4 Spheres all touching each other??












3












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If there are 4 spheres all touching each other and 3 of them have diameters 4, 6 and 12 what is the diameter of the fourth one?
I imagine it like 3 balls on a flat table touching each other and then we are supposed to put another one on top of them but in my imagination the top sphere could be any size basically right?






geometry spheres solid-geometry







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  • $begingroup$
    I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case.
    $endgroup$
    – saulspatz
    9 hours ago






  • 1




    $begingroup$
    There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem
    $endgroup$
    – user376343
    8 hours ago












  • $begingroup$
    If the spheres are, indeed, "lying on a table" beware they are not concentric.
    $endgroup$
    – Oscar Lanzi
    3 hours ago
















3












$begingroup$


If there are 4 spheres all touching each other and 3 of them have diameters 4, 6 and 12 what is the diameter of the fourth one?
I imagine it like 3 balls on a flat table touching each other and then we are supposed to put another one on top of them but in my imagination the top sphere could be any size basically right?






geometry spheres solid-geometry







share|cite|improve this question







New contributor




Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case.
    $endgroup$
    – saulspatz
    9 hours ago






  • 1




    $begingroup$
    There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem
    $endgroup$
    – user376343
    8 hours ago












  • $begingroup$
    If the spheres are, indeed, "lying on a table" beware they are not concentric.
    $endgroup$
    – Oscar Lanzi
    3 hours ago














3












3








3





$begingroup$


If there are 4 spheres all touching each other and 3 of them have diameters 4, 6 and 12 what is the diameter of the fourth one?
I imagine it like 3 balls on a flat table touching each other and then we are supposed to put another one on top of them but in my imagination the top sphere could be any size basically right?






geometry spheres solid-geometry







share|cite|improve this question







New contributor




Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




If there are 4 spheres all touching each other and 3 of them have diameters 4, 6 and 12 what is the diameter of the fourth one?
I imagine it like 3 balls on a flat table touching each other and then we are supposed to put another one on top of them but in my imagination the top sphere could be any size basically right?






geometry spheres solid-geometry




geometry spheres solid-geometry






share|cite|improve this question







New contributor




Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 9 hours ago









Happiness Is The TruthHappiness Is The Truth

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New contributor





Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Happiness Is The Truth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case.
    $endgroup$
    – saulspatz
    9 hours ago






  • 1




    $begingroup$
    There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem
    $endgroup$
    – user376343
    8 hours ago












  • $begingroup$
    If the spheres are, indeed, "lying on a table" beware they are not concentric.
    $endgroup$
    – Oscar Lanzi
    3 hours ago


















  • $begingroup$
    I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case.
    $endgroup$
    – saulspatz
    9 hours ago






  • 1




    $begingroup$
    There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem
    $endgroup$
    – user376343
    8 hours ago












  • $begingroup$
    If the spheres are, indeed, "lying on a table" beware they are not concentric.
    $endgroup$
    – Oscar Lanzi
    3 hours ago
















$begingroup$
I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case.
$endgroup$
– saulspatz
9 hours ago




$begingroup$
I think that's right if the three balls on the table have the same diameter. I wonder if it's true in any other case.
$endgroup$
– saulspatz
9 hours ago




1




1




$begingroup$
There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem
$endgroup$
– user376343
8 hours ago






$begingroup$
There is more than one possibility, if the fourth sphere is not necessarily lying on the table with three others. en.m.wikipedia.org/wiki/Descartes%27_theorem
$endgroup$
– user376343
8 hours ago














$begingroup$
If the spheres are, indeed, "lying on a table" beware they are not concentric.
$endgroup$
– Oscar Lanzi
3 hours ago




$begingroup$
If the spheres are, indeed, "lying on a table" beware they are not concentric.
$endgroup$
– Oscar Lanzi
3 hours ago










4 Answers
4






active

oldest

votes


















6












$begingroup$

Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.



I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
    $endgroup$
    – Ross Millikan
    8 hours ago










  • $begingroup$
    For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
    $endgroup$
    – kasperd
    1 hour ago



















5












$begingroup$

Just complementing the answer just posted. If the new sphere is sufficiently small, it will fit in the hole in the middle, so we have a lower bound for the size of the new sphere.
Similarly, if the new sphere is large enough, it might also make one of the previous spheres' size to be "sufficiently small", thus there might be an upper bound.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
    $endgroup$
    – Joonas Ilmavirta
    8 hours ago










  • $begingroup$
    This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    $endgroup$
    – Javi
    6 hours ago



















3












$begingroup$

Hint



The word cirsphe might refer to a circle, a sphere of a hypersphere.



Descartes' Theorem for $ngeq 2$ dimensions tells us that we need $n+1$ cirsphes in order to be able to determinate the radius of the $n+2$th cirsphe...



Now, how to do that?



