Octahedrons whose faces are congruent quadrangles The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How would you make a (physical) dodecahedron with edges instead of faces?Uniform Polyhedron with 500 congruent right kite faces!How to find out the circumscribed radius of a snub dodecahedron?Coordinates of a rhombohedronPolyhedra with at least 3 pentagonal facesDividing an octahedron into two congruent regionsWhy semiregular polyhedra existsIrregular analogue of cube and octahedron.Obtuse triangle octahedron packingIsohedron splitting into isohedra
Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?
Typeface like Times New Roman but with "tied" percent sign
Do warforged have souls?
How should I replace vector<uint8_t>::const_iterator in an API?
How is simplicity better than precision and clarity in prose?
How are presidential pardons supposed to be used?
Why is superheterodyning better than direct conversion?
Is there a writing software that you can sort scenes like slides in PowerPoint?
Why can't wing-mounted spoilers be used to steepen approaches?
Can a 1st-level character have an ability score above 18?
Cooking pasta in a water boiler
When did F become S in typeography, and why?
how can a perfect fourth interval be considered either consonant or dissonant?
Relations between two reciprocal partial derivatives?
In horse breeding, what is the female equivalent of putting a horse out "to stud"?
How to delete random line from file using Unix command?
Keeping a retro style to sci-fi spaceships?
How do I add random spotting to the same face in cycles?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Finding the path in a graph from A to B then back to A with a minimum of shared edges
How can I protect witches in combat who wear limited clothing?
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
What information about me do stores get via my credit card?
Difference between "generating set" and free product?
Octahedrons whose faces are congruent quadrangles
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How would you make a (physical) dodecahedron with edges instead of faces?Uniform Polyhedron with 500 congruent right kite faces!How to find out the circumscribed radius of a snub dodecahedron?Coordinates of a rhombohedronPolyhedra with at least 3 pentagonal facesDividing an octahedron into two congruent regionsWhy semiregular polyhedra existsIrregular analogue of cube and octahedron.Obtuse triangle octahedron packingIsohedron splitting into isohedra
$begingroup$
There is a octahedron which the faces are all congruent quadrangles. Let $M$ the set of length of edges of faces of the octahedron.
Prove that $|M| le 3$.
Prove that all faces have two edges with equal length which meets at one point.
My approach. I think I have to solve this problem with the numbers of faces, edges, and points.
There are $frac4 times 82 = 16$ edges and 8 faces, but we don't know the numbers of points.
By Euler theorem, $v-e+f=2$ and $e=16, f=8$. So $v=10$.
So, the degrees of the points is $(5, 3, 3, ..., 3)$ or $(4, 4, 3, 3, ..., 3)$. (Because the degrees are at least 3.)
Now what do we do?
solid-geometry
$endgroup$
add a comment |
$begingroup$
There is a octahedron which the faces are all congruent quadrangles. Let $M$ the set of length of edges of faces of the octahedron.
Prove that $|M| le 3$.
Prove that all faces have two edges with equal length which meets at one point.
My approach. I think I have to solve this problem with the numbers of faces, edges, and points.
There are $frac4 times 82 = 16$ edges and 8 faces, but we don't know the numbers of points.
By Euler theorem, $v-e+f=2$ and $e=16, f=8$. So $v=10$.
So, the degrees of the points is $(5, 3, 3, ..., 3)$ or $(4, 4, 3, 3, ..., 3)$. (Because the degrees are at least 3.)
Now what do we do?
solid-geometry
$endgroup$
$begingroup$
Have you neglected to include in the problem statement that all the faces are quadrilaterals?
$endgroup$
– Gerry Myerson
Mar 25 at 9:32
1
$begingroup$
Oh, sorry. I did.
$endgroup$
– coding1101
Mar 25 at 9:33
$begingroup$
For (1) Suppose the side-lengths are $a$, $b$, $c$, $d$, in order, and consider three quadrangles that meet at a vertex of degree three. Show that at least two of the edge-lengths must match. Property (2) follows from the same consideration.
$endgroup$
– Blue
Mar 25 at 10:38
$begingroup$
Where does this question come from? I note Problem 39 in Steinhaus, One Hundred Problems in Elementary Mathematics: "Is it possible to construct an octahedron whose faces are congruent quadrangles?" A solution is given on pages 101-102. But where did you find this question, please?
