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Math Help - Three-dimensional pythagora's theorem

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Three-dimensional Pythagoras' theorem

    Suppose you have a right-angled tetrahedron, i.e. a tetrahedron with a vertex where three faces meet at right angles. Let A be the area of the face opposite this vertex, and B, C, D the areas of the remaining faces. Show that A^2=B^2+C^2+D^2.

    Edit : I had first posted a more general problem but I'll put it on hold for the time being.
    Last edited by Bruno J.; February 9th 2010 at 08:13 PM.
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  2. #2
    Member Black's Avatar
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    Spoiler:
    Let

    sides of A:x,y,z
    sides of B: \alpha, \gamma, y
    sides of C: \alpha, \beta, x
    sides of D: \beta, \gamma, z.

    Then we have

    B=\frac{1}{2}\alpha \gamma, \, C=\frac{1}{2}\alpha \beta , \, D=\frac{1}{2}\beta \gamma

    and

    x^2=\alpha^2+\beta^2
    y^2=\alpha^2+\gamma^2
    z^2=\beta^2+\gamma^2.

    Using Heron's formula, we have

    A^2=\frac{1}{16}(x+y+z)(y+z-x)(x+y-z)(z+x-y)

    =\frac{1}{16}((y+z)+x)((y+z)-x)(x+(y-z))(x-(y-z))

    =\frac{1}{16}((y+z)^2-x^2)(x^2-(y-z)^2)

    =\frac{1}{16}(2yz-(x^2-y^2-z^2))(2yz+(x^2-y^2-z^2))

    =\frac{1}{16}(4y^2z^2-(x^2-y^2-z^2)^2)

    =\frac{1}{16}\left[4(\alpha^2+\gamma^2)(\beta^2+\gamma^2)-((\alpha^2+\beta^2)-(\alpha^2+\gamma^2)-(\beta^2+\gamma^2))^2\right]

    =\frac{1}{16}\left[4(\alpha^2\beta^2+\beta^2\gamma^2+\alpha^2\gamma^2  +\gamma^4)-4\gamma^4\right]

    =\frac{1}{4}\alpha^2\beta^2+\frac{1}{4}\beta^2\gam  ma^2+\frac{1}{4}\alpha^2\gamma^2=C^2+D^2+B^2.
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  3. #3
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    Here's my solution

    Spoiler:
    Couldn't you just put your tetrahedron in the xyz planes where the xy, yz \;\text{and}\; xz plane are three of the four sides. Let the face A be described by the equation ax + by + cz = 1. If so then the area of the sides B, C and D are \frac{1}{2} \frac{1}{ab}, \frac{1}{2} \frac{1}{ac} \; \text{and}\; \frac{1}{2} \frac{1}{bc}. So

     <br />
B^2+C^2+D^2 = \frac{1}{4} \left( \frac{1}{a^2b^2} + \frac{1}{a^2c^2} + \frac{1}{b^2c^2} \right)<br />

    The area of face A is \iint \limits_R \sqrt{1 + z_x^2 + z_y^2}dA = \sqrt{1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}} \frac{1}{2ab}

    and so A^2 = \left(1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}\right) \frac{1}{4a^2b^2} = B^2 + C^2 + D^2
    Last edited by Jester; February 14th 2010 at 08:05 AM. Reason: added spoiler
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  4. #4
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    Let

     A = A(a,0,0)   , B = B(0,b,0) , C = C(0,0,c)

    together with the origin , they form a right-angled tetrahedron

    we have  CA = (a,0,-c) , CB = (0,b,-c)

    and their cross product is

     CA \times CB = (a,0,-c) \times (0,b,-c)

     = ( bc , ac , ab )

    we also know that the area of the face opposite this vertex  S is

    the half of the modulus of the cross product so

     S^2 =  \frac{(ab)^2 + (bc)^2 + (ac)^2 }{4}

     = S^2_1 + S^2_2 + S^2_3

    where  S_1 , S_2 , S_3 are the area of the lateral faces of the tetrahedron .
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