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Slicing a corner off a square gives a right-angled triangle, as shown in the diagram below. The lengths of the sides of this triangle are related by Pythagoras’s theorem: $a^ 2 + b^ 2 = c^ 2$ . Show that this two-dimensional setup generalises to three dimensions in the following way. Slice a corner off a cube, as shown in the diagram below. This gives a tetrahedron in which three of the faces are right-angled triangles, while the fourth is not. Let’s call the areas of the three right-angled faces $A, B, C$ and the area of the fourth face $D$.

$A ^2 + B^ 2 + C^ 2 = D^2$ .enter image description here

Can anyone explain to me what formula/how to go about doing this question? thank you.

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Let $XYZT$ be our tetrahedron, where $TX\perp TY$, $TX\perp TZ$, $TY\perp TZ$, $TX=x$, $TY=y$ and $TZ=z$.

Thus, $XY=\sqrt{x^2+y^2}$, $XZ=\sqrt{x^2+z^2},$ $YZ=\sqrt{y^2+z^2}$ and $$S_{\Delta XYZ}=\frac{1}{4}\sqrt{\sum_{cyc}(2(x^2+y^2)(x^2+z^2)-(x^2+y^2)^2)}=$$ $$=\frac{1}{4}\sqrt{\sum_{cyc}(2x^4+6x^2y^2-2x^4-2x^2y^2)}=\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}$$ and since $$\left(\frac{1}{2}xy\right)^2+\left(\frac{1}{2}xz\right)^2+\left(\frac{1}{2}yz\right)^2=\left(\frac{1}{2}\sqrt{x^2y^2+x^2z^2+y^2z^2}\right)^2,$$ we are done!

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Let $OXYZ$ be the tetrahedron where $a:=OX$, $b:=OY$, and $c:=OZ$ are orthogonal vectors in $\mathbb{R}^3$ with respect to its usual inner product. Then, note that $$A:=\frac{1}{2}\,\|b\times c\|\,,\,\,B:=\frac{1}{2}\,\|c\times a\|\,,\text{ and }C:=\frac{1}{2}\,\|a\times b\|$$ are the areas of the triangle $OYZ$, $OZX$, and $OXY$, respectively. Here, $\times$ is the usual cross product and $\|\_\|$ is the Euclidean norm (induced by the standard inner product) of $\mathbb{R}^3$. Prove that the area of the triangle $XYZ$ is $$D:=\frac{1}{2}\,\big\|(b-a)\times (c-a)\big\|=\frac{1}{2}\,\|b\times c+c\times a+a\times b\|\,.$$ Finally, prove that $$b\times c\,,\,\,c\times a\,,\text{ and }a\times b$$ are mutually orthogonal (i.e., $a\parallel b\times c$, $b\parallel c\times a$, and $c\parallel a\times b$). It follows immediately that $$A^2+B^2+C^2=D^2\,,$$ since $\|p+q+r\|=\sqrt{p^2+q^2+r^2}$ for mutually orthogonal vectors $p,q,r\in\mathbb{R}^3$.


In general, let $n>1$ be an integer and consider the $n$-simplex $OX_1X_2\ldots X_n$ in $\mathbb{R}^{n}$, where $a_i:=OX_i$ for $i=1,2,\ldots,n$ are mutually orthogonal vectors in $\mathbb{R}^n$ with respect to its standard inner product. Suppose that $e_1,e_2,\ldots,e_n$ are standard basis vectors of $\mathbb{R}^n$. Identify the exterior power $\bigwedge^{n-1}\mathbb{R}^n$ as $\mathbb{R}^n$ via the identification $$e_1\wedge e_2 \wedge\ldots \wedge e_{i-1} \wedge e_{i+1} \wedge \ldots \wedge e_n = (-1)^{i+1}e_i\,.$$ (See this link for more detail.) This identification induces an isometric isomorphism $\bigwedge^{n-1}\mathbb{R}^n\cong\mathbb{R}^n$. From now on, we just say that $\bigwedge^{n-1}\mathbb{R}^n=\mathbb{R}^n$, via the identification above. Hence, $\bigwedge^{n-1}\mathbb{R}^n$ inherits the Euclidean norm $\|\_\|$ from $\mathbb{R}^n$.

For $i=1,2,\ldots,n$, the $(n-1)$-volume of the $(n-1)$-simplex $$S_i:=OX_1X_2\ldots X_{i-1}X_{i+1}\ldots X_n$$ is given by $$v_i:=\frac{1}{(n-1)!}\,\left\| A_i\right\|\,,$$ where $$A_i:=a_1\wedge a_2\wedge \ldots \wedge a_{i-1} \wedge a_{i+1} \wedge \ldots \wedge a_n\in{\bigwedge}{^{n-1}}\mathbb{R}^n=\mathbb{R}^n\,,$$ where $\wedge$ is the exterior product. It can be easily seen that the $(n-1)$-volume of the $(n-1)$-simplex $$S:=X_1X_2\ldots X_n$$ is equal to $$v:=\frac{1}{(n-1)!}\,\Big\|(a_2-a_1)\wedge (a_3-a_1)\wedge \ldots \wedge (a_n-a_1)\Big\|\,.$$ With some algebraic manipulations, we get $$v=\frac{1}{(n-1)!}\,\big\|A_1+A_2+\ldots+A_n\big\|\,.$$

As $A_1,A_2,\ldots,A_n$ are mutually orthogonal elements of ${\bigwedge}{^{n-1}}\mathbb{R}^n=\mathbb{R}^n$, we conclude that $$v^2=v_1^2+v_2^2+\ldots+v_n^2\,.$$ This result is known as the $n$-Dimensional Pythagorean Theorem. See also here.

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    $\begingroup$ As a remark, the identification $\bigwedge^2\mathbb{R}^3=\mathbb{R}^3$ in this answer makes the exterior product identical to the cross product. Recall that $e_2\wedge e_3=e_1$, $e_1\wedge e_3=-e_2$, and $e_1\wedge e_2=e_3$ under this identification. Therefore, the exterior product $\wedge$ (together with this particular identification $\bigwedge^2\mathbb{R}^3=\mathbb{R}^3$) is the same as the cross product $\times$ for $\mathbb{R}^3$. $\endgroup$ – Batominovski Oct 10 '18 at 22:02
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Let the sliced corner be at the origin and the fourth face lie in the plane $x/a + y/b + z/c = 1$. The foot of the altitude drawn to the fourth face is at $\lambda (1/a, 1/b, 1/c), \,\lambda = 1/(1/a^2 + 1/b^2 + 1/c^2)$. Therefore $$9 V^2 = A^2 a^2 = B^2 b^2 = C^2 c^2 = D^2 h^2, \\ A^2 + B^2 + C^2 = D^2 h^2 \left( \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \right) = D^2.$$

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