$endgroup$
– Gerry Myerson
Mar 26 at 3:47
$begingroup$
This problem is from an academy, and I was curious about it.
$endgroup$
– coding1101
Mar 26 at 8:23
add a comment |
$begingroup$
There is a octahedron which the faces are all congruent quadrangles. Let $M$ the set of length of edges of faces of the octahedron.
Prove that $|M| le 3$.
Prove that all faces have two edges with equal length which meets at one point.
My approach. I think I have to solve this problem with the numbers of faces, edges, and points.
There are $frac4 times 82 = 16$ edges and 8 faces, but we don't know the numbers of points.
By Euler theorem, $v-e+f=2$ and $e=16, f=8$. So $v=10$.
So, the degrees of the points is $(5, 3, 3, ..., 3)$ or $(4, 4, 3, 3, ..., 3)$. (Because the degrees are at least 3.)
Now what do we do?
solid-geometry
$endgroup$
There is a octahedron which the faces are all congruent quadrangles. Let $M$ the set of length of edges of faces of the octahedron.
Prove that $|M| le 3$.
Prove that all faces have two edges with equal length which meets at one point.
My approach. I think I have to solve this problem with the numbers of faces, edges, and points.
There are $frac4 times 82 = 16$ edges and 8 faces, but we don't know the numbers of points.
By Euler theorem, $v-e+f=2$ and $e=16, f=8$. So $v=10$.
So, the degrees of the points is $(5, 3, 3, ..., 3)$ or $(4, 4, 3, 3, ..., 3)$. (Because the degrees are at least 3.)
Now what do we do?
solid-geometry
solid-geometry
edited Mar 25 at 9:53
coding1101
asked Mar 25 at 9:29
coding1101coding1101
1246
1246
$begingroup$
Have you neglected to include in the problem statement that all the faces are quadrilaterals?
$endgroup$
– Gerry Myerson
Mar 25 at 9:32
1
$begingroup$
Oh, sorry. I did.
$endgroup$
– coding1101
Mar 25 at 9:33
$begingroup$
For (1) Suppose the side-lengths are $a$, $b$, $c$, $d$, in order, and consider three quadrangles that meet at a vertex of degree three. Show that at least two of the edge-lengths must match. Property (2) follows from the same consideration.
$endgroup$
– Blue
Mar 25 at 10:38
$begingroup$
Where does this question come from? I note Problem 39 in Steinhaus, One Hundred Problems in Elementary Mathematics: "Is it possible to construct an octahedron whose faces are congruent quadrangles?" A solution is given on pages 101-102. But where did you find this question, please?
$endgroup$
– Gerry Myerson
Mar 26 at 3:47
$begingroup$
This problem is from an academy, and I was curious about it.
$endgroup$
– coding1101
Mar 26 at 8:23
add a comment |
$begingroup$
Have you neglected to include in the problem statement that all the faces are quadrilaterals?
$endgroup$
– Gerry Myerson
Mar 25 at 9:32
1
$begingroup$
Oh, sorry. I did.
$endgroup$
– coding1101
Mar 25 at 9:33
$begingroup$
For (1) Suppose the side-lengths are $a$, $b$, $c$, $d$, in order, and consider three quadrangles that meet at a vertex of degree three. Show that at least two of the edge-lengths must match. Property (2) follows from the same consideration.
$endgroup$
– Blue
Mar 25 at 10:38
$begingroup$
Where does this question come from? I note Problem 39 in Steinhaus, One Hundred Problems in Elementary Mathematics: "Is it possible to construct an octahedron whose faces are congruent quadrangles?" A solution is given on pages 101-102. But where did you find this question, please?
$endgroup$
– Gerry Myerson
Mar 26 at 3:47
$begingroup$
This problem is from an academy, and I was curious about it.
$endgroup$
– coding1101
Mar 26 at 8:23
$begingroup$
Have you neglected to include in the problem statement that all the faces are quadrilaterals?
$endgroup$
– Gerry Myerson
Mar 25 at 9:32
$begingroup$
Have you neglected to include in the problem statement that all the faces are quadrilaterals?
$endgroup$
– Gerry Myerson
Mar 25 at 9:32
1
1
$begingroup$
Oh, sorry. I did.
$endgroup$
– coding1101
Mar 25 at 9:33
$begingroup$
Oh, sorry. I did.