Define the curvature $k_d$ of the $d$th cirsphe with radius $r_d$ as $$k_d=pmfrac{1}{r_d}$$ (wheter plus or minus dependes on wheter the cirsphe is externally or internally tangent)



Descartes' Theorem for higher dimensions tells us now that




$$bigg(sum _{d=0}^{n+2}k_dbigg)^2=2·sum_{d=0}^{n+2}k_d^2$$




And knowing $k_1, k_2,...,k_{n+1}$ you can determine the curvature of the $n+2$th cirsphe and hence its radius.



There's even a poem regarding this formula!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    just call it a n-sphere
    $endgroup$
    – qwr
    2 hours ago



















0












$begingroup$

Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).



If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...






share|cite|improve this answer











$endgroup$













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.



    I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
      $endgroup$
      – Ross Millikan
      8 hours ago










    • $begingroup$
      For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
      $endgroup$
      – kasperd
      1 hour ago
















    6












    $begingroup$

    Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.



    I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
      $endgroup$
      – Ross Millikan
      8 hours ago










    • $begingroup$
      For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
      $endgroup$
      – kasperd
      1 hour ago














    6












    6








    6





    $begingroup$

    Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.



    I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.






    share|cite|improve this answer









    $endgroup$



    Not any size. If the fourth sphere is sufficiently small, it can fit in the hole in the middle, also resting on the table, without touching any of the other three spheres.



    I’m guessing this problem is asking for the minimum diameter of the fourth sphere which guarantees contact with all of the other three.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 9 hours ago









    MPWMPW

    30.5k12157




    30.5k12157












    • $begingroup$
      The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
      $endgroup$
      – Ross Millikan
      8 hours ago










    • $begingroup$
      For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
      $endgroup$
      – kasperd
      1 hour ago


















    • $begingroup$
      The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
      $endgroup$
      – Ross Millikan
      8 hours ago










    • $begingroup$
      For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
      $endgroup$
      – kasperd
      1 hour ago
















    $begingroup$
    The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
    $endgroup$
    – Ross Millikan
    8 hours ago




    $begingroup$
    The limits can be computed from the four kissing circles problem. If the spheres are at the limit, all the tangent points will be in a plane.
    $endgroup$
    – Ross Millikan
    8 hours ago












    $begingroup$
    For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
    $endgroup$
    – kasperd
    1 hour ago




    $begingroup$
    For some combination of sizes for the initial three spheres it's also possible for the fourth sphere to be too big. I haven't calculated whether the fourth sphere can be too big given the specific combination of the initial three.
    $endgroup$
    – kasperd
    1 hour ago











    5












    $begingroup$

    Just complementing the answer just posted. If the new sphere is sufficiently small, it will fit in the hole in the middle, so we have a lower bound for the size of the new sphere.
    Similarly, if the new sphere is large enough, it might also make one of the previous spheres' size to be "sufficiently small", thus there might be an upper bound.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
      $endgroup$
      – Joonas Ilmavirta
      8 hours ago










    • $begingroup$
      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
      $endgroup$
      – Javi
      6 hours ago
















    5












    $begingroup$

    Just complementing the answer just posted. If the new sphere is sufficiently small, it will fit in the hole in the middle, so we have a lower bound for the size of the new sphere.
    Similarly, if the new sphere is large enough, it might also make one of the previous spheres' size to be "sufficiently small", thus there might be an upper bound.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
      $endgroup$
      – Joonas Ilmavirta
      8 hours ago










    • $begingroup$
      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
      $endgroup$
      – Javi
      6 hours ago














    5












    5








    5





    $begingroup$

    Just complementing the answer just posted. If the new sphere is sufficiently small, it will fit in the hole in the middle, so we have a lower bound for the size of the new sphere.
    Similarly, if the new sphere is large enough, it might also make one of the previous spheres' size to be "sufficiently small", thus there might be an upper bound.






    share|cite|improve this answer









    $endgroup$



    Just complementing the answer just posted. If the new sphere is sufficiently small, it will fit in the hole in the middle, so we have a lower bound for the size of the new sphere.
    Similarly, if the new sphere is large enough, it might also make one of the previous spheres' size to be "sufficiently small", thus there might be an upper bound.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 9 hours ago









    Scott GuanScott Guan

    664




    664












    • $begingroup$
      There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
      $endgroup$
      – Joonas Ilmavirta
      8 hours ago










    • $begingroup$
      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
      $endgroup$
      – Javi
      6 hours ago


















    • $begingroup$
      There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
      $endgroup$
      – Joonas Ilmavirta
      8 hours ago










    • $begingroup$
      This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
      $endgroup$
      – Javi
      6 hours ago
















    $begingroup$
    There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
    $endgroup$
    – Joonas Ilmavirta
    8 hours ago




    $begingroup$
    There certainly is an upper bound for that reason for some initial three radii, but I don't think it'll happen with the OP's data. I guess the condition is roughly whether there's no plane tangent to the three first balls.
    $endgroup$
    – Joonas Ilmavirta
    8 hours ago












    $begingroup$
    This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    $endgroup$
    – Javi
    6 hours ago




    $begingroup$
    This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
    $endgroup$
    – Javi
    6 hours ago











    3












    $begingroup$

    Hint



    The word cirsphe might refer to a circle, a sphere of a hypersphere.



