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Math Help - Tetrahedron

  1. #1
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    Tetrahedron

    A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point p is selected at random inside the circumscribed sphere. The probability that p lies inside one of the five shperes is closest to

    a) 0 b) 0.1 c) 0.2 d) 0.3 e) 0.4

    The answer is c.

    And I am clueless how i could get any numbers at all.
    Could someone take a look at my attached horrible drawing?
    I just want to make sure I have the correct idea.

    Vicky.
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  2. #2
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    Quote Originally Posted by Vicky1997 View Post
    A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point p is selected at random inside the circumscribed sphere. The probability that p lies inside one of the five shperes is closest to

    a) 0 b) 0.1 c) 0.2 d) 0.3 e) 0.4

    The answer is c.

    And I am clueless how i could get any numbers at all.
    Could someone take a look at my attached horrible drawing?
    I just want to make sure I have the correct idea.

    Vicky.
    As far as I understand the question your sketch isn't correct. I've made a sketch but it doesn't take into account that we are dealing with a 3D object.

    Probably () the probability could be calculated by:

    p(P\ in\ spheres) = \dfrac{\text{sum of the 5 volumes}}{\text{volume of the circumscribing sphere}}

    Let a denote the length of one edge of the tetrahedron, then

    R_c = \frac23 \cdot h = a\cdot \frac2{9} \cdot \sqrt{6}

    R_i = \frac12 \cdot R_c

    r = \frac12 \cdot R_i

    That means the radii are proportional like:

    r : R_i : R_c = 1 : 2 : 4

    Can you take it from here?
    Attached Thumbnails Attached Thumbnails Tetrahedron-beruehrkrse_4eder.png  
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  3. #3
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    Quote Originally Posted by earboth View Post
    As far as I understand the question your sketch isn't correct. I've made a sketch but it doesn't take into account that we are dealing with a 3D object.

    Probably () the probability could be calculated by:

    p(P\ in\ spheres) = \dfrac{\text{sum of the 5 volumes}}{\text{volume of the circumscribing sphere}}

    Let a denote the length of one edge of the tetrahedron, then

    R_c = \frac23 \cdot h = a\cdot \frac2{9} \cdot \sqrt{6}

    R_i = \frac12 \cdot R_c

    r = \frac12 \cdot R_i

    That means the radii are proportional like:

    r : R_i : R_c = 1 : 2 : 4

    Can you take it from here?


    I think I got the answer.


    4/3pi 8r cubed + 4 X 4/3pi r cubed over 4/3 pi 64 r cubed = 0.1875

    Thanks a lot for your help.

    [COLOR=#ff0000]But I still can't totally drop the idea that the spheres should be similar in size, if not equal. I like making geo-gamis and one of my favorite shapes is actually the tetrahedron. I can pretty much visualize what it will look like. I made my own tetrahedron and spheres so could you please take a look at my attachment. It is quite accurate in proportion.

    I don't know how to get the proportion of the radii, so I would be grateful if you could try what you did with my idea. And if it doesn't work please tell me what I have wrong.


    Vicky.
    Attached Thumbnails Attached Thumbnails Tetrahedron-0106000956.jpg  
    Last edited by Vicky1997; January 6th 2010 at 06:28 PM.
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  4. #4
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    Quote Originally Posted by Vicky1997 View Post
    [/COLOR]

    I think I got the answer.


    4/3pi 8r cubed + 4 X 4/3pi r cubed over 4/3 pi 64 r cubed = 0.1875 <<<<<< correct

    Thanks a lot for your help.

    [COLOR=#ff0000]But I still can't totally drop the idea that the spheres should be similar in size, if not equal. I like making geo-gamis and one of my favorite shapes is actually the tetrahedron. I can pretty much visualize what it will look like. I made my own tetrahedron and spheres so could you please take a look at my attachment. It is quite accurate in proportion. <<<<<<<< unfortunately no. The circles which are part of the circumscribing sphere must pass through 2 vertices and the midpoint of the opposite edge. Your circles are much to large.

    I don't know how to get the proportion of the radii, so I would be grateful if you could try what you did with my idea. And if it doesn't work please tell me what I have wrong.


    Vicky.
    I've modified your foto a little bit. Have a look.
    Attached Thumbnails Attached Thumbnails Tetrahedron-tetrahedron_beruehrkrs2.png  
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  5. #5
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    Quote Originally Posted by earboth View Post
    I've modified your foto a little bit. Have a look.
    Thanks!!!

    Now I understand what was wrong.
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