$endgroup$
– coding1101
Mar 25 at 9:33
$begingroup$
For (1) Suppose the side-lengths are $a$, $b$, $c$, $d$, in order, and consider three quadrangles that meet at a vertex of degree three. Show that at least two of the edge-lengths must match. Property (2) follows from the same consideration.
$endgroup$
– Blue
Mar 25 at 10:38
$begingroup$
For (1) Suppose the side-lengths are $a$, $b$, $c$, $d$, in order, and consider three quadrangles that meet at a vertex of degree three. Show that at least two of the edge-lengths must match. Property (2) follows from the same consideration.
$endgroup$
– Blue
Mar 25 at 10:38
$begingroup$
Where does this question come from? I note Problem 39 in Steinhaus, One Hundred Problems in Elementary Mathematics: "Is it possible to construct an octahedron whose faces are congruent quadrangles?" A solution is given on pages 101-102. But where did you find this question, please?
$endgroup$
– Gerry Myerson
Mar 26 at 3:47
$begingroup$
Where does this question come from? I note Problem 39 in Steinhaus, One Hundred Problems in Elementary Mathematics: "Is it possible to construct an octahedron whose faces are congruent quadrangles?" A solution is given on pages 101-102. But where did you find this question, please?
$endgroup$
– Gerry Myerson
Mar 26 at 3:47
$begingroup$
This problem is from an academy, and I was curious about it.
$endgroup$
– coding1101
Mar 26 at 8:23
$begingroup$
This problem is from an academy, and I was curious about it.
$endgroup$
– coding1101
Mar 26 at 8:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All right, let's clear up some things about the configuration.
First, any planar graph with all faces quadrilaterals (or even polygons) must be bipartite. How do those parts split? If there are six or more vertices in one of the parts, then, since each of those vertices has degree at least $3$ and each edge has only one of its vertices in the part, that's at least $18$ edges. This contradicts our calculation of $16$ edges, so the two parts must have five vertices each.
Now, count degrees at the vertices in each part. With $16$ total edges over five vertices, that must be four vertices of degree $3$ and one of degree $4$. The same is true of the other part.
Now, this splits into two cases for the configuration. Either the two vertices of degree $4$ are directly connected by an edge, or they're not. Let's see if we can build a planar graph in the former case:
We have several edges left to fill in, but the six faces around the two degree-$4$ vertices give us this much. Number the vertices and label as shown; the vertices of one part are shown in red, and the other part in blue.
Next, what can vertex blue #5 ($U_5$) connect to? It has three edges going to red vertices #2,3,4,5 ($R_2$,$R_3$,$R_4$,$R_5$). In particular, that means it must connect to at least one of $R_2$ or $R_3$. WLOG, that's $R_2$.
Then, there's a quadrilateral with consecutive vertices $U_5$, $R_2$, $U_1$. This must be face $C$, and the fourth vertex is $R_4$. That means it closes with an edge between $U_5$ and $R_4$, our second edge drawn in.
Now, we have some edges in the picture that only have one of their connected faces indicated. First, the triad U2 to R2 to U5 has to be part of a face. It can't be $A$, $B$, or $C$ since those faces have all of their vertices known. It can't be $D$, $E$, or $F$ since those faces have three known vertices each that agree with the list of three we have in at most one place. Therefore, it must be a new face; call it $G$.
Second, the edge from $R_3$ to $U_3$ is part of face $B$. What's its second face? It can't be $A$ or $C$ because those don't have either $R_3$ or $U_3$ as a vertex. It can't be $E$ or $G$ since those have three known vertices and can't include both $R_3$ and $U_3$. It can't be $D$ or $F$ since two faces of the polyhedron can't share adjacent edges. All of the old faces are ruled out, so it has to be a new face $H$.
Third, the edge from $R_4$ to $U_5$ is part of face $C$. What's its second face? Not $A$, $B$, $E$, or $F$ because there aren't enough free vertices. Not $D$ or $G$ because it can't share two adjacent edges with $C$. There are only eight faces, so that leaves us with $H$ as the only option.
How can $R_3$, $U_3$, $R_4$, and $U_5$ be part of the same face $H$? We must have edges from $R_3$ to $U_5$ and $R_4$ to $U_3$.
But that doesn't work. There's no way to connect those edges while remaining planar. Merging vertices $R_1$ and $R_2$ through their connection at $U_2$, that's the equivalent of a $K_3,3$ between $(U_1,U_3,U_5)$ and $(R_3,R_4,R_1,2)$.
