Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can you give examples of proofs without words? In particular, can you give examples of proofs without words for non-trivial results?

(One could ask if this is of interest to mathematicians, and I would say yes, in so far as the kind of little gems that usually fall under the title of 'proofs without words' is quite capable of providing the aesthetic rush we all so professionally appreciate. That is why we will sometimes stubbornly stare at one of these mathematical autostereograms with determination until we joyously see it.)

(I'll provide an answer as an example of what I have in mind in a second)

share|improve this question
1  
where possible could people also either note the image source or explain/provide a link to a "how to" for constructing the associated diagram? I think that such would also be helpful for folks` –  Carter Tazio Schonwald Dec 14 '09 at 23:57
3  
I hope I am not alone in being (usually) unable to appreciate "proof by picture"... –  Suvrit Jul 8 '11 at 21:14
5  
@Suvrit: I hope I am not alone in being most often unable to appreciate "proof by word" until I've read it at least twenty times and wrestled with it for many days per page! –  WetSavannaAnimal aka Rod Vance Jul 9 '11 at 12:11
1  
I am actually quite fond of this question, David! I tend to make comments on answers that are not relevant, and they have a tendency to get deleted after that. –  Mariano Suárez-Alvarez Sep 16 '11 at 17:34
24  
My opinion is that almost every proof-without-words is improved by a few well-chosen words. –  Joel David Hamkins Feb 12 '12 at 0:47

63 Answers 63

The area under a cycloid is three times the area of the generating circle. enter image description here

share|improve this answer

Here http://www.maa.org/sites/default/files/269122948517.pdf you can find Grace Lin's proof without Words that The Product of the Perimeter of a Triangle and Its Inradius Is Twice the Area of the Triangle (see the figure below)

                                    Grace Lin's proof

The proof originally appeared in the 1999 October issue of Mathematics Magazine.

share|improve this answer

Trig proof

A nice proof for trigonometric equation:

$sin^2(x)+cos^2(x)=1$

share|improve this answer

This is elementary as well, but one of my favorite ones :)

$1^2 + 2^2 + \dots + n^2 = \frac13n(n+1)(n+\frac12)$

(Author: Man-Keung Siu)

alt text

share|improve this answer
6  
There's an analogous proof that the integral of n^2 from 0 to x is x^3/3. It can be obtained from this proof by smoothing out the stepped pyramids into actual pyramids. –  Michael Lugo Dec 14 '09 at 16:47
18  
I think very few people have enough spatial imagination to figure out what happens exactly in the area where the three pieces come together, or could easily depict the structure seen from the opposite end. For me the picture is not convincing at all (I'd rather say the formula convinces me the picture is correct than the other way round). However maybe playing with an actual model would be quite convincing. –  Marc van Leeuwen Dec 12 '11 at 13:31
1  
@Mark - I think if you just think about the width of each step at each level, you will be able to see that they do all fit together. Just counting back along a given row or column shows you that it all fits. –  Steven Gubkin Feb 15 '12 at 15:10
1  
A variant of Mike's construction for $\sum_{k=1}^n k^2$, easier to visualize (I'm going to try a proof-without-words, without pictures). Take $6$ copies of each parallelepiped of size $k \times k \times 1$. Glue them together so as to make the four lateral walls of a parallelepiped of (external) size $k \times (k+1) \times (2k+1)$. Do this for k from 1 to n, forming a collection of bracelets. Insert each one in the next, like matrioskas, getting a whole parallelepiped of size $n\times(n+1)\times(2n+1)$. –  Pietro Majer Apr 10 '13 at 10:48

                          visual proof

This visual proof of $$\sum\limits_{n=1}^\infty \left (\frac{1}{2}\right)^{\,2n}=\frac{1}{3}$$ is from http://www.cecm.sfu.ca/~loki/Papers/Numbers/ (Visible Structures in Number Theory, by Peter Borwein and Loki Jorgenson, The American Mathematical Monthly, vol. 108, no. 5, 2002, pp. 897-910).

share|improve this answer

A proof of the identity $$1+2+\cdots + (n-1) = \binom{n}{2}$$

alt text

(Adapted from an entry I saw at Wolfram Demonstrations)

share|improve this answer
23  
Wow ! –  Dinakar Muthiah Dec 19 '09 at 22:56
12  
@Johann, people who thing that mathematics is about deducing theorems from axioms have such a mistaken idea of what the mathematical activity is thar their judgment is more or less irrelevant :D –  Mariano Suárez-Alvarez Jun 29 '10 at 13:05
29  
Am I the only one who doesn't understand this "proof" at all? –  mathreader Oct 17 '10 at 17:07
24  
@mathreader - the yellow dots are the sum of the first n numbers. Choosing two of the n+1 blue dots uniquely specifies a yellow dot in a bijective fashion. –  Steven Gubkin Nov 11 '10 at 13:40
25  
This beautiful proof warrants proper attribution. It was discovered by Loren Larson, professor emeritus at St. Olaf College. He included it along with a number of other, more standard, proofs, in "A Discrete Look at 1+2+...+n," published in 1985 in The College Mathematics Journal (vol. 16, no. 5, pp. 369-382). –  Barry Cipra Oct 15 '11 at 2:17

The cover of Peter Winkler's first book is a great proof without words of a statement which I'll leave you to guess, regarding the combinatorics of tiling a heaxagon with rhombi.

