Irregular analogue of cube and octahedron.

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

favorite












If we take a Dodecahedron and remove the constraint that the pentagonal faces have to be regular pentagons, we get a solid called a Tetartoid. If we take the dual of that, we will end up with the corresponding irregular Icosahedron.



enter image description here



Similarly, there seems to be an irregular analogue for the tetrahedron. However, I've never seen one for the cube and octahedron (which happen to be dual of each other). Is it even possible to have irregular analogues of those two? The criterion is that the solid should be convex with all faces identical and the same number of faces meeting at each vertex. The cube analogue would have six faces which were all quadrilaterals while the octahedron analogue would have eight faces which were all scalene triangles.



EDIT: Actually the Icosahedral object might not necessarily be composed of identical triangles.




geometry euclidean-geometry solid-geometry platonic-solids




share|cite|improve this question





















  • See this for an irregular cube: math.stackexchange.com/questions/2144830/…
    – Aretino
    19 hours ago














up vote
4
down vote

favorite












If we take a Dodecahedron and remove the constraint that the pentagonal faces have to be regular pentagons, we get a solid called a Tetartoid. If we take the dual of that, we will end up with the corresponding irregular Icosahedron.



enter image description here



Similarly, there seems to be an irregular analogue for the tetrahedron. However, I've never seen one for the cube and octahedron (which happen to be dual of each other). Is it even possible to have irregular analogues of those two? The criterion is that the solid should be convex with all faces identical and the same number of faces meeting at each vertex. The cube analogue would have six faces which were all quadrilaterals while the octahedron analogue would have eight faces which were all scalene triangles.



EDIT: Actually the Icosahedral object might not necessarily be composed of identical triangles.




geometry euclidean-geometry solid-geometry platonic-solids




share|cite|improve this question





















  • See this for an irregular cube: math.stackexchange.com/questions/2144830/…
    – Aretino
    19 hours ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











If we take a Dodecahedron and remove the constraint that the pentagonal faces have to be regular pentagons, we get a solid called a Tetartoid. If we take the dual of that, we will end up with the corresponding irregular Icosahedron.



enter image description here



Similarly, there seems to be an irregular analogue for the tetrahedron. However, I've never seen one for the cube and octahedron (which happen to be dual of each other). Is it even possible to have irregular analogues of those two? The criterion is that the solid should be convex with all faces identical and the same number of faces meeting at each vertex. The cube analogue would have six faces which were all quadrilaterals while the octahedron analogue would have eight faces which were all scalene triangles.



EDIT: Actually the Icosahedral object might not necessarily be composed of identical triangles.




geometry euclidean-geometry solid-geometry platonic-solids




share|cite|improve this question













If we take a Dodecahedron and remove the constraint that the pentagonal faces have to be regular pentagons, we get a solid called a Tetartoid. If we take the dual of that, we will end up with the corresponding irregular Icosahedron.



enter image description here



Similarly, there seems to be an irregular analogue for the tetrahedron. However, I've never seen one for the cube and octahedron (which happen to be dual of each other). Is it even possible to have irregular analogues of those two? The criterion is that the solid should be convex with all faces identical and the same number of faces meeting at each vertex. The cube analogue would have six faces which were all quadrilaterals while the octahedron analogue would have eight faces which were all scalene triangles.



EDIT: Actually the Icosahedral object might not necessarily be composed of identical triangles.





geometry euclidean-geometry solid-geometry platonic-solids





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 18:47
























asked Jul 29 at 7:19









Rohit Pandey

791718




791718











  • See this for an irregular cube: math.stackexchange.com/questions/2144830/…
    – Aretino
    19 hours ago
















  • See this for an irregular cube: math.stackexchange.com/questions/2144830/…
    – Aretino
    19 hours ago















See this for an irregular cube: math.stackexchange.com/questions/2144830/…
– Aretino
19 hours ago




See this for an irregular cube: math.stackexchange.com/questions/2144830/…
– Aretino
19 hours ago










3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










For the octahedron it is easy. Take a rhombus $R$ in the $x$-$y$-plane. The diagonals divide it into four congruent scalene triangles. The diagonals intersect in $p$, so shift $p$ in positive $z$-direction and obtain the apex of a pyramid $P$ with base $R$. Reflect $P$ at the $x$-$y$-plane and obtain a pyramid $P'$. Then $T = P cup P'$ is the desired object. It is obviously obtained by a distortion of the standard octahedron.



