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Thread: 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 $\displaystyle A$ be the area of the face opposite this vertex, and $\displaystyle B, C, D$ the areas of the remaining faces. Show that $\displaystyle A^2=B^2+C^2+D^2$.

    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.; Feb 9th 2010 at 08:13 PM.
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  2. #2
    Member Black's Avatar
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    Spoiler:
    Let

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

    Then we have

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

    and

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

    Using Heron's formula, we have

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

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

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

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

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

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

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

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

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

    $\displaystyle
    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)
    $

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

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

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

    and their cross product is

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

    $\displaystyle = ( bc , ac , ab ) $

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

    the half of the modulus of the cross product so

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

    $\displaystyle = S^2_1 + S^2_2 + S^2_3 $

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