20 regular hexagons and x numbers of regular pentagons are stitched together to make a soccer ball. Find x












2












$begingroup$


20 regular hexagons and x numbers of regular pentagons are stitched together to make a soccer ball. Find x



I tried making a soccer ball pattern but quickly realized that, due to the average mathematician's poor art skills, there has to be a way around from drawing the pattern. I'd go ahead and try to do 20*6 to get the number of vertices but I know it isn't that simple because those vertices will be shared amongst other shapes, so I'm unsure as to how to proceed with this problem. Also, am I right to assume that this has something to do with V+F-E=2?






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    How many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:35










  • $begingroup$
    This is all the question provides. :(
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:36










  • $begingroup$
    Can you work out how many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:37










  • $begingroup$
    Is it 20(6)+5x?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:38










  • $begingroup$
    Each edge borders two polygons.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:45
















2












$begingroup$


20 regular hexagons and x numbers of regular pentagons are stitched together to make a soccer ball. Find x



I tried making a soccer ball pattern but quickly realized that, due to the average mathematician's poor art skills, there has to be a way around from drawing the pattern. I'd go ahead and try to do 20*6 to get the number of vertices but I know it isn't that simple because those vertices will be shared amongst other shapes, so I'm unsure as to how to proceed with this problem. Also, am I right to assume that this has something to do with V+F-E=2?






graph-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    How many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:35










  • $begingroup$
    This is all the question provides. :(
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:36










  • $begingroup$
    Can you work out how many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:37










  • $begingroup$
    Is it 20(6)+5x?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:38










  • $begingroup$
    Each edge borders two polygons.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:45














2












2








2





$begingroup$


20 regular hexagons and x numbers of regular pentagons are stitched together to make a soccer ball. Find x



I tried making a soccer ball pattern but quickly realized that, due to the average mathematician's poor art skills, there has to be a way around from drawing the pattern. I'd go ahead and try to do 20*6 to get the number of vertices but I know it isn't that simple because those vertices will be shared amongst other shapes, so I'm unsure as to how to proceed with this problem. Also, am I right to assume that this has something to do with V+F-E=2?






graph-theory







share|cite|improve this question











$endgroup$




20 regular hexagons and x numbers of regular pentagons are stitched together to make a soccer ball. Find x



I tried making a soccer ball pattern but quickly realized that, due to the average mathematician's poor art skills, there has to be a way around from drawing the pattern. I'd go ahead and try to do 20*6 to get the number of vertices but I know it isn't that simple because those vertices will be shared amongst other shapes, so I'm unsure as to how to proceed with this problem. Also, am I right to assume that this has something to do with V+F-E=2?






graph-theory




graph-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 2:38







DevAllanPer

















asked Dec 7 '18 at 2:33









DevAllanPerDevAllanPer

1336




1336












  • $begingroup$
    How many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:35










  • $begingroup$
    This is all the question provides. :(
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:36










  • $begingroup$
    Can you work out how many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:37










  • $begingroup$
    Is it 20(6)+5x?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:38










  • $begingroup$
    Each edge borders two polygons.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:45


















  • $begingroup$
    How many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:35










  • $begingroup$
    This is all the question provides. :(
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:36










  • $begingroup$
    Can you work out how many edges?
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:37










  • $begingroup$
    Is it 20(6)+5x?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 2:38










  • $begingroup$
    Each edge borders two polygons.
    $endgroup$
    – Lord Shark the Unknown
    Dec 7 '18 at 2:45
















$begingroup$
How many edges?
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 2:35




$begingroup$
How many edges?
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 2:35












$begingroup$
This is all the question provides. :(
$endgroup$
– DevAllanPer
Dec 7 '18 at 2:36




$begingroup$
This is all the question provides. :(
$endgroup$
– DevAllanPer
Dec 7 '18 at 2:36












$begingroup$
Can you work out how many edges?
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 2:37




$begingroup$
Can you work out how many edges?
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 2:37












$begingroup$
Is it 20(6)+5x?
$endgroup$
– DevAllanPer
Dec 7 '18 at 2:38




$begingroup$
Is it 20(6)+5x?
$endgroup$
– DevAllanPer
Dec 7 '18 at 2:38












$begingroup$
Each edge borders two polygons.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 2:45




$begingroup$
Each edge borders two polygons.
$endgroup$
– Lord Shark the Unknown
Dec 7 '18 at 2:45










1 Answer
1






active

oldest

votes


















4












$begingroup$

You should recognize that no more than three faces can fit at each vertex, because the angles in the polygons are too large to fit four faces together.



Then we have the following:



Faces =$x+20$



Edges = $(5x+120)/2$. Each pentagon has five edges and each hexagon has six; adding up all the polygons counts every edge twice because EA h edge is shared by two faces



Vertices = $(5x+120)/3$. Each Pentagon has give vertices, each hexagon has six; adding up all polygons counts each vertex three times because only three faces meet at each vertex (see above).



