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How many of these lines lie entirely in the interior of the original cube?
How many disconnected graphs of the Rubik's cube exist?Measuring angles in a prismFind the area of an equilateral triangle given the distances from an interior point to the verticesSubdivided icosahedron points do not lie on circumscribed sphereGiven an $n times n$ square grid, how many ways can we remove $k$ vertices with the graph still connected?What is the upper bound of $Runderbrace(3,3,3, ldots,3)_text$k$ times$?geometry/combinatorics question: max # intersections of lines in a triangleRandom point inside a trianglePlane Geometry using Complex NumbersHow many edges are on the graph connecting points of $0, 1^n$ that differ by only one coordinate?
$begingroup$
A portion of a wooden cube is sawed off at each vertex so that a small equilateral triangle is formed at each corner with vertices on the edges of the cube. The $24$ vertices of the new object are all connected to each other by straight lines. How many of these lines (with the exception, of course, of their end-points) lie entirely in the interior of the original cube?
combinatorics geometry
edited 5 hours ago
Parcly Taxel
44.4k1376109
asked 5 hours ago
Mittal GMittal G
1,369516
$endgroup$
add a comment |
$begingroup$
A portion of a wooden cube is sawed off at each vertex so that a small equilateral triangle is formed at each corner with vertices on the edges of the cube. The $24$ vertices of the new object are all connected to each other by straight lines. How many of these lines (with the exception, of course, of their end-points) lie entirely in the interior of the original cube?
combinatorics geometry
edited 5 hours ago
Parcly Taxel
44.4k1376109
asked 5 hours ago
Mittal GMittal G
1,369516
$endgroup$
1
$begingroup$
This is an interesting puzzle. Where does it come from? is this a homework exercise? in what class? What motivates the problem? why are you interested in it? and why should we be interested in it? At the very least, please indicate what you have attempted to do, and perhaps explain what tools (theorems, definitions, etc) you have access to.
$endgroup$
– Xander Henderson
18 mins ago
add a comment |
$begingroup$
A portion of a wooden cube is sawed off at each vertex so that a small equilateral triangle is formed at each corner with vertices on the edges of the cube. The $24$ vertices of the new object are all connected to each other by straight lines. How many of these lines (with the exception, of course, of their end-points) lie entirely in the interior of the original cube?
combinatorics geometry
edited 5 hours ago
Parcly Taxel
44.4k1376109
asked 5 hours ago
Mittal GMittal G
1,369516
$endgroup$
A portion of a wooden cube is sawed off at each vertex so that a small equilateral triangle is formed at each corner with vertices on the edges of the cube. The $24$ vertices of the new object are all connected to each other by straight lines. How many of these lines (with the exception, of course, of their end-points) lie entirely in the interior of the original cube?
combinatorics geometry
combinatorics geometry
edited 5 hours ago
Parcly Taxel
44.4k1376109
asked 5 hours ago
Mittal GMittal G
1,369516
edited 5 hours ago
Parcly Taxel
44.4k1376109
asked 5 hours ago
Mittal GMittal G
1,369516
edited 5 hours ago
Parcly Taxel
44.4k1376109
edited 5 hours ago
Parcly Taxel
44.4k1376109
edited 5 hours ago
Parcly Taxel
44.4k1376109
44.4k1376109
asked 5 hours ago
Mittal GMittal G
1,369516
asked 5 hours ago
Mittal GMittal G
1,369516
asked 5 hours ago
Mittal GMittal G
1,369516
1,369516
1
$begingroup$
This is an interesting puzzle. Where does it come from? is this a homework exercise? in what class? What motivates the problem? why are you interested in it? and why should we be interested in it? At the very least, please indicate what you have attempted to do, and perhaps explain what tools (theorems, definitions, etc) you have access to.
$endgroup$
– Xander Henderson
18 mins ago
add a comment |
1
$begingroup$
This is an interesting puzzle. Where does it come from? is this a homework exercise? in what class? What motivates the problem? why are you interested in it? and why should we be interested in it? At the very least, please indicate what you have attempted to do, and perhaps explain what tools (theorems, definitions, etc) you have access to.
$endgroup$
– Xander Henderson
18 mins ago
1
1
$begingroup$
This is an interesting puzzle. Where does it come from? is this a homework exercise? in what class? What motivates the problem? why are you interested in it? and why should we be interested in it? At the very least, please indicate what you have attempted to do, and perhaps explain what tools (theorems, definitions, etc) you have access to.
