Results 1 to 4 of 4

Math Help - Volume of Sphere Inscribed within a Tetrahedron

  1. #1
    Newbie
    Joined
    Apr 2009
    From
    California
    Posts
    5

    Volume of Sphere Inscribed within a Tetrahedron

    So we have a tetrahedral pyramid with side length a. What's the volume of the inscribed sphere that is tangent to the sides of the pyramid?

    I just need some feedback on a solution, as I have no idea if it is correct. And I should mention that all the faces are equilateral triangles.

    Start by solving for the height of each triangular face, with side length a.
    a^2+(\frac{a}{2})^2=k
    where k is the height of one of our face triangles.
    Solving, k=\frac{a\sqrt{3}}{2}
    Now, we can use k to solve for the height of the pyramid, h; imagine a new triangle cutting the pyramid in half.
    k^2=(\frac{k}{2})^2+h^2
    Solving, h=\frac{k\sqrt{3}}{2}=\frac{3a}{4}
    Using the relationship between a 2-d triangle's height and an inscribed circle's diameter, which is
    height=\frac{3}{2}*diameter,
    diameter=\frac{2}{3}*height,
    we can find the diameter, d, and radius, r, of the inscribed sphere.
    r=\frac{d}{2}=\frac{h}{3}=\frac{3a}{4}*\frac{1}{3}  =\frac{a}{4}
    Plug the radius, r, into the volume of a sphere, V, to get the solution.
    V=\frac{4}{3} \pi r^3=\frac{4}{3}\pi (\frac{a}{4})^3=\frac{4\pi}{3}*\frac{a^3}{64}=\fra  c{\pi a^3}{48}
    ...and that's the solution I've come up with. And I have no idea how to even check to see if it's correct. I'll try to get a picture up so the solution is more visible.
    Last edited by nickerus; July 15th 2009 at 05:38 PM. Reason: details
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,804
    Thanks
    115
    Quote Originally Posted by nickerus View Post
    So we have a tetrahedral pyramid with side length a. What's the volume of the inscribed sphere that is tangent to the sides of the pyramid?

    I just need some feedback on a solution, as I have no idea if it is correct. And I should mention that all the faces are equilateral triangles.

    Start by solving for the height of each triangular face, with side length a.
    a^2+(\frac{a}{2})^2=k
    where k is the height of one of our face triangles.
    Solving, k=\frac{a\sqrt{3}}{2}
    Now, we can use k to solve for the height of the pyramid, h; imagine a new triangle cutting the pyramid in half.
    k^2=(\frac{k}{2})^2+h^2 ..... <<<<< this can't be correct. See attachment.

    ...
    ...

    In my opinion the equation should read:

    k^2=\left(\dfrac13 k\right)^2 + h^2

    or

    a^2=\left(\dfrac23 k\right)^2 + h^2
    Attached Thumbnails Attached Thumbnails Volume of Sphere Inscribed within a Tetrahedron-tetraeder.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2009
    From
    California
    Posts
    5
    Yeah I think that's right; I don't know why I went with k/2 hehe. Then, the height of the pyramid would be
    h=\frac{2k\sqrt{2}}{3}=\frac{2\sqrt{2}}{3}*\frac{a  \sqrt{3}}{2}=\frac{a\sqrt{6}}{3},
    and the radius of the sphere would be
    r=\frac{h}{3}=\frac{1}{3}*\frac{a\sqrt{6}}{3}=\fra  c{a\sqrt{6}}{9}.
    But then the volume works out to a pretty gnarly fraction,
    V=\frac{8\pi a^3 \sqrt{6}}{729}
    On wikipedia, it lists the radius of the insphere of a tetrahedron as \frac{a}{\sqrt{24}}, so how did they get that..? I could always just be doing my algebra wrong.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member alunw's Avatar
    Joined
    May 2009
    Posts
    188
    I've tried working things like this out before now and its usually not that easy.
    Here's how I'd go about it for a tetrahedron.
    If you take an ordinary cube, and pick two vertices diagonally opposite each other on one face, and the other two vertices diagonally opposite each other on the opposite face you get a regular tetrahedron.
    This shows that the shape with vertices at say (1,1,1),(1,-1,-1),(-1,-1,1),(-1,1,-1) is a regular tetrahedron. The side of the cube is 2, and the side of the tetrahedron is 2*sqrt(2) = sqrt(8). The point of tangency of the sphere to one of the faces is easy to calculate
    as well. For example one of them is 1/3 of the way along the line from (1,0,0) (the midpoint of a side ) to (-1,1,-1) (one of the two opposite vertices), i.e at (1/3,1/3,-1/3) since the sphere is surely tangent at the centre of the face. Now we certainly hope the centre of the inscribed sphere is at (0,0,0) and by considering all the other possible tetrahedra we could inscribe in the cube we see this must be so.

    So the radius of the sphere is sqrt(1/3) and sqrt(1/3)/sqrt(8) is sqrt(1/24).
    Last edited by alunw; July 17th 2009 at 05:10 AM. Reason: typo corrected and explanation improved
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: November 16th 2011, 09:11 AM
  2. Sphere inscribed in a regular tetrahedron
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 10th 2010, 05:22 AM
  3. Replies: 0
    Last Post: March 7th 2010, 07:47 PM
  4. Sphere in a tetrahedron
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 26th 2009, 03:09 AM
  5. Tetrahedron in a unit sphere
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 21st 2007, 07:31 AM

Search Tags


/mathhelpforum @mathhelpforum