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Math Help - Tetrahedron proof

  1. #1
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    Tetrahedron proof

    Could I have a clue how to start this?

    The edges of a tetrahedron meet at a vertex so that a right angle is formed between each pair of edges. Prove that the base triangle cannot be right angled.
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  2. #2
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    Re: Tetrahedron proof

    Quote Originally Posted by Stuck Man View Post
    Could I have a clue how to start this?

    The edges of a tetrahedron meet at a vertex so that a right angle is formed between each pair of edges. Prove that the base triangle cannot be right angled.
    A tetrahedron has six edges. Not all edges meet at the same vertex. This question does not make sense to me.
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  3. #3
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    Re: Tetrahedron proof

    Hello, Stuck Man!

    Did you make a sketch?


    Three edges of a tetrahedron meet at a vertex so that a right angle
    is formed between each pair of edges.
    Prove that the base triangle cannot be right-angled.

    Code:
                *
               * * * c
              *   *   *
            a*     *b    *
            *       *   *e
           *         * *
          * * * * * * *
                d
    The sides of the three right angles are a,b,c.
    The sides of the base are d,e,f.

    We have: . \begin{Bmatrix}a^2+b^2 &=& d^2 \\ b^2+c^2 &=& e^2 \\ c^2+a^2 &=& f^2 \end{Bmatrix}


    Suppose the base triangle has a right angle,
    . . then we have (for example): . d^2+e^2 \:=\:f^2

    Substitute: . (a^2+b^2) + (b^2+c^2) \:=\;c^2+a^2

    And we have: . 2b^2 \:=\:0 \quad\Rightarrow\quad b \:=\:0

    . . There is no tetrahedron.

    Therefore, the base triangle canNOT be right-angled.
    Thanks from johng
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