1. ## 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.

2. 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

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

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

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

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

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

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

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

$\displaystyle =\frac{1}{4}\alpha^2\beta^2+\frac{1}{4}\beta^2\gam ma^2+\frac{1}{4}\alpha^2\gamma^2=C^2+D^2+B^2.$

3. 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 $\displaystyle A^2 = \left(1 + \frac{a^2}{c^2} + \frac{b^2}{c^2}\right) \frac{1}{4a^2b^2} = B^2 + C^2 + D^2$

4. 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 .