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3

1

$begingroup$

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

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edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

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put on hold as off-topic by Saad, Alex Provost, Lord Shark the Unknown, Eevee Trainer, YiFan Mar 20 at 6:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Alex Provost, Lord Shark the Unknown, Eevee Trainer, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

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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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3

1

$begingroup$

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

share|cite|improve this question

edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

$endgroup$

put on hold as off-topic by Saad, Alex Provost, Lord Shark the Unknown, Eevee Trainer, YiFan Mar 20 at 6:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Alex Provost, Lord Shark the Unknown, Eevee Trainer, YiFan

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 14 at 19:07
  

  
  
  
  

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

share|cite|improve this question

edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

$endgroup$

share|cite|improve this question

edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

share|cite|improve this question

edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

edited Mar 14 at 13:48

Parcly Taxel

44.7k1376110

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

asked Mar 14 at 13:35

Mittal GMittal G

1,379516

3 Answers
3

active

oldest

votes

5

$begingroup$

Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:

$hspace{3cm}$

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: $8cdot 10=color{red}{80}$.

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: $color{red}{2}$ 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: $4cdot 5=color{red}{20}$. Overall, $color{red}{22}$ 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 $color{red}{2}$ 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: $4cdot 4=color{red}{16}$. Overall, $color{red}{18}$ 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: $color{red}{80}+color{red}{22}+color{red}{18}=120$.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  something is missing, let me double check
  $endgroup$
  – farruhota
  Mar 14 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  now it is fixed.
  $endgroup$
  – farruhota
  Mar 14 at 16:36
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  David Z, I assume that is what Parcly Taxel did after finding the number of inner connections. I just wanted to see through all inner connections from different perspectives. Cheers.
  $endgroup$
  – farruhota
  Mar 14 at 19:34
  

  
  
  
  

add a comment | 

14

$begingroup$

In the resulting truncated cube, each vertex $v$ is incident to one triangle and two 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 $frac{24cdot10}2=120$ interior lines.

share|cite|improve this answer

edited Mar 15 at 1:02

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
  $endgroup$
  – antkam
  Mar 14 at 18:21
  

  
  
  
  


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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...
  $endgroup$
  – TonyK
  Mar 14 at 18:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
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  @TonyK It's a typo, that's all!
  $endgroup$
  – Parcly Taxel
  Mar 15 at 1:02
  

  
  
  
  

add a comment | 

7

$begingroup$

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 $6{8 choose 2}=168$ 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.

share|cite|improve this answer

edited Mar 14 at 13:42

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

300k24200374

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're double-counting octagon-octagon edges.
  $endgroup$
  – Parcly Taxel
  Mar 14 at 13:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ParclyTaxel: Good point. Fixed. Thanks.
  $endgroup$
  – Ross Millikan
  Mar 14 at 13:45
  

  
  
  
  

add a comment | 

5

$begingroup$

Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:

$hspace{3cm}$

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: $8cdot 10=color{red}{80}$.

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: $color{red}{2}$ 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: $4cdot 5=color{red}{20}$. Overall, $color{red}{22}$ 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 $color{red}{2}$ 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: $4cdot 4=color{red}{16}$. Overall, $color{red}{18}$ 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: $color{red}{80}+color{red}{22}+color{red}{18}=120$.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  something is missing, let me double check
  $endgroup$
  – farruhota
  Mar 14 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  now it is fixed.
  $endgroup$
  – farruhota
  Mar 14 at 16:36
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  David Z, I assume that is what Parcly Taxel did after finding the number of inner connections. I just wanted to see through all inner connections from different perspectives. Cheers.
  $endgroup$
  – farruhota
  Mar 14 at 19:34
  

  
  
  
  

add a comment | 

5

$begingroup$

Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:

$hspace{3cm}$

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: $8cdot 10=color{red}{80}$.

