File:Hexaedro Estrellado de Joel Ampliado.gif
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Español: Hexaedro Estrellado de Joel Ampliado: Es un poliedro cóncavo estrellado hueco, que está compuesto por 24 caras que tienes formas de triángulos isósceles de Jose, 48 caras que poseen forma de triángulos escalenos, para una suma total de 72 caras triangulares. Además posee 8 vértices intermedios, 24 vértices interiores y 6 vértices exteriores para un total de 38 vértices. Posee 12 aristas intermedias, 72 aristas interiores, y 24 aristas exteriores, para una suma total de 108 aristas.
El hexaedro estrellado de Joel ampliado, pertenece al conjunto de las Sucesiones Poliédricas Triangulares, ocupando la posición # 23, (L=23), A=3L+3, C=2L+2, V=L+3, siendo A= Cantidad de arista del poliedro de caras triangulares, C= cantidad de caras triangulares y V= cantidad de vértices que posee el poliedro de caras triangulares. Esta formulas fueron descubiertas por el profesor Jose Joel Leonardo en el año 2012. Hexaedro estrellado de Joel ampliado área y Volumen. Arista intermedia de la base = L, L= 4, Arista intermedia Lateral = La, La = 16, arista interior de la base = Li, Li = 2.83, arista interior lateral Lil, Lil= 15.87450786638754, altura de la pirámide cuadrada semis hueca de Leonardo ampliada = hp, hp = 15.87450786638754, altura de uno de los triángulos isósceles de la base = h, h=2. K4=Li/L, sustituyendo Le y L: K4= 2.83/4 = 0.7075, K4=0.7075. Li = KL= 0.7075L, Li = 0.7075L K5 = h/L = 2/4 = 0.5 K5 = 0.5 Luego, h=K5L, sustituyendo K5: h = 0.5L. Área de uno de los 24 triángulos isósceles de Jose, que están colocados en la base que forman el hexaedro estrellado de Joel ampliado. Athej=Lh/2, sustituyendo h: Athej= 0.5L (L)/2 =0.25 (L elevada a potencia 2). Athej= 0.25 (L elevada a potencia 2). Área de uno de los triángulos escalenos rectángulo que forman el hexaedro estrellado de Joel ampliado = Atheja. Lil= 15.87450786638754, Lil= altura del triángulo escaleno rectángulo = hr =15.87450786638754 K6= hr/L = 15.87450786638754/4 =3.968626966596885 Luego hr = 3.968626966596885L Li = 0.7075L Atheja = (hr) (Li)/2 = (3.968626966596885 L) (0.7075 L) /2 = 2.807803578867296 (L elevada a potencia 2)/2 = 1.403901789433648 (L elevada a potencia 2). Atheja = 1.403901789433648 (L elevada a potencia 2). Área del hexaedro estrellado de Joel ampliado = Aheja. Área del hexaedro estrellado de Joel= Aheja. Aheja = 48Atheja + 24Athej=48(1.403901789433648 (L elevada a potencia 2)) + 24 (0.25 (L elevada a potencia 2)) = 67.3872858928151 (L elevada a potencia 2)+ 6 (L elevada a potencia 2) = 73.3872858928151 (L elevada a potencia 2). Aheja = 73.3872858928151 (L elevada a potencia 2). Volumen del hexaedro estrellado de Joel ampliado = Vheja. Vheja = 24 (Atheja) E + (L elevada a potencia 3) = 24(1.403901789433648 (L elevada a potencia 2)) E + (L elevada a potencia 3) = 33.69364294640755 (L elevada a potencia 2) E+ (L elevada a potencia 3). Vheja = 33.69364294640755 (L elevada a potencia 2) E+ (L elevada a potencia 3). En el mismo se cumple la fórmula de poliedro de Leonhard Euler: C+V-A=2, sustituyendo en la formula C=72, V=38, A=108, (C+V-A=2)= 72+38-108=2, hemos comprobado el cumplimiento de la fórmula de Euler.English: Joel Starred Hexahedron Expanded: It is a hollow starry concave polyhedron, which is composed of 24 faces that are shaped like Jose's isosceles triangles, 48 faces that have the shape of scalene triangles, for a total sum of 72 triangular faces. It also has 8 intermediate vertices, 24 interior vertices and 6 exterior vertices for a total of 38 vertices. It has 12 intermediate edges, 72 interior edges, and 24 exterior edges, for a total sum of 108 edges.
