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Loomis famously wrote hundreds of proofs of Pythagoras' Theorem (reference below), but these are all basically proofs "from below". Today on Twitter @panlepan mentioned Carnot's theorem which has Pythagoras' Theorem as a special case and this got me thinking. Which theorems have Pythagoras' Theorem as a special case?

@BOOK{Loomis1968,
  author =       {Loomis, E.},
  title =        {The {Pythagorean} Proposition},
  publisher =    {National Council of Teachers of Mathematics},
  year =         {1968},
  address =      {Washington},
}
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    $\begingroup$ Moreover, saying that Pithagoras is a special case of Theorem X means that it is not used (not even indirectly) in the proof of $X$. $\endgroup$ May 5 at 12:31
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    $\begingroup$ @FrancescoPolizzi Why? To me it just means that the statement of Theorem X becomes Pithagoras theorem's statement if you strengthen the assumptions. $\endgroup$ May 5 at 12:44
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    $\begingroup$ @AlessandroDellaCorte: It is a matter of definitions. What is a "special case"? Maybe you are right and yours is the common meaning. However, when I think of Theorem Y as a "special case" of Theorem X, I imagine that I first prove X and then, specializing the assumptions, I get Y. But If I strongly used Y in order to prove X, it seems strange to me to call it a "special case". In fact, in this situation X and Y are equivalent statements. $\endgroup$ May 5 at 12:52
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    $\begingroup$ I think one has to be careful describing statements as equivalent—of course, any two true (and any two false!) statements are equivalent. One often "knows what it means" to say that two statements are equivalent, or that one implies the other, but in the pure logical sense I think that there's no rigorous ground for distinguishing these two cases. (Of course more interesting is, perhaps like @Dirk's answer, a property of a class of spaces that may or may not hold, and that holds exactly when Pythagoras does … in whatever sense that should be interpreted.) $\endgroup$
    – LSpice
    May 5 at 13:42
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    $\begingroup$ I assumed by Carnot's theorem you meant the thermodynamical one and was interested to see how that implied Pythagoras' theorem. I can see how such a thing might happen, but it turns out there is a different Carnot's theorem, named after the father of the thermo guy. $\endgroup$ May 5 at 14:02

16 Answers 16

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The Law of cosines is the first that comes to my mind:

$$c^{2}=a^{2}+b^{2}-2ab\cos \gamma$$

enter image description here
(source: Wikipedia)

If $\gamma$ is a right angle, its cosine is 0 and all that remains is Pythagoras' Theorem.

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    $\begingroup$ But Trigonometry is based on the Pithagorean rule $\cos^2 x + \sin ^2 x=1$. So, saying that Carnot implies Pithagoras seems to me a sort of circular argument. $\endgroup$ May 5 at 12:27
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    $\begingroup$ @FrancescoPolizzi Well, there are multiple reasons one may want a theorem with another as a special case. If the reason is to look for a more fundamental proof then your objection is significant, but if the reason is because there is a problem where the original theorem almost applies but doesn't quite, then the law of cosines is a great choice. $\endgroup$
    – Will Sawin
    May 5 at 12:54
  • $\begingroup$ @WillSawin: yes, of course. My objection was mainly of foundational nature. $\endgroup$ May 5 at 12:56
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    $\begingroup$ @StevenLandsburg me neither, but that wasn't posted when I wrote this. And I learned about the Law of cosines in (Dutch) middle/high school but the dot product and vector spaces are only taught in the first year of university mathematics. $\endgroup$
    – Glorfindel
    May 5 at 19:45
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    $\begingroup$ @FrancescoPolizzi. "Circular argument" haha $\endgroup$
    – wnx
    May 6 at 11:37
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Parseval identities in the theory of Fourier series and integrals.

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The parallelogram law says that if $\mathcal{V}$ is an inner product space and $\mathbf{v},\mathbf{w} \in \mathcal{V}$, then $$ 2\|\mathbf{v}\|^2 + 2\|\mathbf{w}\|^2 = \|\mathbf{v} + \mathbf{w}\|^2 + \|\mathbf{v} - \mathbf{w}\|^2, $$ where the norm here is the one induced by the inner product (in fact, a result of John von Neumann says that a norm is induced by an inner product if and only if the above parallelogram law holds). Its name comes from the fact that, if $\mathcal{V} = \mathbb{R}^2$ then the equality above simply tells us something about the sides lengths of a parallelogram compared to the lengths of its diagonals.

