# Show that $A^2 +B^2+C^2=D^2$ using the following diagram (tetrahedron)

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 $$D$$.

$$A ^2 + B^ 2 + C^ 2 = D^2$$ . Can anyone explain to me what formula/how to go about doing this question? thank you.

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!

• Where does the formula in the third row above for SΔxyz come from? Dec 29, 2018 at 20:26
• @twa14 It's the Heron's formula. See here: en.wikipedia.org/wiki/Heron%27s_formula Dec 30, 2018 at 1:52

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.

• 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 $\mathbb{R}^3$. Oct 10, 2018 at 22:02

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 $$9 V^2 = A^2 a^2 = B^2 b^2 = C^2 c^2 = D^2 h^2, \\ A^2 + B^2 + C^2 = D^2 h^2 \left( \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \right) = D^2.$$