The expanded heelhedron of Joel, belongs to the set of Triangular Polyhedral Successions, occupying the position # 23, (L = 23), A = 3L + 3, C = 2L + 2, V = L + 3, being A = Quantity of polyhedron edge of triangular faces, C = amount of triangular faces and V = number of vertices that the polyhedron of triangular faces possesses. These formulas were discovered by Professor Jose Joel Leonardo in 2012. Joel's starry hexahedron expanded area and volume. Intermediate edge of the base = L, L = 4, Intermediate edge Lateral = La, La = 16, interior edge of the base = Li, Li = 2.83, interior lateral edge Lil, Lil = 15.87450786638754, height of the hollow semis square pyramid of Leonardo extended = hp, hp = 15.87450786638754, height of one of the isosceles triangles of the base = h, h = 2. K4 = Li / L, substituting Le and L: K4 = 2.83 / 4 = 0.7075, K4 = 0.7075. Li = KL = 0.7075L, Li = 0.7075L K5 = h / L = 2/4 = 0.5 K5 = 0.5 Then, h = K5L, substituting K5: h = 0.5L. Area of one of Jose's 24 isosceles triangles, which are placed at the base that form Joel's starry hexahedron. Athej = Lh / 2, substituting h: Athej = 0.5L (L) / 2 = 0.25 (L raised to power 2). Athej = 0.25 (L raised to power 2). Area of one of the right scalene triangles that form Joel's starry hexahedron enlarged = Atheja. Lil = 15.87450786638754, Lil = height of the right scalene triangle = hr = 15.87450786638754 K6 = hr / L = 15.87450786638754 / 4 = 3.968626966596885 Then hr = 3.968626966596885L Li = 0.7075L Atheja = (hr) (Li) / 2 = (3.968626966596885 L) (0.7075 L) / 2 = 2.807803578867296 (L raised to power 2) / 2 = 1.403901789433648 (L raised to power 2). Atheja = 1.403901789433648 (L raised to power 2). Joel's starry hexahedron area expanded = Aheja. Joel starry hexahedron area = Aheja. Aheja = 48Atheja + 24Athej = 48 (1.403901789433648 (L raised to power 2)) + 24 (0.25 (L raised to power 2)) = 67.3872858928151 (L raised to power 2) + 6 (L raised to power 2) = 73.3872858928151 ( L raised to power 2). Aheja = 73.3872858928151 (L raised to power 2). Volume of Joel's starry hexahedron expanded = Vheja. Vheja = 24 (Atheja) E + (L raised to power 3) = 24 (1.403901789433648 (L raised to power 2)) E + (L raised to power 3) = 33.69364294640755 (L raised to power 2) E + (L raised to power 3). Vheja = 33.69364294640755 (L raised to power 2) E + (L raised to power 3). In it, Leonhard Euler's polyhedron formula is met: C + VA = 2, substituting in formula C = 72, V = 38, A = 108, (C + VA = 2) = 72 + 38-108 = 2 , we have verified compliance with Euler's formula.Français : Joel Starred Hexahedron Expanded: Il s'agit d'un polyèdre concave étoilé creux, composé de 24 faces en forme de triangles isocèles de Jose, 48 faces en forme de triangles scalènes, pour un total de 72 faces triangulaires. Il possède également 8 sommets intermédiaires, 24 sommets intérieurs et 6 sommets extérieurs pour un total de 38 sommets. Il a 12 bords intermédiaires, 72 bords intérieurs et 24 bords extérieurs, pour une somme totale de 108 bords.