In the special case when $\langle \mathbf{v},\mathbf{w}\rangle = 0$, we have $\|\mathbf{v} + \mathbf{w}\|^2 = \|\mathbf{v} - \mathbf{w}\|^2$, so the parallelogram law simplifies to $$ \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 = \|\mathbf{v} + \mathbf{w}\|^2, $$ which is the Pythagorean theorem (if $\mathcal{V} = \mathbb{R}^2$).

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    $\begingroup$ More generally we can consider any of the polarization identities. $\endgroup$
    – flawr
    May 6 at 13:04
  • $\begingroup$ Yes. I was looking for this. This is this Linear Algebra 101. $\endgroup$
    – BCLC
    May 6 at 16:51
  • $\begingroup$ Same spirit as Alessandro's answer. $\endgroup$
    – wlad
    May 8 at 7:28
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The Pythagorean theorem is a limit of the general formula for spherical or hyperbolic space: $\cos(a\sqrt{\kappa})\cos(b\sqrt{\kappa})=\cos(c\sqrt{\kappa})$, where $\kappa$ is the curvature.

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So far no one has mentioned the original generalization!

Early in Euclid's Elements, the Pythagorean theorem is stated by comparing square areas:

Book I, Proposition 47: In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.

enter image description here

Later on, the Elements generalizes this far beyond squares. First it defines similar figures, meaning any similar figures with straight line segments:

Book VI, Definition 1: Similar rectilinear figures are such as have their angles severally equal and the sides about the equal angles proportional.

So the generalization that appears in the Elements is the Pythageorean theorem for any similar figures, as known some 2500 years ago:

Book VI, Proposition 31: In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle.

enter image description here

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    $\begingroup$ Perhaps the most interesting choice for such shapes - albeit not for the present question - is triangles similar (or on the hypotenuse, congruent) to the original, because these provide a proof of the Pythagorean theorem, as the sum-of-areas results is otherwise geometrically obvious. $\endgroup$
    – J.G.
    May 6 at 21:56
  • $\begingroup$ Yes, indeed, and Polya makes a lot of this in his "Induction and Analogy" book. $\endgroup$ May 7 at 8:53
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This is probably the simplest: $(a+b)^2=a^2+b^2+2ab$, if you take $a,b$ elements of some inner product vector space and $(\cdot)^2$ means inner product with itself.

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  • $\begingroup$ Same spirit as Nathaniel's answer $\endgroup$
    – wlad
    May 8 at 8:31
  • $\begingroup$ @wlad yes, although not exactly the same: there are inner spaces in which the parallelogram law doesn’t hold. I don’t remember which answer was posted first, btw, but below there is an answer on the law if cosine which is closer to this one. Also there, however, I’d say that there is some difference, as this does not rely on Euclidean geometry and can be deduced only by the axioms verified by any inner product vector space. $\endgroup$ May 8 at 9:13
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One of the most attractive generalisations is de Gua's theorem: If a tetrahedron $ABCD$ is rectangular at $A$, then $$ |BCD|^2=|ABC|^2+|ACD|^2+|ABD|^2$$ where the absolute value signs denote area.

This can easily be proved analytically by the $p,q$ method. One assumes,as one can, that the vertices are $(0,0,0)$, $(1,0,0)$, $(0,q,0)$ and $(0,0,t)$--the calculations are then done in a matter of minutes.

Given the objection below, here is the hierarchy of results: The square of the measure of a measurable set in $m$-dimensional affine subspace of euclidean $n$-space is the sum of the squares of the measures of the $n \choose m$ orthogonal projections on all the $m$-dimensional coordinate subspaces of $n$-space with respect to a fixed ONB basis of the latter (Conant and Beyer).