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: $color{red}{2}$ 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: $4cdot 5=color{red}{20}$. Overall, $color{red}{22}$ 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 $color{red}{2}$ 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: $4cdot 4=color{red}{16}$. Overall, $color{red}{18}$ 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: $color{red}{80}+color{red}{22}+color{red}{18}=120$.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  something is missing, let me double check
  $endgroup$
  – farruhota
  Mar 14 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  now it is fixed.
  $endgroup$
  – farruhota
  Mar 14 at 16:36
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 19:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  David Z, I assume that is what Parcly Taxel did after finding the number of inner connections. I just wanted to see through all inner connections from different perspectives. Cheers.
  $endgroup$
  – farruhota
  Mar 14 at 19:34
  

  
  
  
  

add a comment | 

$begingroup$

Alternatively, consider four-storey building with $8,4,4,8$ points labeled $1$ through $24$:

$hspace{3cm}$

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: $8cdot 10=color{red}{80}$.

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: $color{red}{2}$ 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: $4cdot 5=color{red}{20}$. Overall, $color{red}{22}$ 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 $color{red}{2}$ 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: $4cdot 4=color{red}{16}$. Overall, $color{red}{18}$ 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: $color{red}{80}+color{red}{22}+color{red}{18}=120$.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

$endgroup$

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

answered Mar 14 at 16:20

farruhotafarruhota

21.5k2842

14

$begingroup$

In the resulting truncated cube, each vertex $v$ is incident to one triangle and two 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 $frac{24cdot10}2=120$ interior lines.

share|cite|improve this answer

edited Mar 15 at 1:02

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
  $endgroup$
  – antkam
  Mar 14 at 18:21
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 18:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK It's a typo, that's all!
  $endgroup$
  – Parcly Taxel
  Mar 15 at 1:02
  

  
  
  
  

add a comment | 

14

$begingroup$

In the resulting truncated cube, each vertex $v$ is incident to one triangle and two 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 $frac{24cdot10}2=120$ interior lines.

share|cite|improve this answer

edited Mar 15 at 1:02

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 IMHO the most natural answer (also quickest, also least prone to counting error) since every vertex is the same.
  $endgroup$
  – antkam
  Mar 14 at 18:21
  

  
  
  
  


• 
  
  
  

  
  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
  Mar 14 at 18:31
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK It's a typo, that's all!
  $endgroup$
  – Parcly Taxel
  Mar 15 at 1:02
  

  
  
  
  

add a comment | 

$begingroup$

In the resulting truncated cube, each vertex $v$ is incident to one triangle and two 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 $frac{24cdot10}2=120$ interior lines.

share|cite|improve this answer

edited Mar 15 at 1:02

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

$endgroup$

share|cite|improve this answer

edited Mar 15 at 1:02

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

answered Mar 14 at 13:42

Parcly TaxelParcly Taxel

44.7k1376110

7

$begingroup$

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 $6{8 choose 2}=168$ 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.

share|cite|improve this answer

edited Mar 14 at 13:42

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

300k24200374

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're double-counting octagon-octagon edges.
  $endgroup$
  – Parcly Taxel
  Mar 14 at 13:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ParclyTaxel: Good point. Fixed. Thanks.
  $endgroup$
  – Ross Millikan
  Mar 14 at 13:45
  

  
  
  
  

add a comment | 

7

$begingroup$

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 $6{8 choose 2}=168$ 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.

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edited Mar 14 at 13:42

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

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  You're double-counting octagon-octagon edges.
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  – Parcly Taxel
  Mar 14 at 13:42
  

  
  
  
  


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  @ParclyTaxel: Good point. Fixed. Thanks.
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  – Ross Millikan
  Mar 14 at 13:45
  

  
  
  
  

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$begingroup$

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 $6{8 choose 2}=168$ 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.

share|cite|improve this answer

edited Mar 14 at 13:42

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

300k24200374

$endgroup$

share|cite|improve this answer

edited Mar 14 at 13:42

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

300k24200374

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

300k24200374

answered Mar 14 at 13:40

Ross MillikanRoss Millikan

300k24200374