$endgroup$
– Xander Henderson
18 mins ago
$begingroup$
This is an interesting puzzle. Where does it come from? is this a homework exercise? in what class? What motivates the problem? why are you interested in it? and why should we be interested in it? At the very least, please indicate what you have attempted to do, and perhaps explain what tools (theorems, definitions, etc) you have access to.
$endgroup$
– Xander Henderson
18 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:
$hspace3cm$
First floor: Number $1$ can connect with none on the first floor. $1$ can connect with $11,12$ on the second floor, with $15,16$ on the third floor and with $19,20,21,22,23,24$ on the fourth floor. So, $1$ can connect with $2+2+6=10$ points, so does each of $2-8$, hence: $8cdot 10=colorred80$.
Second floor: Number $9$ can connect with none on the first floor. $9$ can connect with $11$ on the second floor, so does $10$ with $12$. Hence: $colorred2$ connections on the second floor only. $9$ can connect with $15$ on the third floor and with $19,20,21,22$ on the fourth floor. So, $9$ can connect with $1+4=5$ points on the third and fourth floors, so does each each of $10,11,12$, hence: $4cdot 5=colorred20$. Overall, $colorred22$ connections.
Third floor: $13$ can connect with none on the first or second floor. $13$ can connect with $15$ on the third floor, so does $14$ with $16$, hence $colorred2$ connections on the third floor only. $13$ can connect with $19,20,21,22$ on the fourth floor. So, $13$ can connect with $4$ points on the fourth floor, so does each of $14,15,16$, hence: $4cdot 4=colorred16$. Overall, $colorred18$ connections.
Fourth floor: $17$ can connect with none on the first, second, third or fourth floor. So does each other on the fourth floor.
Hence, the number of inner connections is: $colorred80+colorred22+colorred18=120$.
answered 3 hours ago
farruhotafarruhota
21k2841
$endgroup$
1
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
1
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
add a comment |
$begingroup$
In the resulting truncated cube, each vertex $v$ is incident to one triangle and three octagons, which together have $3+8+8-3-3=13$ vertices between them excluding $v$. Lines from $v$ to these $13$ vertices will not be completely in the cube; lines to the other $24-13-1=10$ vertices will. Since each connecting line is incident to two vertices, there are $frac24cdot102=120$ interior lines.
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
$endgroup$
$begingroup$
+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
$endgroup$
– antkam
1 hour ago
1
$begingroup$
One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
$endgroup$
– TonyK
53 mins ago
add a comment |
$begingroup$
Every one of them that does not run along a face.
Each octahedral face of the object, of which there are $6,$ contains $8$ vertices, so there are $68 choose 2=168$ segments along those faces. The segments that are part of the $12$ original edges of the cube are counted twice, so we subtract $12$, leaving $156$ segments on the octagonal faces of the object.
The triangular faces have segments running along them, but we have already counted those because they also run along an octahedral face. There are $24 choose 2=276$ line segments, so there are $276-156=120$ running through the body of the object.
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
You're double-counting octagon-octagon edges.
$endgroup$
– Parcly Taxel
5 hours ago
$begingroup$
@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
The object you describe is called the truncated cube, it's an Archimedian solid.
See for example:
https://en.wikipedia.org/wiki/Truncated_cube
All of its edges lie on the surface of the orginal cube.
answered 5 hours ago
quaraguequarague
436210
$endgroup$
3
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:
$hspace3cm$
First floor: Number $1$ can connect with none on the first floor. $1$ can connect with $11,12$ on the second floor, with $15,16$ on the third floor and with $19,20,21,22,23,24$ on the fourth floor. So, $1$ can connect with $2+2+6=10$ points, so does each of $2-8$, hence: $8cdot 10=colorred80$.
Second floor: Number $9$ can connect with none on the first floor. $9$ can connect with $11$ on the second floor, so does $10$ with $12$. Hence: $colorred2$ connections on the second floor only. $9$ can connect with $15$ on the third floor and with $19,20,21,22$ on the fourth floor. So, $9$ can connect with $1+4=5$ points on the third and fourth floors, so does each each of $10,11,12$, hence: $4cdot 5=colorred20$. Overall, $colorred22$ connections.
Third floor: $13$ can connect with none on the first or second floor. $13$ can connect with $15$ on the third floor, so does $14$ with $16$, hence $colorred2$ connections on the third floor only. $13$ can connect with $19,20,21,22$ on the fourth floor. So, $13$ can connect with $4$ points on the fourth floor, so does each of $14,15,16$, hence: $4cdot 4=colorred16$. Overall, $colorred18$ connections.