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    $\begingroup$ How is Pythagoras a special case of this? $\endgroup$
    – Wojowu
    May 5 at 15:05
  • $\begingroup$ @Wojowu, re, couldn't one shrink one of the faces to $0$? $\endgroup$
    – LSpice
    May 5 at 15:35
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    $\begingroup$ @NoahSchweber, you are right. I also don't understand what I had in mind. $\endgroup$
    – LSpice
    May 6 at 3:35
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    $\begingroup$ @Wojowu - Assuming one vertex is at the origin and the others on the axes, dividing by the non-zero coordinate of one vertex and taking the limit as it goes out to infinity - that would yield PT. $\endgroup$ May 8 at 12:38
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    $\begingroup$ It is a generalisation in that it states for 3 dimensions the (!) equivalent of what Pythagoras states for 2 dimensions. $\endgroup$
    – Wolfgang
    May 10 at 20:25
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Pythagoras' theorem is a special case of the three point identity for Bregman distances: Let $h$ be convex and lower semi-continuous on a Banach space - further assume differentiability of $h$ for simplicity. Then $h$ induces a Bregman distance $$D_h(u,v) = h(u)-h(v) - \langle \nabla h(v), u-v\rangle $$ and this Bregman distance fulfills the mentioned three point equality $$\langle \nabla h(u) - \nabla h(v),v-w\rangle = D_h(w,u) + D_h(w,v) + D_h(v,u).$$ The special case appears for $h(u) = \tfrac12\|u\|^2$ on a Hilbert space $X$ (or on $X = \mathbb{R}^2$). Here we get $\nabla h(u) = u$ and $D_h(u,v) = \tfrac12\|x-y\|^2$ and you get Pythagoras.

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  • $\begingroup$ I think you have some terms on the wrong side of the equations? Looking at my own proof and e.g. math.stackexchange.com/questions/3668286/… or math.stackexchange.com/questions/2766399/… $\endgroup$
    – usul
    May 7 at 5:03
  • $\begingroup$ More specifically, I believe the "generic" Pythagorean special case is that when the vector $w-v$ is orthogonal to $\nabla h(u) - \nabla h(v)$, we should get $D_h(w,v)+D_h(v,u)=D_h(w,u)$. Choosing $h(u) = \tfrac{1}{2}\|u\|^2$, and in particular choosing $v = $ the origin, gets us to the most basic Pythagoras as the condition collapses to $w \perp u$. $\endgroup$
    – usul
    May 7 at 5:18
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$$1=\cos^2 x+\sin^2x$$ (which can be proven without using Pythagoras) holds for arbitrary $x$ in $\mathbb C$ and yields Pythagoras for real $x$.

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    $\begingroup$ This is because in $a^2+b^2=c^2$ you divide both sides by $c^2$ ? $\endgroup$
    – BCLC
    May 6 at 16:53
  • $\begingroup$ Wait ...is pythagoras' theorem perhaps equivalent to this? $\endgroup$
    – BCLC
    May 8 at 10:05
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    $\begingroup$ @BCLC yes I think so because this is the special case when triangle's hypotenuse=1, you just scale up the triangle by whatever hypotenuse length you'd like after that. $\endgroup$ May 8 at 16:34
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    $\begingroup$ I don't want to turn this into an answer, but it's also a special case of the angle sum formulas from trigonometry. Special case of angle sum formulas (derivable visually and using linear algebra): $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$. now just set $y=-x$ to get the above identity in this answer and you're done, since $\sin^2(x)+\cos^2(x)=1$. $\endgroup$ May 8 at 16:39
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The discrete form of the parallel axes theorem for the second moment of area for $\,n\,$ points $\,A_k\,$ with centroid $\,G\,$ and an arbitrary point $\,P\,$ is $\,\sum_{k=1}^n PA_k^2 = n \cdot PG^2 + \sum_{k=1}^n GA_k^2 \;\;(*)\,$.

