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Geometry tetrahedron problem

  1. Dec 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Volume of tetrahedron T.ABC = V
    Point P is on the middle of TA, Q is on the expansion of AB making AQ = 2AB
    A shape is made through PQ which is parallel to BC so that it cuts the tetahedron into 2 pieces.
    What is the volume of the biggest piece?

    3. The attempt at a solution

    I sketch the problem

    http://www.sumopaint.com/images/temp/xzohqlsrafmmqrck.png

    Then, I have no idea how to continue
     
  2. jcsd
  3. Dec 24, 2015 #2

    BvU

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    Hi Terry,

    I am missing something ! : 2. Relevant equations
    For this part of the template, you could think of this or this
    Deleting (even just a part of) the template irritates the spirits that watch over us, so you really don't want to do that. A litte googling usually helps if your notes or textbook don't have anything useful. And in this case the second link might have given you an idea...

    I notice you've drawn PM parallalel to AC instead of NM parallel to BC. Perhaps you could clarify ?

    And, for the record, I don't have the answer, so we still need help. :frown:

    --
     
  4. Dec 24, 2015 #3

    Ray Vickson

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    Nevertheless, you need to make an attempt; those are the PF rules.
     
  5. Dec 24, 2015 #4
    I'm sorry. Now, I want to edit the thread but it's disabled now :frown:
    Okay, I just add them in this post

    - Relevant equations

    V = Base Area * height

    - My Attempts

    I see the answer in the book, but it's very messy and I don't understand.

    2j3phy8.jpg

    Please help

    I don't understand why it is TN/NB = TM/MC (I think it should be TN/TB = TM/TC)
    Then, TP/TA*TN/TB*TM/TC comes out which gets me more confused
    I think the book has lots of typos, but I'm pretty sure the book's got the idea

    (The book is written in Bahasa, this is the translation :
    P is the center point of TA,
    Q is on the expansion of AB => AQ = 2AB
    The shape through PQ//BC that cuts the tetrahedron into two pieces is)
     
    Last edited: Dec 24, 2015
  6. Dec 25, 2015 #5

    haruspex

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    TN/TB=TN/(TN+NB)=1/(1+NB/TN).
    Likewise, TM/TC=1/(1+MC/TM).
    So if TN/TB=TM/TC then TN/NB=TM/MC.
     
  7. Dec 25, 2015 #6
    Why is it 2/1 ? And, how does TP/TA*TN/TB*TM/TC come from? I really get nothing from the book answer
     
    Last edited: Dec 25, 2015
  8. Dec 25, 2015 #7

    Samy_A

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    I don't know if it is the easiest way, but the theorem of Menelaus, applied to the triangle TAB and the line PNQ, shows that TN/NB = 2.
     
  9. Dec 25, 2015 #8

    haruspex

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    Consider the centroid of triangle TAQ.
    The volume of a tetrahedron Is 1/6 of the triple scalar product of the vectors representing ithree sides adjacent to one vertex. So if you keep the angles fixed and vary the lengths, it is proportional to the product of the three lengths.
     
  10. Dec 25, 2015 #9
    Isn't the volume of tetrahedron is a^3/(6√2) where a is the length of the edge??
    But, how to know that the angle is fixed?? The shape is a bit different from the big initial tetrahedron I think.
     
  11. Dec 25, 2015 #10

    haruspex

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    For a regular tetrahedron, yes, but this is not.
    We're keeping the three angles at T fixed.
    Think of T as the origin. The volume of TABC is one sixth the triple scalar product of vectors TA, TB, TC. Likewise, the volume of TMNP is one sixth the triple scalar product of the vectors TM, TN, TP.
    The triple scalar product of three vectors is the product of the magnitudes of the vectors, multiplied by a function of the angles between them.
     
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