Fourth floor: $17$ can connect with none on the first, second, third or fourth floor. So does each other on the fourth floor.
Hence, the number of inner connections is: $colorred80+colorred22+colorred18=120$.
answered 3 hours ago
farruhotafarruhota
21k2841
$endgroup$
1
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
1
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
add a comment |
$begingroup$
Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:
$hspace3cm$
First floor: Number $1$ can connect with none on the first floor. $1$ can connect with $11,12$ on the second floor, with $15,16$ on the third floor and with $19,20,21,22,23,24$ on the fourth floor. So, $1$ can connect with $2+2+6=10$ points, so does each of $2-8$, hence: $8cdot 10=colorred80$.
Second floor: Number $9$ can connect with none on the first floor. $9$ can connect with $11$ on the second floor, so does $10$ with $12$. Hence: $colorred2$ connections on the second floor only. $9$ can connect with $15$ on the third floor and with $19,20,21,22$ on the fourth floor. So, $9$ can connect with $1+4=5$ points on the third and fourth floors, so does each each of $10,11,12$, hence: $4cdot 5=colorred20$. Overall, $colorred22$ connections.
Third floor: $13$ can connect with none on the first or second floor. $13$ can connect with $15$ on the third floor, so does $14$ with $16$, hence $colorred2$ connections on the third floor only. $13$ can connect with $19,20,21,22$ on the fourth floor. So, $13$ can connect with $4$ points on the fourth floor, so does each of $14,15,16$, hence: $4cdot 4=colorred16$. Overall, $colorred18$ connections.
Fourth floor: $17$ can connect with none on the first, second, third or fourth floor. So does each other on the fourth floor.
Hence, the number of inner connections is: $colorred80+colorred22+colorred18=120$.
answered 3 hours ago
farruhotafarruhota
21k2841
$endgroup$
1
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
1
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
add a comment |
$begingroup$
Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:
$hspace3cm$
First floor: Number $1$ can connect with none on the first floor. $1$ can connect with $11,12$ on the second floor, with $15,16$ on the third floor and with $19,20,21,22,23,24$ on the fourth floor. So, $1$ can connect with $2+2+6=10$ points, so does each of $2-8$, hence: $8cdot 10=colorred80$.
Second floor: Number $9$ can connect with none on the first floor. $9$ can connect with $11$ on the second floor, so does $10$ with $12$. Hence: $colorred2$ connections on the second floor only. $9$ can connect with $15$ on the third floor and with $19,20,21,22$ on the fourth floor. So, $9$ can connect with $1+4=5$ points on the third and fourth floors, so does each each of $10,11,12$, hence: $4cdot 5=colorred20$. Overall, $colorred22$ connections.
Third floor: $13$ can connect with none on the first or second floor. $13$ can connect with $15$ on the third floor, so does $14$ with $16$, hence $colorred2$ connections on the third floor only. $13$ can connect with $19,20,21,22$ on the fourth floor. So, $13$ can connect with $4$ points on the fourth floor, so does each of $14,15,16$, hence: $4cdot 4=colorred16$. Overall, $colorred18$ connections.
Fourth floor: $17$ can connect with none on the first, second, third or fourth floor. So does each other on the fourth floor.
Hence, the number of inner connections is: $colorred80+colorred22+colorred18=120$.
answered 3 hours ago
farruhotafarruhota
21k2841
$endgroup$
Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:
$hspace3cm$
First floor: Number $1$ can connect with none on the first floor. $1$ can connect with $11,12$ on the second floor, with $15,16$ on the third floor and with $19,20,21,22,23,24$ on the fourth floor. So, $1$ can connect with $2+2+6=10$ points, so does each of $2-8$, hence: $8cdot 10=colorred80$.
Second floor: Number $9$ can connect with none on the first floor. $9$ can connect with $11$ on the second floor, so does $10$ with $12$. Hence: $colorred2$ connections on the second floor only. $9$ can connect with $15$ on the third floor and with $19,20,21,22$ on the fourth floor. So, $9$ can connect with $1+4=5$ points on the third and fourth floors, so does each each of $10,11,12$, hence: $4cdot 5=colorred20$. Overall, $colorred22$ connections.