Let $\,\triangle ABC\,$ be a right triangle and $\,O\,$ the midpoint of hypotenuse $\,BC\,$, known to also be its circumcenter, so $\,OA = OB = OC = \frac{1}{2} BC\,$. Then$\,(*)\,$ for $\,A_1A_2A_3 \equiv BOC\,$, $\,G\equiv O\,$ and $\,P \equiv A\,$ reduces to Pythagoras' theorem:

$$ \require{cancel} \begin{align} AB^2+AO^2+AC^2 = 3 \, AO^2 + OB^2+\bcancel{0}+OC^2 \iff AB^2+AC^2 &=4 AO^2 = BC^2 \end{align} $$

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    $\begingroup$ Oh wow did you just use a physics theorem to prove a mathematics theorem? $\endgroup$
    – BCLC
    May 6 at 16:52
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    $\begingroup$ @BCLC The special case used here is equivalent to the median length (Apollonius') theorem. The general case can also be framed as a quadratic minimization problem. But it felt more in the spirit of the question to root it into a physics theorem. $\endgroup$
    – dxiv
    May 6 at 21:59
  • $\begingroup$ dxiv translation: i'm too humble/modest to admit I'm the only answerer here to use a theorem outside maths to prove a maths theorem. / ok but re your spirit of the question i don't get what's in the tweet that makes it physics? $\endgroup$
    – BCLC
    May 6 at 22:35
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    $\begingroup$ @BCLC Not so much about physics, but the question seems to invite answers that have a touch of "unexpected" in them. $\endgroup$
    – dxiv
    May 6 at 23:18
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I like to think about Pythagoras theorem as a corollary/special case of the following theorem:

Theorem: Let $X$ be a finite dimensional real Banach space such that the group of linear isometries (that is, isometries which fix $0$) of $X$ acts transitively on the unit sphere of $X$. Then $X$ is a Hilbert space.

That is, having enough rotations forces the norm to come from an inner product, and from this Pythagoras theorem follows

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    $\begingroup$ Fun fact: it is open whether this holds for separable Banach spaces (but it is known that it fails for nonseparable ones) $\endgroup$ May 8 at 7:10
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The Binet-Cauchy formula says that if $A$ and $B$ are a $n\times m$ and $B$ a $m\times n$ real matrices, respectively, and for $s\subseteq\{1,\ldots ,m\}$ with $|s|=n$ we denote by $A_s$ the $n\times n$ submatrix of $A$ obtained deleting the columns not in $s$, and by $B^s$ the $n\times n$ submatrix of $B$ obtained deleting the rows not in $s$, then

$$\det(AB)=\sum_{\begin{array}{c} s\subseteq\{1,\ldots ,m\}\\ |s=n|\end{array}} \det(A_s)\det(B^s)$$

In the case $B = A^T$, the transpose of $A$, since $B_s = (A^T)_s=(A_s)^T$ the formula gives

$$\det(AA^T) = \sum_{\begin{array}{c} s\subseteq\{1,\ldots ,m\}\\ |s=n|\end{array}} \det(A_s)\det((A_s)^T)=\sum_{\begin{array}{c} s\subseteq\{1,\ldots ,m\}\\ |s=n|\end{array}} (\det(A_s))^2$$

Since the parallelotope in $\mathbb{R}^n$ generated by the $n$ row vectors of $A$ has measure $\sqrt{\det(AA^T)}$, then the formula says that the square of the $n$-dimensional measure of an $n$-dimensional parallelotope equals the sum of the squares of the measures of its projections onto all possible $n$-dimensional coordinate hyperplanes. If $n = 1$ this reduces to the Pythagorean theorem.

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In a less “higher” maths fashion : This Numberphile video somewhat says that Pythagoras theorem is a special case of Ptolemy’s theorem which is a more general view of properties of a cyclic quadrilateral. But navigating between the 2 always seems some sort of tautology to me …

There is also Casey’s theorem which reduces to Ptolemy’s (so which could reduce to Pythagoras’).

Quoting the Ptolemy’s theorem Wikipedia article :

More generally, if the quadrilateral is a rectangle with sides a and b and diagonal d then Ptolemy's theorem reduces to the Pythagorean theorem. In this case the center of the circle coincides with the point of intersection of the diagonals. The product of the diagonals is then d2, the right hand side of Ptolemy's relation is the sum a2 + b2.

Later on, it is cited as a corollary theorem.

