# Analogues and Generalizations of the Pythagorean Theorem

Pythagorean Theorem is one of the most fundamental results of Mathematics. Using the theorem we define what's known as euclidean distance dist_{2}. This notion of distance extends to spaces with scalar product - Hilbert spaces.

Proofs ##13, 17, and 18 gave us plane generalizations of the theorem. Below I consider an *analogue* that holds in the $3$-dimensional space $\mathbb{R}^{3}.$ The statement leads to the definition of euclidean distance in $\mathbb{R}^{3}.$ Afterwards, there is an additional and unexpected analog of the theorem in $\mathbb{R}^{3}.$

It's convenient to think of the Pythagorean Theorem as defining the length of the diagonal in a rectangle when its two sides are given. Now consider a parallelepiped with sides $a,$ $b,$ and $c.$ Incidently, the diagonal in question serves as the hypotenuse of the right triangle formed by the edge $c$ and the diagonal of the face $ab.$ The latter, by the Pythagorean Theorem, equals $\sqrt{a^{2}+b^{2}}.$ Applying it the second time gives the length of the diagonal as $\sqrt{a^{2}+b^{2}+c^{2}}.$

When we moved from a $2$-dimensional space to a $3$-dimensional space the formula for the diagonal of a shape built on orthogonal segments remained virtually the same except the number of terms grew from $2$ to $3,$ as appropriate. However, in both cases squared were line segments. Pythagorean Theorem has an analog where squared are areas of triangles.

The theorem applies to a special kind of tetrahedra in which all three edges emanating from one of the vertices are perpendicular to each other. One can obtain such a pyramid by cutting a corner from a parallelepiped. Let's introduce the areas $A,B,C$ of the faces that house right angles, and let $D$ be the area of the remaining face. We have

$A = qr/2,$ $B = rp/2,$ $C = pq/2.$

What I want to show is that $A^{2} + B^{2} + C^{2} = D^{2}.$

Draw a plane through $p$ perpendicular to $a.$ Then both $k$ and $h$ will be perpendicular to $a.$ We find that $D = ha/2,$ while $h^{2} = k^{2} + p^{2}.$ Also $A = ak/2.$ Therefore,

$\begin{align} 4D^{2} &= a^{2}h^{2}\\ &= a^{2}(k^{2} + p^{2})\\ &= 4A^{2} + a^{2}p^{2}\\ &= 4A^{2} + (r^{2} + q^{2})p^{2}\\ &= 4A^{2} + (rp)^{2} + (pq)^{2}\\ &= 4A^{2} + 4B^{2} + 4C^{2} \end{align}$

Q.E.D.

Oops, I almost forgot the Cosine Law which is a clear generalization of the Pythagorean Theorem. For a triangle with sides $a,$ $b,$ and $c$ and angle $C$ opposite the side $c,$ one has

$c^{2} = a^{2} + b^{2} - 2ab\cdot \cos(C)$

which, in turn, admits a generalization to higher dimensional spaces.

The fact expressed nowadays by the single identity appears in Euclid's *Elements* as two separate propositions: II.12 for obtuse-angled triangles and
II.13 for the acute ones.

Dr. Scott Brodie from the Mount Sinai School of Medicine, NY, sent me a proof of the theorem and a dynamic Geometer's SketchPad illustration.

### References

- G. Polya,
*Mathematical Discovery*, John Wiley and Sons, 1981.

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