Le heelhedron étendu de Joel, appartient à l'ensemble des successions polyédriques triangulaires, occupant la position # 23, (L = 23), A = 3L + 3, C = 2L + 2, V = L + 3, étant A = quantité du bord du polyèdre des faces triangulaires, C = quantité de faces triangulaires et V = nombre de sommets que possède le polyèdre des faces triangulaires. Ces formules ont été découvertes par le professeur Jose Joel Leonardo en 2012. L'hexaèdre étoilé de Joel a étendu la surface et le volume. Bord intermédiaire de la base = L, L = 4, Bord intermédiaire latéral = La, La = 16, bord intérieur de la base = Li, Li = 2,83, bord latéral intérieur Lil, Lil = 15,87450786638754, hauteur de la pyramide carrée semi-creuse de Leonardo étendu = hp, hp = 15.87450786638754, hauteur de l'un des triangles isocèles de la base = h, h = 2. K4 = Li / L, en remplaçant Le et L: K4 = 2,83 / 4 = 0,7075, K4 = 0,7075. Li = KL = 0,7075L, Li = 0,7075L K5 = h / L = 2/4 = 0,5 K5 = 0,5 Ensuite, h = K5L, en remplaçant K5: h = 0,5L. Aire d'un des 24 triangles isocèles de Jose, qui sont placés à la base qui forme l'hexaèdre étoilé de Joel. Athej = Lh / 2, en remplaçant h: Athej = 0,5L (L) / 2 = 0,25 (L élevé à la puissance 2). Athej = 0,25 (L élevé à la puissance 2). Aire de l'un des triangles scalènes droits qui forme l'hexaèdre étoilé de Joel agrandi = Atheja. Lil = 15.87450786638754, Lil = hauteur du triangle scalène droit = h = 15.87450786638754 K6 = h / L = 15,87450786638754 / 4 = 3,968626966596885 Alors h = 3,968626966596885L Li = 0,7075L Atheja = (h) (Li) / 2 = (3,968626966596885 L) (0,7075 L) / 2 = 2,807803578867296 (L élevé à la puissance 2) / 2 = 1,403901789433648 (L élevé à la puissance 2). Atheja = 1,403901789433648 (L élevé à la puissance 2). La zone hexaèdre étoilée de Joel s'est étendue = Aheja. Zone hexaèdre étoilée de Joel = Aheja. Aheja = 48Atheja + 24Athej = 48 (1.403901789433648 (L élevé à la puissance 2)) + 24 (0,25 (L élevé à la puissance 2)) = 67.3872858928151 (L élevé à la puissance 2) + 6 (L élevé à la puissance 2) = 73.3872858928151 ( L élevé au pouvoir 2). Aheja = 73,3872858928151 (L élevé à la puissance 2). Volume de l'hexaèdre étoilé de Joel augmenté = Vheja. Vheja = 24 (Atheja) E + (L élevé à la puissance 3) = 24 (1.403901789433648 (L élevé à la puissance 2)) E + (L élevé à la puissance 3) = 33.69364294640755 (L élevé à la puissance 2) E + (L élevé à la puissance 2) puissance 3). Vheja = 33,69364294640755 (L élevé à la puissance 2) E + (L élevé à la puissance 3). Dans ce document, la formule du polyèdre de Leonhard Euler est remplie: C + VA = 2, en remplaçant dans la formule C = 72, V = 38, A = 108, (C + VA = 2) = 72 + 38-108 = 2 , nous avons vérifié la conformité avec la formule d'Euler. |
| Date | |
| Source | Own work |
| Author | Jose J. Leonard |
https://commons.wikimedia.org/wiki/File:Estelaciones_del_Cubo.jpg, https://commons.wikimedia.org/wiki/File:Poliedro_Hueco.jpg, https://commons.wikimedia.org/wiki/File:Sucesiones_Poliedricas_Triangulares.jpg https://commons.wikimedia.org/wiki/File:Hexaedro_estrellado_de_Joel.gif https://www.geogebra.org/m/uFEGhbXD
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