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There’s an “n-dimensional Pythagorean theorem” https://billcookmath.com/papers/2012-06_nD_pythag.pdf, saying that the square of the $k$-dimensional area of a $k$-dimensional parallelogram $P$ in $n$-space is equal to the sum of the squares of the $k$-dimensional areas of the projections of $P$ onto all $k$-dimensional planes spanned by the coordinate axes.

This theorem is in particular relevant when introducing differential forms; I learned about it from this MSE question.

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There is a generalisation of Pythagoras to the case of a non-right angled triangle. Inscribe an isosceles triangle $\triangle ADE$ in $\triangle ABC$ so that $[DE]\subseteq[BC]$ and $|\angle BAC|=|\angle ADC|=|\angle AEB|$. Set $r=|BE|$ and $s=|DC|$. (See a picture here.) Then by observing that $\triangle ABC$, $\triangle DAC$ and $\triangle EBA$ are similar, we get $\frac{a}{b}=\frac{b}{s}$ and $\frac{a}{c}=\frac{c}{r}$, which yields $$b^2+c^2=a(r+s).$$

If $\angle BAC$ is acute, then $[BE]$ and $[DC]$ overlap, so that $r+s>a$, while if $\angle BAC$ is obtuse we have $r+s<a$. And of course if $\angle BAC$ is a right angle, we get $r+s=a$ and recover Pythagoras' Theorem.

Wikipedia attributes this to Thābit ibn Qurra.

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The law of total variance says that if $X,Y$ are real-valued random variables and $\operatorname E(X^2)<+\infty,$ then $$ \operatorname{var}(X) = \overbrace{\operatorname{var}(\operatorname E(X\mid Y))}^{\begin{smallmatrix} \text{explained} \\ \text{component} \\ \text{of the variance} \end{smallmatrix}} {} + {} \overbrace{ \operatorname E(\operatorname{var}(X\mid Y)) }^{\begin{smallmatrix} \text{unexplained} \\ \text{component} \\ \text{of the variance} \end{smallmatrix}} $$

A more concrete version is that in the analysis of variance, the sum of squares of residuals ("residuals" are not to be confused with "errors") plus the sum of squares due to regression equals the total corrected sum of squares (that last being the sum of squares of deviations from the sample mean).

Most statistics texts tell you that the proportion of the total variance that is "explained" is the square of the correlation between $X$ and $Y.$ (For that you need to assume $Y$ also has a finite second moment. But the validity of the identity above does not depend on $Y$ being real-valued at all, and in cases where $Y$ takes values in a set with no structure or in $\mathbb R^n$ or something else, it is still standard to call the explained proportion of the total variance $\text{“}R^2\text{”},$ even though in such cases there is no quantity called $R.$) Some textbooks go on to say that that is also the square of the cosine of an angle (one of the angles of the right triangle that is involved), and then deduce certain bounds on the correlation between $X$ and $Z$ given those between $X$ and $Y$ and between $Y$ and $Z.$

A further generalization is Brillinger's law of total cumulance, of which the following is the case $n=4{:}$

\begin{align} \text{joint cumulant} = {} & \kappa(X_1,X_2,X_3,X_4) \\[8pt] = {} & \kappa(\kappa(X_1,X_2,X_3,X_4\mid Y)) \\[6pt] & {} + {} \underbrace{ \kappa(\kappa(X_1,X_2,X_3\mid Y), \kappa(X_4\mid Y)) + \cdots}_\text{4 terms} \\[2pt] & {} + {} \underbrace{ \kappa(\kappa(X_1,X_2\mid Y),\kappa(X_3,X_4\mid Y)) + \cdots }_\text{3 terms} \\[6pt] & {} + {} \underbrace{\kappa(\kappa(X_1,X_2\mid Y),\kappa(X_3\mid Y), \kappa(X_4\mid Y)) + \cdots}_\text{6 terms} \\[6pt] & {} + \kappa(\kappa(X_1\mid Y),\kappa(X_2\mid Y), \kappa(X_3\mid Y), \kappa(X_4\mid Y)) \end{align} where the sum is over the set of all partitions of the set $\{X_1,\ldots,X_n\}$ of random variables.

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