This configuration has failed, leaving only one possibility - the two degree-4 vertices aren't connected by an edge, and indeed don't share a face. That configuration works as a polyhedron. A flattened version:
I'm maintaining the color choices; the two caps will meet at corresponding colors when we stack them and open it up.
And that's the configuration you'll need to work with in figuring out how many different edge lengths there can be.
$endgroup$
add a comment |
$begingroup$
Thanks @Blue.
I solved this problem in a very simple way, thanks to you.
1) Solved this by contradiction.
If $|M| ge 4$, the lengths of edges of quadrangles are all different.
Let the four lengths $a, b, c, d$.
Then, it seems to be a contradiction because,
2) It is simmilar from above.
Let the four lengths $a, b, a, c$.
Then, it seems to be a contradiction because,
$endgroup$
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
add a comment |
Your Answer
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161559%2foctahedrons-whose-faces-are-congruent-quadrangles%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All right, let's clear up some things about the configuration.
First, any planar graph with all faces quadrilaterals (or even polygons) must be bipartite. How do those parts split? If there are six or more vertices in one of the parts, then, since each of those vertices has degree at least $3$ and each edge has only one of its vertices in the part, that's at least $18$ edges. This contradicts our calculation of $16$ edges, so the two parts must have five vertices each.
Now, count degrees at the vertices in each part. With $16$ total edges over five vertices, that must be four vertices of degree $3$ and one of degree $4$. The same is true of the other part.
Now, this splits into two cases for the configuration. Either the two vertices of degree $4$ are directly connected by an edge, or they're not. Let's see if we can build a planar graph in the former case:
We have several edges left to fill in, but the six faces around the two degree-$4$ vertices give us this much. Number the vertices and label as shown; the vertices of one part are shown in red, and the other part in blue.
Next, what can vertex blue #5 ($U_5$) connect to? It has three edges going to red vertices #2,3,4,5 ($R_2$,$R_3$,$R_4$,$R_5$). In particular, that means it must connect to at least one of $R_2$ or $R_3$. WLOG, that's $R_2$.
Then, there's a quadrilateral with consecutive vertices $U_5$, $R_2$, $U_1$. This must be face $C$, and the fourth vertex is $R_4$. That means it closes with an edge between $U_5$ and $R_4$, our second edge drawn in.
Now, we have some edges in the picture that only have one of their connected faces indicated. First, the triad U2 to R2 to U5 has to be part of a face. It can't be $A$, $B$, or $C$ since those faces have all of their vertices known. It can't be $D$, $E$, or $F$ since those faces have three known vertices each that agree with the list of three we have in at most one place. Therefore, it must be a new face; call it $G$.
Second, the edge from $R_3$ to $U_3$ is part of face $B$. What's its second face? It can't be $A$ or $C$ because those don't have either $R_3$ or $U_3$ as a vertex. It can't be $E$ or $G$ since those have three known vertices and can't include both $R_3$ and $U_3$. It can't be $D$ or $F$ since two faces of the polyhedron can't share adjacent edges. All of the old faces are ruled out, so it has to be a new face $H$.
Third, the edge from $R_4$ to $U_5$ is part of face $C$. What's its second face? Not $A$, $B$, $E$, or $F$ because there aren't enough free vertices. Not $D$ or $G$ because it can't share two adjacent edges with $C$. There are only eight faces, so that leaves us with $H$ as the only option.
How can $R_3$, $U_3$, $R_4$, and $U_5$ be part of the same face $H$? We must have edges from $R_3$ to $U_5$ and $R_4$ to $U_3$.
But that doesn't work. There's no way to connect those edges while remaining planar. Merging vertices $R_1$ and $R_2$ through their connection at $U_2$, that's the equivalent of a $K_3,3$ between $(U_1,U_3,U_5)$ and $(R_3,R_4,R_1,2)$.
This configuration has failed, leaving only one possibility - the two degree-4 vertices aren't connected by an edge, and indeed don't share a face. That configuration works as a polyhedron. A flattened version:
I'm maintaining the color choices; the two caps will meet at corresponding colors when we stack them and open it up.
And that's the configuration you'll need to work with in figuring out how many different edge lengths there can be.
$endgroup$
add a comment |
$begingroup$
All right, let's clear up some things about the configuration.