Third floor: $13$ can connect with none on the first or second floor. $13$ can connect with $15$ on the third floor, so does $14$ with $16$, hence $colorred2$ connections on the third floor only. $13$ can connect with $19,20,21,22$ on the fourth floor. So, $13$ can connect with $4$ points on the fourth floor, so does each of $14,15,16$, hence: $4cdot 4=colorred16$. Overall, $colorred18$ connections.
Fourth floor: $17$ can connect with none on the first, second, third or fourth floor. So does each other on the fourth floor.
Hence, the number of inner connections is: $colorred80+colorred22+colorred18=120$.
answered 3 hours ago
farruhotafarruhota
21k2841
edited 2 hours ago
answered 3 hours ago
farruhotafarruhota
21k2841
answered 3 hours ago
farruhotafarruhota
21k2841
answered 3 hours ago
farruhotafarruhota
21k2841
21k2841
1
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
1
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
add a comment |
1
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
1
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
1
1
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
something is missing, let me double check
$endgroup$
– farruhota
3 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
$begingroup$
now it is fixed.
$endgroup$
– farruhota
2 hours ago
1
1
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
$begingroup$
Another way of looking at it: since the symmetry group of the truncated cube lets you rotate any vertex on to any other vertex, once you count how many connections a single one can make (10), you can multiply that by 24 to get the number of directed connections, then divide by two because that figure counts each connection twice (once in each direction), and you get the same answer of 120. I don't know if this would be independent enough to make a good separate answer.
$endgroup$
– David Z
15 mins ago
add a comment |
$begingroup$
In the resulting truncated cube, each vertex $v$ is incident to one triangle and three octagons, which together have $3+8+8-3-3=13$ vertices between them excluding $v$. Lines from $v$ to these $13$ vertices will not be completely in the cube; lines to the other $24-13-1=10$ vertices will. Since each connecting line is incident to two vertices, there are $frac24cdot102=120$ interior lines.
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
$endgroup$
$begingroup$
+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
$endgroup$
– antkam
1 hour ago
1
$begingroup$
One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
$endgroup$
– TonyK
53 mins ago
add a comment |
$begingroup$
In the resulting truncated cube, each vertex $v$ is incident to one triangle and three octagons, which together have $3+8+8-3-3=13$ vertices between them excluding $v$. Lines from $v$ to these $13$ vertices will not be completely in the cube; lines to the other $24-13-1=10$ vertices will. Since each connecting line is incident to two vertices, there are $frac24cdot102=120$ interior lines.
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
$endgroup$
$begingroup$
+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
$endgroup$
– antkam
1 hour ago
1
$begingroup$
One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
$endgroup$
– TonyK
53 mins ago
add a comment |
$begingroup$
In the resulting truncated cube, each vertex $v$ is incident to one triangle and three octagons, which together have $3+8+8-3-3=13$ vertices between them excluding $v$. Lines from $v$ to these $13$ vertices will not be completely in the cube; lines to the other $24-13-1=10$ vertices will. Since each connecting line is incident to two vertices, there are $frac24cdot102=120$ interior lines.
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
$endgroup$
In the resulting truncated cube, each vertex $v$ is incident to one triangle and three octagons, which together have $3+8+8-3-3=13$ vertices between them excluding $v$. Lines from $v$ to these $13$ vertices will not be completely in the cube; lines to the other $24-13-1=10$ vertices will. Since each connecting line is incident to two vertices, there are $frac24cdot102=120$ interior lines.
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
answered 5 hours ago
Parcly TaxelParcly Taxel
44.4k1376109
44.4k1376109
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+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
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– antkam
1 hour ago
1
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One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
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– TonyK
53 mins ago
add a comment |
$begingroup$
+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
$endgroup$
– antkam
1 hour ago
1
$begingroup$
One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
$endgroup$
– TonyK
53 mins ago
$begingroup$
+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
$endgroup$
– antkam
1 hour ago
$begingroup$
+1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
$endgroup$
– antkam
1 hour ago
1
1
$begingroup$
One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
$endgroup$
– TonyK
53 mins ago
$begingroup$
One triangle and two octagons. Given this mistake, can you please explain your $3+8+8-3-3$? It gets the right answer, but it looks like sheer luck...
$endgroup$
– TonyK
53 mins ago
add a comment |
$begingroup$
Every one of them that does not run along a face.
Each octahedral face of the object, of which there are $6,$ contains $8$ vertices, so there are $68 choose 2=168$ segments along those faces. The segments that are part of the $12$ original edges of the cube are counted twice, so we subtract $12$, leaving $156$ segments on the octagonal faces of the object.