Now try $V+F-E=2$ with the above renderings for each term and solve for $x$.



You might say that with this method you can get answers to this kind of problem by the dozen.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the help! so X=12?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 3:46










  • $begingroup$
    Look at a soccer ball and see :-) .
    $endgroup$
    – Oscar Lanzi
    Dec 7 '18 at 9:44











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

You should recognize that no more than three faces can fit at each vertex, because the angles in the polygons are too large to fit four faces together.



Then we have the following:



Faces =$x+20$



Edges = $(5x+120)/2$. Each pentagon has five edges and each hexagon has six; adding up all the polygons counts every edge twice because EA h edge is shared by two faces



Vertices = $(5x+120)/3$. Each Pentagon has give vertices, each hexagon has six; adding up all polygons counts each vertex three times because only three faces meet at each vertex (see above).



Now try $V+F-E=2$ with the above renderings for each term and solve for $x$.



You might say that with this method you can get answers to this kind of problem by the dozen.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the help! so X=12?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 3:46










  • $begingroup$
    Look at a soccer ball and see :-) .
    $endgroup$
    – Oscar Lanzi
    Dec 7 '18 at 9:44
















4












$begingroup$

You should recognize that no more than three faces can fit at each vertex, because the angles in the polygons are too large to fit four faces together.



Then we have the following:



Faces =$x+20$



Edges = $(5x+120)/2$. Each pentagon has five edges and each hexagon has six; adding up all the polygons counts every edge twice because EA h edge is shared by two faces



Vertices = $(5x+120)/3$. Each Pentagon has give vertices, each hexagon has six; adding up all polygons counts each vertex three times because only three faces meet at each vertex (see above).



Now try $V+F-E=2$ with the above renderings for each term and solve for $x$.



You might say that with this method you can get answers to this kind of problem by the dozen.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the help! so X=12?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 3:46










  • $begingroup$
    Look at a soccer ball and see :-) .
    $endgroup$
    – Oscar Lanzi
    Dec 7 '18 at 9:44














4












4








4





$begingroup$

You should recognize that no more than three faces can fit at each vertex, because the angles in the polygons are too large to fit four faces together.



Then we have the following:



Faces =$x+20$



Edges = $(5x+120)/2$. Each pentagon has five edges and each hexagon has six; adding up all the polygons counts every edge twice because EA h edge is shared by two faces



Vertices = $(5x+120)/3$. Each Pentagon has give vertices, each hexagon has six; adding up all polygons counts each vertex three times because only three faces meet at each vertex (see above).



Now try $V+F-E=2$ with the above renderings for each term and solve for $x$.



You might say that with this method you can get answers to this kind of problem by the dozen.






share|cite|improve this answer











$endgroup$



You should recognize that no more than three faces can fit at each vertex, because the angles in the polygons are too large to fit four faces together.



Then we have the following:



Faces =$x+20$



Edges = $(5x+120)/2$. Each pentagon has five edges and each hexagon has six; adding up all the polygons counts every edge twice because EA h edge is shared by two faces



Vertices = $(5x+120)/3$. Each Pentagon has give vertices, each hexagon has six; adding up all polygons counts each vertex three times because only three faces meet at each vertex (see above).



Now try $V+F-E=2$ with the above renderings for each term and solve for $x$.



You might say that with this method you can get answers to this kind of problem by the dozen.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 7 '18 at 9:54

























answered Dec 7 '18 at 2:47









Oscar LanziOscar Lanzi

12.4k12036




12.4k12036












  • $begingroup$
    Thanks for the help! so X=12?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 3:46










  • $begingroup$
    Look at a soccer ball and see :-) .
    $endgroup$
    – Oscar Lanzi
    Dec 7 '18 at 9:44


















  • $begingroup$
    Thanks for the help! so X=12?
    $endgroup$
    – DevAllanPer
    Dec 7 '18 at 3:46










  • $begingroup$
    Look at a soccer ball and see :-) .
    $endgroup$
    – Oscar Lanzi
    Dec 7 '18 at 9:44
















$begingroup$
Thanks for the help! so X=12?
$endgroup$
– DevAllanPer
Dec 7 '18 at 3:46




$begingroup$
Thanks for the help! so X=12?
$endgroup$
– DevAllanPer
Dec 7 '18 at 3:46












$begingroup$
Look at a soccer ball and see :-) .
$endgroup$
– Oscar Lanzi
Dec 7 '18 at 9:44




$begingroup$
Look at a soccer ball and see :-) .
$endgroup$
– Oscar Lanzi
Dec 7 '18 at 9:44


















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