Elongated triangular orthobicupola: Difference between revisions
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{{Short description|Johnson solid with 20 faces}} |
{{Short description|Johnson solid with 20 faces}} |
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{{Infobox polyhedron |
{{Infobox polyhedron |
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|image= |
| image = Elongated triangular orthobicupola.png |
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|type=[[Johnson solid|Johnson]]<br>{{math|[[pentagonal orthobirotunda|''J''{{sub|34}}]] – '''''J''{{sub|35}}''' – [[elongated triangular gyrobicupola|''J''{{sub|36}}]]}} |
| type = [[Johnson solid|Johnson]]<br>{{math|[[pentagonal orthobirotunda|''J''{{sub|34}}]] – '''''J''{{sub|35}}''' – [[elongated triangular gyrobicupola|''J''{{sub|36}}]]}} |
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|faces= |
| faces = 8 [[triangle]]s<br>12 [[Square (geometry)|square]]s |
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|edges=36 |
| edges = 36 |
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|vertices=18 |
| vertices = 18 |
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|symmetry= |
| symmetry = <math> D_{3h} </math> |
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|vertex_config= |
| vertex_config = <math> \begin{align} |
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&6 \times (3 \times 4 \times 3 \times 4) + \\ |
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|dual=- |
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&12 \times (3 \times 4^3) |
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\end{align} </math> |
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}} |
}} |
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In [[geometry]], the '''elongated triangular orthobicupola |
In [[geometry]], the '''elongated triangular orthobicupola''' is a polyhedron constructed by attaching two regular [[triangular cupola]] into the base of a regular [[hexagonal prism]]. It is an example of [[Johnson solid]]. |
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== Construction == |
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{{Johnson solid}} |
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The elongated triangular orthobicupola can be constructed from a [[hexagonal prism]] by attaching two regular [[triangular cupola]]e onto its base, covering its hexagonal faces.{{r|rajwade}} This construction process known as [[Elongation (geometry)|elongation]], giving the resulting polyhedron has 8 [[equilateral triangle]]s and 12 squares.{{r|berman}} A [[Convex set|convex]] polyhedron in which all faces are [[Regular polygon|regular]] is [[Johnson solid]], and the elongated triangular orthobicupola is one among them, enumerated as 35th Johnson solid <math> J_{35} </math>.{{r|francis}} |
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== |
== Properties == |
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An elongated triangular orthobicupola with a given edge length <math> a </math> has a surface area, by adding the area of all regular faces:{{r|berman}} |
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The volume of ''J''<sub>35</sub> can be calculated as follows: |
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<math display="block"> \left(12 + 2\sqrt{3}\right)a^2 \approx 15.464a^2. </math> |
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Its volume can be calculated by cutting it off into two triangular cupolae and a hexagonal prism with regular faces, and then adding their volumes up:{{r|berman}} |
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It has the same [[Point groups in three dimensions|three-dimensional symmetry groups]] as the [[triangular orthobicupola]], the dihedral group <math> D_{3h} </math> of order 12. Its [[dihedral angle]] can be calculated by adding the angle of the triangular cupola and hexagonal prism. The dihedral angle of a hexagonal prism between two adjacent squares is the [[internal angle]] of a regular hexagon <math> 120^\circ = 2\pi/3</math>, and that between its base and square face is <math> \pi/2 = 90^\circ </math>. The dihedral angle of a regular triangular cupola between each triangle and the hexagon is approximately <math> 70.5^\circ </math>, that between each square and the hexagon is <math> 54.7^\circ </math>, and that between square and triangle is <math> 125.3^\circ </math>. The dihedral angle of an elongated triangular orthobicupola between the triangle-to-square and square-to-square, on the edge where the triangular cupola and the prism is attached, is respectively:{{r|johnson}} |
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''J''<sub>35</sub> consists of 2 cupolae and hexagonal prism. |
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<math display="block"> \begin{align} |
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\frac{\pi}{2} + 70.5^\circ &\approx 160.5^\circ, \\ |
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\frac{\pi}{2} + 54.7^\circ &\approx 144.7^\circ. |
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\end{align} </math> |
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The two cupolae makes 1 cuboctahedron = 8 tetrahedra + 6 half-octahedra. |
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1 octahedron = 4 tetrahedra, so total we have 20 tetrahedra. |
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What is the volume of a tetrahedron? |
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Construct a tetrahedron having vertices in common with alternate vertices of a |
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cube (of side <math>\textstyle\frac{1}{\sqrt{2}}</math>, if tetrahedron has unit edges). The 4 triangular |
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pyramids left if the tetrahedron is removed from the cube form half an |
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octahedron = 2 tetrahedra. So |
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:<math>V_\mathrm{tetrahedron} = \frac{1}{3} V_\mathrm{cube} = \frac{1}{3} \frac{1}{{\sqrt{2}}^3} = \frac{\sqrt{2}}{12}</math> |
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{{reflist|refs= |
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<ref name="berman">{{cite journal |
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The hexagonal prism is more straightforward. The hexagon has area <math>\textstyle 6 \frac{\sqrt{3}}{4}</math>, so |
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| last = Berman | first = Martin |
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| year = 1971 |
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| title = Regular-faced convex polyhedra |
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| journal = Journal of the Franklin Institute |
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| volume = 291 |
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| issue = 5 |
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| pages = 329–352 |
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| doi = 10.1016/0016-0032(71)90071-8 |