For the cube we can proceed similarly and obtain an object with rhombical faces. I do not know whether it is possible to find something with more irregular faces.






share|cite|improve this answer























  • Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
    – Rohit Pandey
    Jul 29 at 17:52






  • 1




    For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
    – Paul Frost
    Jul 29 at 18:40

















up vote
2
down vote













A cube could be elongated in either of its directions of face normals. this makes some of its faces to rectangles instead.



Alternatively a pair of faces (bases, if considered a square prism) could be sheered wrt each other. This makes some of its faces into parallelograms.



Finally you could squash a cube diametrally (along its body diagonal). This makes all faces into rhombs.



And, for sure, you could combine all theese operations.



--- rk






share|cite|improve this answer

















  • 2




    The OP wants all faces identical.
    – Ethan Bolker
    Jul 29 at 13:18

















up vote
2
down vote













See the Wikipedia entry on the rhombohedron for images:



         
rhombohedron

Note the 2nd shape: $6$ identical rhombi.




share|cite|improve this answer

















  • 1




    Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
    – Rohit Pandey
    Jul 29 at 18:44










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865842%2firregular-analogue-of-cube-and-octahedron%23new-answer', 'question_page');

);

Post as a guest

















3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










For the octahedron it is easy. Take a rhombus $R$ in the $x$-$y$-plane. The diagonals divide it into four congruent scalene triangles. The diagonals intersect in $p$, so shift $p$ in positive $z$-direction and obtain the apex of a pyramid $P$ with base $R$. Reflect $P$ at the $x$-$y$-plane and obtain a pyramid $P'$. Then $T = P cup P'$ is the desired object. It is obviously obtained by a distortion of the standard octahedron.



For the cube we can proceed similarly and obtain an object with rhombical faces. I do not know whether it is possible to find something with more irregular faces.






share|cite|improve this answer























  • Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
    – Rohit Pandey
    Jul 29 at 17:52






  • 1




    For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
    – Paul Frost
    Jul 29 at 18:40














up vote
3
down vote



accepted










For the octahedron it is easy. Take a rhombus $R$ in the $x$-$y$-plane. The diagonals divide it into four congruent scalene triangles. The diagonals intersect in $p$, so shift $p$ in positive $z$-direction and obtain the apex of a pyramid $P$ with base $R$. Reflect $P$ at the $x$-$y$-plane and obtain a pyramid $P'$. Then $T = P cup P'$ is the desired object. It is obviously obtained by a distortion of the standard octahedron.



For the cube we can proceed similarly and obtain an object with rhombical faces. I do not know whether it is possible to find something with more irregular faces.






share|cite|improve this answer























  • Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
    – Rohit Pandey
    Jul 29 at 17:52






  • 1




    For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
    – Paul Frost
    Jul 29 at 18:40












up vote
3
down vote



accepted







up vote
3
down vote



accepted






For the octahedron it is easy. Take a rhombus $R$ in the $x$-$y$-plane. The diagonals divide it into four congruent scalene triangles. The diagonals intersect in $p$, so shift $p$ in positive $z$-direction and obtain the apex of a pyramid $P$ with base $R$. Reflect $P$ at the $x$-$y$-plane and obtain a pyramid $P'$. Then $T = P cup P'$ is the desired object. It is obviously obtained by a distortion of the standard octahedron.



For the cube we can proceed similarly and obtain an object with rhombical faces. I do not know whether it is possible to find something with more irregular faces.






share|cite|improve this answer















For the octahedron it is easy. Take a rhombus $R$ in the $x$-$y$-plane. The diagonals divide it into four congruent scalene triangles. The diagonals intersect in $p$, so shift $p$ in positive $z$-direction and obtain the apex of a pyramid $P$ with base $R$. Reflect $P$ at the $x$-$y$-plane and obtain a pyramid $P'$. Then $T = P cup P'$ is the desired object. It is obviously obtained by a distortion of the standard octahedron.