    Descartes' Theorem for $ngeq 2$ dimensions tells us that we need $n+1$ cirsphes in order to be able to determinate the radius of the $n+2$th cirsphe...



    Now, how to do that?



    Define the curvature $k_d$ of the $d$th cirsphe with radius $r_d$ as $$k_d=pmfrac{1}{r_d}$$ (wheter plus or minus dependes on wheter the cirsphe is externally or internally tangent)



    Descartes' Theorem for higher dimensions tells us now that




    $$bigg(sum _{d=0}^{n+2}k_dbigg)^2=2·sum_{d=0}^{n+2}k_d^2$$




    And knowing $k_1, k_2,...,k_{n+1}$ you can determine the curvature of the $n+2$th cirsphe and hence its radius.



    There's even a poem regarding this formula!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      just call it a n-sphere
      $endgroup$
      – qwr
      2 hours ago
















    3












    $begingroup$

    Hint



    The word cirsphe might refer to a circle, a sphere of a hypersphere.



    Descartes' Theorem for $ngeq 2$ dimensions tells us that we need $n+1$ cirsphes in order to be able to determinate the radius of the $n+2$th cirsphe...



    Now, how to do that?



    Define the curvature $k_d$ of the $d$th cirsphe with radius $r_d$ as $$k_d=pmfrac{1}{r_d}$$ (wheter plus or minus dependes on wheter the cirsphe is externally or internally tangent)



    Descartes' Theorem for higher dimensions tells us now that




    $$bigg(sum _{d=0}^{n+2}k_dbigg)^2=2·sum_{d=0}^{n+2}k_d^2$$




    And knowing $k_1, k_2,...,k_{n+1}$ you can determine the curvature of the $n+2$th cirsphe and hence its radius.



    There's even a poem regarding this formula!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      just call it a n-sphere
      $endgroup$
      – qwr
      2 hours ago














    3












    3








    3





    $begingroup$

    Hint



    The word cirsphe might refer to a circle, a sphere of a hypersphere.



    Descartes' Theorem for $ngeq 2$ dimensions tells us that we need $n+1$ cirsphes in order to be able to determinate the radius of the $n+2$th cirsphe...



    Now, how to do that?



    Define the curvature $k_d$ of the $d$th cirsphe with radius $r_d$ as $$k_d=pmfrac{1}{r_d}$$ (wheter plus or minus dependes on wheter the cirsphe is externally or internally tangent)



    Descartes' Theorem for higher dimensions tells us now that




    $$bigg(sum _{d=0}^{n+2}k_dbigg)^2=2·sum_{d=0}^{n+2}k_d^2$$




    And knowing $k_1, k_2,...,k_{n+1}$ you can determine the curvature of the $n+2$th cirsphe and hence its radius.



    There's even a poem regarding this formula!






    share|cite|improve this answer









    $endgroup$



    Hint



    The word cirsphe might refer to a circle, a sphere of a hypersphere.



    Descartes' Theorem for $ngeq 2$ dimensions tells us that we need $n+1$ cirsphes in order to be able to determinate the radius of the $n+2$th cirsphe...



    Now, how to do that?



    Define the curvature $k_d$ of the $d$th cirsphe with radius $r_d$ as $$k_d=pmfrac{1}{r_d}$$ (wheter plus or minus dependes on wheter the cirsphe is externally or internally tangent)



    Descartes' Theorem for higher dimensions tells us now that




    $$bigg(sum _{d=0}^{n+2}k_dbigg)^2=2·sum_{d=0}^{n+2}k_d^2$$




    And knowing $k_1, k_2,...,k_{n+1}$ you can determine the curvature of the $n+2$th cirsphe and hence its radius.



    There's even a poem regarding this formula!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 4 hours ago









    Dr. MathvaDr. Mathva

    2,086324




    2,086324












    • $begingroup$
      just call it a n-sphere
      $endgroup$
      – qwr
      2 hours ago


















    • $begingroup$
      just call it a n-sphere
      $endgroup$
      – qwr
      2 hours ago
















    $begingroup$
    just call it a n-sphere
    $endgroup$
    – qwr
    2 hours ago




    $begingroup$
    just call it a n-sphere
    $endgroup$
    – qwr
    2 hours ago











    0












    $begingroup$

    Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).



    If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).



      If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...






      share|cite|improve this answer











      $endgroup$
















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        $begingroup$

        Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).



        If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...






        share|cite|improve this answer











        $endgroup$



        Imagine all three balls touch each other. The fourth one must exactly fit in the middle of these three balls. Only then each ball touches every other ball. (of course the assumption here is, that it's all in 2D).



        If we say, that speres are allowed inside other spheres or these are more than 2D spheres, then there are many other possibilities, too...







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 4 hours ago









        yaryar

        1034




        1034






















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