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| mr = 290245 |
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}}</ref> |
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<ref name="francis">{{cite journal |
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:<math>V_\mathrm{prism} = \frac{3 \sqrt{3}}{2}</math> |
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| last = Francis | first = Darryl |
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| title = Johnson solids & their acronyms |
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| journal = Word Ways |
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| date = August 2013 |
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| volume = 46 | issue = 3 | page = 177 |
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| url = https://go.gale.com/ps/i.do?id=GALE%7CA340298118 |
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}}</ref> |
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<ref name="johnson">{{cite journal |
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Finally |
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| last = Johnson | first = Norman W. | authorlink = Norman W. Johnson |
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| year = 1966 |
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| title = Convex polyhedra with regular faces |
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| journal = [[Canadian Journal of Mathematics]] |
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| volume = 18 |
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| pages = 169–200 |
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| doi = 10.4153/cjm-1966-021-8 |
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| mr = 0185507 |
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| s2cid = 122006114 |
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| zbl = 0132.14603| doi-access = free |
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}}</ref> |
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<ref name="rajwade">{{cite book |
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:<math>V_{J_{35}} = 20 V_\mathrm{tetrahedron} + V_\mathrm{prism} = |
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| last = Rajwade | first = A. R. |
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| title = Convex Polyhedra with Regularity Conditions and Hilbert's Third Problem |
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| series = Texts and Readings in Mathematics |
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| year = 2001 |
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| url = https://books.google.com/books?id=afJdDwAAQBAJ&pg=PA84 |
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| page = 84–89 |
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| publisher = Hindustan Book Agency |
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| isbn = 978-93-86279-06-4 |
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| doi = 10.1007/978-93-86279-06-4 |
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}}</ref> |
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}} |
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numerical value: |
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:<math>V_{J_{35}} = 4.9550988153084743549606507192748</math> |
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{{Reflist}} |
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==External links== |
==External links== |
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* {{MathWorld2|title2=Johnson solid|urlname2=JohnsonSolid| urlname=ElongatedTriangularOrthobicupola | title=Elongated triangular orthobicupola}} |
* {{MathWorld2|title2=Johnson solid|urlname2=JohnsonSolid| urlname=ElongatedTriangularOrthobicupola | title=Elongated triangular orthobicupola}} |
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{{Johnson solids navigator}} |
{{Johnson solids navigator}} |
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{{Polyhedron-stub}} |
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[[Category:Johnson solids]] |
[[Category:Johnson solids]] |
Latest revision as of 02:15, 1 April 2024
Elongated triangular orthobicupola | |
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Type | Johnson J34 – J35 – J36 |
Faces | 8 triangles 12 squares |
Edges | 36 |
Vertices | 18 |
Vertex configuration | |
Symmetry group | |
Properties | convex |
Net | |
In geometry, the elongated triangular orthobicupola is a polyhedron constructed by attaching two regular triangular cupola into the base of a regular hexagonal prism. It is an example of Johnson solid.
Construction
[edit]The elongated triangular orthobicupola can be constructed from a hexagonal prism by attaching two regular triangular cupolae onto its base, covering its hexagonal faces.[1] This construction process known as elongation, giving the resulting polyhedron has 8 equilateral triangles and 12 squares.[2] A convex polyhedron in which all faces are regular is Johnson solid, and the elongated triangular orthobicupola is one among them, enumerated as 35th Johnson solid .[3]
Properties
[edit]An elongated triangular orthobicupola with a given edge length has a surface area, by adding the area of all regular faces:[2] Its volume can be calculated by cutting it off into two triangular cupolae and a hexagonal prism with regular faces, and then adding their volumes up:[2]
It has the same three-dimensional symmetry groups as the triangular orthobicupola, the dihedral group of order 12. Its dihedral angle can be calculated by adding the angle of the triangular cupola and hexagonal prism. The dihedral angle of a hexagonal prism between two adjacent squares is the internal angle of a regular hexagon , and that between its base and square face is . The dihedral angle of a regular triangular cupola between each triangle and the hexagon is approximately , that between each square and the hexagon is , and that between square and triangle is . The dihedral angle of an elongated triangular orthobicupola between the triangle-to-square and square-to-square, on the edge where the triangular cupola and the prism is attached, is respectively:[4]
Related polyhedra and honeycombs
[edit]The elongated triangular orthobicupola forms space-filling honeycombs with tetrahedra and square pyramids.[5]
References
[edit]- ^ Rajwade, A. R. (2001). Convex Polyhedra with Regularity Conditions and Hilbert's Third Problem. Texts and Readings in Mathematics. Hindustan Book Agency. p. 84–89. doi:10.1007/978-93-86279-06-4. ISBN 978-93-86279-06-4.
- ^ a b c Berman, Martin (1971). "Regular-faced convex polyhedra". Journal of the Franklin Institute. 291 (5): 329–352. doi:10.1016/0016-0032(71)90071-8. MR 0290245.
- ^ Francis, Darryl (August 2013). "Johnson solids & their acronyms". Word Ways. 46 (3): 177.
- ^ Johnson, Norman W. (1966). "Convex polyhedra with regular faces". Canadian Journal of Mathematics. 18: 169–200. doi:10.4153/cjm-1966-021-8. MR 0185507. S2CID 122006114. Zbl 0132.14603.
- ^ "J35 honeycomb".