For the cube we can proceed similarly and obtain an object with rhombical faces. I do not know whether it is possible to find something with more irregular faces.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 29 at 13:15


























answered Jul 29 at 10:33









Paul Frost

3,583420




3,583420











  • Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
    – Rohit Pandey
    Jul 29 at 17:52






  • 1




    For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
    – Paul Frost
    Jul 29 at 18:40
















  • Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
    – Rohit Pandey
    Jul 29 at 17:52






  • 1




    For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
    – Paul Frost
    Jul 29 at 18:40















Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
– Rohit Pandey
Jul 29 at 17:52




Thanks. It was very surprising to me that the dual solid of the irregular Octahedron doesn't satisfy my requirements. I don't understand how you would go about the cube analogue. What 2-D shape would you start with?
– Rohit Pandey
Jul 29 at 17:52




1




1




For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
– Paul Frost
Jul 29 at 18:40




For the cube anlaogue I would start with a rhombus in the $x$-$y$-plane and add congruent rhombi along the edges. It is inelegant to describe this with words. The result is known as the trigonal trapezohedron. See en.wikipedia.org/wiki/Rhombohedron.
– Paul Frost
Jul 29 at 18:40










up vote
2
down vote













A cube could be elongated in either of its directions of face normals. this makes some of its faces to rectangles instead.



Alternatively a pair of faces (bases, if considered a square prism) could be sheered wrt each other. This makes some of its faces into parallelograms.



Finally you could squash a cube diametrally (along its body diagonal). This makes all faces into rhombs.



And, for sure, you could combine all theese operations.



--- rk






share|cite|improve this answer

















  • 2




    The OP wants all faces identical.
    – Ethan Bolker
    Jul 29 at 13:18














up vote
2
down vote













A cube could be elongated in either of its directions of face normals. this makes some of its faces to rectangles instead.



Alternatively a pair of faces (bases, if considered a square prism) could be sheered wrt each other. This makes some of its faces into parallelograms.



Finally you could squash a cube diametrally (along its body diagonal). This makes all faces into rhombs.



And, for sure, you could combine all theese operations.



--- rk






share|cite|improve this answer

















  • 2




    The OP wants all faces identical.
    – Ethan Bolker
    Jul 29 at 13:18












up vote
2
down vote










up vote
2
down vote









A cube could be elongated in either of its directions of face normals. this makes some of its faces to rectangles instead.



Alternatively a pair of faces (bases, if considered a square prism) could be sheered wrt each other. This makes some of its faces into parallelograms.



Finally you could squash a cube diametrally (along its body diagonal). This makes all faces into rhombs.



And, for sure, you could combine all theese operations.



--- rk






share|cite|improve this answer













A cube could be elongated in either of its directions of face normals. this makes some of its faces to rectangles instead.



Alternatively a pair of faces (bases, if considered a square prism) could be sheered wrt each other. This makes some of its faces into parallelograms.



Finally you could squash a cube diametrally (along its body diagonal). This makes all faces into rhombs.



And, for sure, you could combine all theese operations.



--- rk







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 29 at 11:49









Dr. Richard Klitzing

7286




7286







  • 2




    The OP wants all faces identical.
    – Ethan Bolker
    Jul 29 at 13:18












  • 2




    The OP wants all faces identical.
    – Ethan Bolker
    Jul 29 at 13:18







2




2




The OP wants all faces identical.
– Ethan Bolker
Jul 29 at 13:18




The OP wants all faces identical.
– Ethan Bolker
Jul 29 at 13:18










up vote
2
down vote













See the Wikipedia entry on the rhombohedron for images:



         
rhombohedron

Note the 2nd shape: $6$ identical rhombi.




share|cite|improve this answer

















  • 1




    Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
    – Rohit Pandey
    Jul 29 at 18:44














up vote
2
down vote













See the Wikipedia entry on the rhombohedron for images:



         
rhombohedron

Note the 2nd shape: $6$ identical rhombi.




share|cite|improve this answer

















  • 1




    Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
    – Rohit Pandey
    Jul 29 at 18:44












up vote
2
down vote










up vote
2
down vote









See the Wikipedia entry on the rhombohedron for images:



         
rhombohedron

Note the 2nd shape: $6$ identical rhombi.




share|cite|improve this answer













See the Wikipedia entry on the rhombohedron for images:



         
rhombohedron

Note the 2nd shape: $6$ identical rhombi.





share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 29 at 18:26









Joseph O'Rourke

17.1k248103




17.1k248103







  • 1




    Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
    – Rohit Pandey
    Jul 29 at 18:44












  • 1




    Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
    – Rohit Pandey
    Jul 29 at 18:44







1




1




Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
– Rohit Pandey
Jul 29 at 18:44




Thanks, +1. And in this article - en.wikipedia.org/wiki/Trigonal_trapezohedron they also mention quadrilaterals that have two adjacent sides equal can be used to form a cube analogue (not just Rhombi).
– Rohit Pandey
Jul 29 at 18:44












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865842%2firregular-analogue-of-cube-and-octahedron%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more