First, any planar graph with all faces quadrilaterals (or even polygons) must be bipartite. How do those parts split? If there are six or more vertices in one of the parts, then, since each of those vertices has degree at least $3$ and each edge has only one of its vertices in the part, that's at least $18$ edges. This contradicts our calculation of $16$ edges, so the two parts must have five vertices each.
Now, count degrees at the vertices in each part. With $16$ total edges over five vertices, that must be four vertices of degree $3$ and one of degree $4$. The same is true of the other part.
Now, this splits into two cases for the configuration. Either the two vertices of degree $4$ are directly connected by an edge, or they're not. Let's see if we can build a planar graph in the former case:
We have several edges left to fill in, but the six faces around the two degree-$4$ vertices give us this much. Number the vertices and label as shown; the vertices of one part are shown in red, and the other part in blue.
Next, what can vertex blue #5 ($U_5$) connect to? It has three edges going to red vertices #2,3,4,5 ($R_2$,$R_3$,$R_4$,$R_5$). In particular, that means it must connect to at least one of $R_2$ or $R_3$. WLOG, that's $R_2$.
Then, there's a quadrilateral with consecutive vertices $U_5$, $R_2$, $U_1$. This must be face $C$, and the fourth vertex is $R_4$. That means it closes with an edge between $U_5$ and $R_4$, our second edge drawn in.
Now, we have some edges in the picture that only have one of their connected faces indicated. First, the triad U2 to R2 to U5 has to be part of a face. It can't be $A$, $B$, or $C$ since those faces have all of their vertices known. It can't be $D$, $E$, or $F$ since those faces have three known vertices each that agree with the list of three we have in at most one place. Therefore, it must be a new face; call it $G$.
Second, the edge from $R_3$ to $U_3$ is part of face $B$. What's its second face? It can't be $A$ or $C$ because those don't have either $R_3$ or $U_3$ as a vertex. It can't be $E$ or $G$ since those have three known vertices and can't include both $R_3$ and $U_3$. It can't be $D$ or $F$ since two faces of the polyhedron can't share adjacent edges. All of the old faces are ruled out, so it has to be a new face $H$.
Third, the edge from $R_4$ to $U_5$ is part of face $C$. What's its second face? Not $A$, $B$, $E$, or $F$ because there aren't enough free vertices. Not $D$ or $G$ because it can't share two adjacent edges with $C$. There are only eight faces, so that leaves us with $H$ as the only option.
How can $R_3$, $U_3$, $R_4$, and $U_5$ be part of the same face $H$? We must have edges from $R_3$ to $U_5$ and $R_4$ to $U_3$.
But that doesn't work. There's no way to connect those edges while remaining planar. Merging vertices $R_1$ and $R_2$ through their connection at $U_2$, that's the equivalent of a $K_3,3$ between $(U_1,U_3,U_5)$ and $(R_3,R_4,R_1,2)$.
This configuration has failed, leaving only one possibility - the two degree-4 vertices aren't connected by an edge, and indeed don't share a face. That configuration works as a polyhedron. A flattened version:
I'm maintaining the color choices; the two caps will meet at corresponding colors when we stack them and open it up.
And that's the configuration you'll need to work with in figuring out how many different edge lengths there can be.
$endgroup$
add a comment |
$begingroup$
All right, let's clear up some things about the configuration.
First, any planar graph with all faces quadrilaterals (or even polygons) must be bipartite. How do those parts split? If there are six or more vertices in one of the parts, then, since each of those vertices has degree at least $3$ and each edge has only one of its vertices in the part, that's at least $18$ edges. This contradicts our calculation of $16$ edges, so the two parts must have five vertices each.
Now, count degrees at the vertices in each part. With $16$ total edges over five vertices, that must be four vertices of degree $3$ and one of degree $4$. The same is true of the other part.
Now, this splits into two cases for the configuration. Either the two vertices of degree $4$ are directly connected by an edge, or they're not. Let's see if we can build a planar graph in the former case:
We have several edges left to fill in, but the six faces around the two degree-$4$ vertices give us this much. Number the vertices and label as shown; the vertices of one part are shown in red, and the other part in blue.
Next, what can vertex blue #5 ($U_5$) connect to? It has three edges going to red vertices #2,3,4,5 ($R_2$,$R_3$,$R_4$,$R_5$). In particular, that means it must connect to at least one of $R_2$ or $R_3$. WLOG, that's $R_2$.
