772), T=(0,0,-2,041241452319315). On utilise les douze premiers triplets cartésiens (A, B, C, D, E, F, G, H, I, J, K, L), du point A au point L, qui définissent le tétraèdre d'Archimède tronqué. Puis du point M, au point T, définir le centre de chacune des faces du polytope d'Archimède. En combinant les points (M, N, O, P, Q, R, S, T), nous déterminons le tétraèdre étoilé Davincien, démontrant la dualité entre les deux solides géométriques.
Italiano: Con otto punte centralizzate, sulle otto facce poliedriche del tetraedro archimedeo troncato, definiamo perfettamente il tetraedro stellato davinciano. Per definire il grafico che mostra i vertici 20: A = (1,6666666666666667, -2,886751345948129, -2,041241452319315), B = (1.6666666666666667,2.886751345948129, -2,041241452319315), C = (- 1,6666666666666667, -2,886751345948129, -2,041241452319315), D = (- 1,797762114336972 , 2,886751345948129, -2,041241452319316), E = (3.3333333333333335,0, -2,041241452319315), F = (- 3.3333333333333335,0, -2,041241452319315), G = (3.3333333333333335,1.924500897298752,0.680413817439772), H = (- 3.3333333333333335,1.924500897298752,0.680413817439772) , I = (0, -3.849001794597505,0.680413817439772), J = (- 1.6666666666666667,0.962250448649376,3.402069087198858), K = (1.6666666666666667,0.962250448649376,3.402069087198858), L = (0, -1.924500897298753,3.402069087198858), M = (0,0 , 3,402069087198858), N = (- 2.777777777777778,1.603750747748961, -1,134023029066286), O = (2.777777777777778,1.603750747748961, -1,134023029066286), P = (0, -3,207501495497921, -1,134023029066286), Q = (- 1,666666666666666, -0.962250448649377,0.680413817439772) , R=(1.6666666666666667 ,-0.962250448649377,0.680413817439772), S=(0,1.924500897298753,0.680413817439772), T=(0.0,-2.041241452319315). Utilizziamo le prime dodici terzine cartesiane (A, B, C, D, E, F, G, H, I, J, K, L), dal punto A al punto L, che definiscono il tetraedro archimede troncato. Quindi dal punto M, al punto T, definire il centro di ciascuna delle facce del politopo di Archimede. Combinando i punti (M, N, O, P, Q, R, S, T), determiniamo il tetraedro della stella davinciana, dimostrando la dualità tra i due solidi geometrici.
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Author Jose J. Leonard

File:Área_y_Volumen_del_Tetraedro_Estrellado_Davinciano.gif File:Tetraedro_Estrellado_Davinciano.gif https://www.geogebra.org/m/ehnngveg

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