The triangular faces have segments running along them, but we have already counted those because they also run along an octahedral face. There are $24 choose 2=276$ line segments, so there are $276-156=120$ running through the body of the object.
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
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You're double-counting octagon-octagon edges.
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– Parcly Taxel
5 hours ago
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@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
Every one of them that does not run along a face.
Each octahedral face of the object, of which there are $6,$ contains $8$ vertices, so there are $68 choose 2=168$ segments along those faces. The segments that are part of the $12$ original edges of the cube are counted twice, so we subtract $12$, leaving $156$ segments on the octagonal faces of the object.
The triangular faces have segments running along them, but we have already counted those because they also run along an octahedral face. There are $24 choose 2=276$ line segments, so there are $276-156=120$ running through the body of the object.
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
You're double-counting octagon-octagon edges.
$endgroup$
– Parcly Taxel
5 hours ago
$begingroup$
@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
Every one of them that does not run along a face.
Each octahedral face of the object, of which there are $6,$ contains $8$ vertices, so there are $68 choose 2=168$ segments along those faces. The segments that are part of the $12$ original edges of the cube are counted twice, so we subtract $12$, leaving $156$ segments on the octagonal faces of the object.
The triangular faces have segments running along them, but we have already counted those because they also run along an octahedral face. There are $24 choose 2=276$ line segments, so there are $276-156=120$ running through the body of the object.
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
Every one of them that does not run along a face.
Each octahedral face of the object, of which there are $6,$ contains $8$ vertices, so there are $68 choose 2=168$ segments along those faces. The segments that are part of the $12$ original edges of the cube are counted twice, so we subtract $12$, leaving $156$ segments on the octagonal faces of the object.
The triangular faces have segments running along them, but we have already counted those because they also run along an octahedral face. There are $24 choose 2=276$ line segments, so there are $276-156=120$ running through the body of the object.
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
edited 5 hours ago
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
answered 5 hours ago
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
You're double-counting octagon-octagon edges.
$endgroup$
– Parcly Taxel
5 hours ago
$begingroup$
@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
You're double-counting octagon-octagon edges.
$endgroup$
– Parcly Taxel
5 hours ago
$begingroup$
@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
$begingroup$
You're double-counting octagon-octagon edges.
$endgroup$
– Parcly Taxel
5 hours ago
$begingroup$
You're double-counting octagon-octagon edges.
$endgroup$
– Parcly Taxel
5 hours ago
$begingroup$
@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
$begingroup$
@ParclyTaxel: Good point. Fixed. Thanks.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
The object you describe is called the truncated cube, it's an Archimedian solid.
See for example:
https://en.wikipedia.org/wiki/Truncated_cube
All of its edges lie on the surface of the orginal cube.
answered 5 hours ago
quaraguequarague
436210
$endgroup$
3
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
The object you describe is called the truncated cube, it's an Archimedian solid.
See for example:
https://en.wikipedia.org/wiki/Truncated_cube
All of its edges lie on the surface of the orginal cube.
answered 5 hours ago
quaraguequarague
436210
$endgroup$
3
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
$begingroup$
The object you describe is called the truncated cube, it's an Archimedian solid.
See for example:
https://en.wikipedia.org/wiki/Truncated_cube
All of its edges lie on the surface of the orginal cube.
answered 5 hours ago
quaraguequarague
436210
$endgroup$
The object you describe is called the truncated cube, it's an Archimedian solid.
See for example:
https://en.wikipedia.org/wiki/Truncated_cube
All of its edges lie on the surface of the orginal cube.
answered 5 hours ago
quaraguequarague
436210
answered 5 hours ago
quaraguequarague
436210
answered 5 hours ago
quaraguequarague
436210
answered 5 hours ago
quaraguequarague
436210
436210
3
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
3
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
3
3
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
$begingroup$
But there are segments that run through the body as well. If we put the cube in the first quadrant, there are $9$ segments from the corners near $(0,0,0)$ to corners near $(1,1,1)$ that just have endpoints on the exterior. Those are the ones we want to count.
$endgroup$
– Ross Millikan
5 hours ago
add a comment |
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This is an interesting puzzle. Where does it come from? is this a homework exercise? in what class? What motivates the problem? why are you interested in it? and why should we be interested in it? At the very least, please indicate what you have attempted to do, and perhaps explain what tools (theorems, definitions, etc) you have access to.
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– Xander Henderson
18 mins ago