Then, there's a quadrilateral with consecutive vertices $U_5$, $R_2$, $U_1$. This must be face $C$, and the fourth vertex is $R_4$. That means it closes with an edge between $U_5$ and $R_4$, our second edge drawn in.
Now, we have some edges in the picture that only have one of their connected faces indicated. First, the triad U2 to R2 to U5 has to be part of a face. It can't be $A$, $B$, or $C$ since those faces have all of their vertices known. It can't be $D$, $E$, or $F$ since those faces have three known vertices each that agree with the list of three we have in at most one place. Therefore, it must be a new face; call it $G$.
Second, the edge from $R_3$ to $U_3$ is part of face $B$. What's its second face? It can't be $A$ or $C$ because those don't have either $R_3$ or $U_3$ as a vertex. It can't be $E$ or $G$ since those have three known vertices and can't include both $R_3$ and $U_3$. It can't be $D$ or $F$ since two faces of the polyhedron can't share adjacent edges. All of the old faces are ruled out, so it has to be a new face $H$.
Third, the edge from $R_4$ to $U_5$ is part of face $C$. What's its second face? Not $A$, $B$, $E$, or $F$ because there aren't enough free vertices. Not $D$ or $G$ because it can't share two adjacent edges with $C$. There are only eight faces, so that leaves us with $H$ as the only option.
How can $R_3$, $U_3$, $R_4$, and $U_5$ be part of the same face $H$? We must have edges from $R_3$ to $U_5$ and $R_4$ to $U_3$.
But that doesn't work. There's no way to connect those edges while remaining planar. Merging vertices $R_1$ and $R_2$ through their connection at $U_2$, that's the equivalent of a $K_3,3$ between $(U_1,U_3,U_5)$ and $(R_3,R_4,R_1,2)$.
This configuration has failed, leaving only one possibility - the two degree-4 vertices aren't connected by an edge, and indeed don't share a face. That configuration works as a polyhedron. A flattened version:
I'm maintaining the color choices; the two caps will meet at corresponding colors when we stack them and open it up.
And that's the configuration you'll need to work with in figuring out how many different edge lengths there can be.
$endgroup$
All right, let's clear up some things about the configuration.
First, any planar graph with all faces quadrilaterals (or even polygons) must be bipartite. How do those parts split? If there are six or more vertices in one of the parts, then, since each of those vertices has degree at least $3$ and each edge has only one of its vertices in the part, that's at least $18$ edges. This contradicts our calculation of $16$ edges, so the two parts must have five vertices each.
Now, count degrees at the vertices in each part. With $16$ total edges over five vertices, that must be four vertices of degree $3$ and one of degree $4$. The same is true of the other part.
Now, this splits into two cases for the configuration. Either the two vertices of degree $4$ are directly connected by an edge, or they're not. Let's see if we can build a planar graph in the former case:
We have several edges left to fill in, but the six faces around the two degree-$4$ vertices give us this much. Number the vertices and label as shown; the vertices of one part are shown in red, and the other part in blue.
Next, what can vertex blue #5 ($U_5$) connect to? It has three edges going to red vertices #2,3,4,5 ($R_2$,$R_3$,$R_4$,$R_5$). In particular, that means it must connect to at least one of $R_2$ or $R_3$. WLOG, that's $R_2$.
Then, there's a quadrilateral with consecutive vertices $U_5$, $R_2$, $U_1$. This must be face $C$, and the fourth vertex is $R_4$. That means it closes with an edge between $U_5$ and $R_4$, our second edge drawn in.
Now, we have some edges in the picture that only have one of their connected faces indicated. First, the triad U2 to R2 to U5 has to be part of a face. It can't be $A$, $B$, or $C$ since those faces have all of their vertices known. It can't be $D$, $E$, or $F$ since those faces have three known vertices each that agree with the list of three we have in at most one place. Therefore, it must be a new face; call it $G$.
Second, the edge from $R_3$ to $U_3$ is part of face $B$. What's its second face? It can't be $A$ or $C$ because those don't have either $R_3$ or $U_3$ as a vertex. It can't be $E$ or $G$ since those have three known vertices and can't include both $R_3$ and $U_3$. It can't be $D$ or $F$ since two faces of the polyhedron can't share adjacent edges. All of the old faces are ruled out, so it has to be a new face $H$.
Third, the edge from $R_4$ to $U_5$ is part of face $C$. What's its second face? Not $A$, $B$, $E$, or $F$ because there aren't enough free vertices. Not $D$ or $G$ because it can't share two adjacent edges with $C$. There are only eight faces, so that leaves us with $H$ as the only option.
How can $R_3$, $U_3$, $R_4$, and $U_5$ be part of the same face $H$? We must have edges from $R_3$ to $U_5$ and $R_4$ to $U_3$.
But that doesn't work. There's no way to connect those edges while remaining planar. Merging vertices $R_1$ and $R_2$ through their connection at $U_2$, that's the equivalent of a $K_3,3$ between $(U_1,U_3,U_5)$ and $(R_3,R_4,R_1,2)$.
This configuration has failed, leaving only one possibility - the two degree-4 vertices aren't connected by an edge, and indeed don't share a face. That configuration works as a polyhedron. A flattened version:
I'm maintaining the color choices; the two caps will meet at corresponding colors when we stack them and open it up.
And that's the configuration you'll need to work with in figuring out how many different edge lengths there can be.
answered Mar 25 at 23:46
jmerryjmerry
17.1k11633
17.1k11633
add a comment |
add a comment |
$begingroup$
Thanks @Blue.
I solved this problem in a very simple way, thanks to you.
1) Solved this by contradiction.
If $|M| ge 4$, the lengths of edges of quadrangles are all different.
Let the four lengths $a, b, c, d$.
Then, it seems to be a contradiction because,
2) It is simmilar from above.
Let the four lengths $a, b, a, c$.
Then, it seems to be a contradiction because,
$endgroup$
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
add a comment |
$begingroup$
Thanks @Blue.
I solved this problem in a very simple way, thanks to you.
1) Solved this by contradiction.
If $|M| ge 4$, the lengths of edges of quadrangles are all different.
Let the four lengths $a, b, c, d$.
Then, it seems to be a contradiction because,
2) It is simmilar from above.
Let the four lengths $a, b, a, c$.
Then, it seems to be a contradiction because,
$endgroup$
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
add a comment |
$begingroup$
Thanks @Blue.
I solved this problem in a very simple way, thanks to you.
1) Solved this by contradiction.
If $|M| ge 4$, the lengths of edges of quadrangles are all different.
Let the four lengths $a, b, c, d$.
Then, it seems to be a contradiction because,
2) It is simmilar from above.
Let the four lengths $a, b, a, c$.
Then, it seems to be a contradiction because,
$endgroup$
Thanks @Blue.
I solved this problem in a very simple way, thanks to you.
1) Solved this by contradiction.
If $|M| ge 4$, the lengths of edges of quadrangles are all different.
Let the four lengths $a, b, c, d$.
Then, it seems to be a contradiction because,
2) It is simmilar from above.
Let the four lengths $a, b, a, c$.
Then, it seems to be a contradiction because,
edited Mar 27 at 7:33
answered Mar 26 at 9:44
coding1101coding1101
1246
1246
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
add a comment |
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
I think you have your two diagrams each where the other should be.
$endgroup$
– Gerry Myerson
Mar 26 at 23:40
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
$begingroup$
Thank you. My mistake.
$endgroup$
– coding1101
Mar 27 at 7:33
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3161559%2foctahedrons-whose-faces-are-congruent-quadrangles%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Have you neglected to include in the problem statement that all the faces are quadrilaterals?
$endgroup$
– Gerry Myerson
Mar 25 at 9:32
1
$begingroup$
Oh, sorry. I did.
$endgroup$
– coding1101
Mar 25 at 9:33
$begingroup$
For (1) Suppose the side-lengths are $a$, $b$, $c$, $d$, in order, and consider three quadrangles that meet at a vertex of degree three. Show that at least two of the edge-lengths must match. Property (2) follows from the same consideration.
$endgroup$
– Blue
Mar 25 at 10:38
$begingroup$
Where does this question come from? I note Problem 39 in Steinhaus, One Hundred Problems in Elementary Mathematics: "Is it possible to construct an octahedron whose faces are congruent quadrangles?" A solution is given on pages 101-102. But where did you find this question, please?
$endgroup$
– Gerry Myerson
Mar 26 at 3:47
$begingroup$
This problem is from an academy, and I was curious about it.
$endgroup$
– coding1101
Mar 26 at 8:23