EDIT: I think the guessing game isn't helpful. The statement is that when tiling a perfect hexagon with the appropriate kind of rhombi of various orientations, the number of tiles in each orientation is the same. The image is slightly misleading in its use of color; there ought to be just three colors, corresponding to the three orientations.

share|improve this answer
2  
I'd be more impressed by this if I knew what statement was supposedly being proven by this illustration. That rhombus tilings are in 1-1 correspondence with 3d orthogonal surfaces (Thurston 1990, dx.doi.org/10.2307/2324578)? –  David Eppstein Dec 14 '09 at 23:06
1  
Also, there are equal numbers of rhombi of each orientation in any tiling, and in fact, any tiling can be obtained from any other one by rotating "unit" hexagons formed by three rhombi. –  Darsh Ranjan Dec 15 '09 at 2:35
1  
What do the colors represent? In particular, there are two colors for "upward-facing" rhombi (red and light gray) and two colors for "right-facing" rhombi (brown and dark gray), and I don't see why. –  Michael Lugo Dec 15 '09 at 3:03

A classic one, from the late 19th century, that surprized Peano's contemporaries.

Question : "A curve that fills a plane ? You must be kidding"

Answer :

enter image description here

Well, of course a formal proof was necessary, but it is still one of my favorites.

share|improve this answer
20  
How can you be sure that you're eventually covering all the points with irrational or transcendental coordinates? And giving a sequence of curves which fill more and more of the plane isn't the same as giving a single curve that does it all at once - it's not clear that such a limiting curve exists just looking at the pictures. –  Michael Burge Sep 14 '10 at 8:47
12  
Existence of the limits object is something that is very often forgotten. For example most Introductions to fractals give geometric descriptions of Koch's snowflake etc. via such an iteration but don't prove that there exists a limit of this iteration. –  Johannes Hahn Sep 14 '10 at 9:22
7  
Project: Fill the square one pixel at a time by following (an approximation to) this curve; then find some suitable baroque music accompaniment; then upload it to youtube. –  Michael Hardy Nov 16 '10 at 21:51
24  
If you look at the picture in detail you can see that you are defining a sequence of continuous functions that converge uniformly. It's also clear from the picture that the image is dense. Therefore the limiting function exists and its image (being dense and compact) is the whole square. Of course, this proof isn't 100% visual but the non-visual part -- the basic facts about uniform convergence and compactness -- can be regarded as background knowledge. So I think it's a nice example. –  gowers Apr 10 '11 at 20:18
2  
Remarkably, no picture nor mention to it was made in Peano's article, the construction being completely based on ternary expansions. The picture of a sequence converging to a square-filling curve appeared one year later in the paper by Hilbert. –  Pietro Majer Nov 17 '11 at 14:14

The sequence of pictures

intersection of 3 diangles intersection of 2 diangles intersection of 2 diangles intersection of 2 diangles

proves the area formula for spherical triangles $A=\hat{ABC}+\hat{BCA}+\hat{CAB}-\pi$.

share|improve this answer
3  
Thomas Harriot first proved this formula in 1603, apparently by a similar argument, though I have not seen his picture(s). –  John Stillwell Feb 22 '10 at 22:31
4  
Haha, I'm happy to see these illustrations useful to someone! I created them some years ago, mainly to crystalize what I saw in my minds eye after finding some simple proofs of this identity online. The words accompanying these images can be found at planetmath.org/encyclopedia/AreaOfASphericalTriangle.html Also, original MetaPost source can be obtained from this unfortunately obscure link: images.planetmath.org:8080/cache/objects/5841/src/sph-tri.mp –  Igor Khavkine Apr 26 '10 at 20:55
2  
There is an analogous proof using the fact that although the hyperbolic plane has infinite area, a triply asymptotic triangle has finite area, so once you pick one of the two triply asymptotic triangles containing your triangle, you're in business. The relevant picture's in my answer posted separately (I posted it before I had the reputation to leave comments): mathoverflow.net/questions/8846/proofs-without-words/… –  Vaughn Climenhaga May 18 '10 at 19:04

from Steven Strogatz's column: http://opinionator.blogs.nytimes.com/2010/04/04/take-it-to-the-limit/

pi

share|improve this answer
8  
Nice, but that reminds me of the "proof" of $2=\pi$ by approximating a straight line of length 2 by starting with a circle with this line as diameter, then two circles with one half of the line as diameter each, then for circles with on quarter of the line as diameter, ... One still has to find an argument that a geometric process converges at all and converges to the desired result. Both cannot be deduced purely from looking at a picture. –  Johannes Hahn Nov 8 '10 at 11:27
1  
Hmm, not sure, the point behind a proof by picture is that you do "get it," i.e., you see how the argument works in its full rigor. Now, either you do or you don't, but in this case I think it's all there. With circular arcs approximating a straight line you might notice upon observation that the arc length is independent of the iterations, which immediately discounts convergence... –  AndrewLMarshall Nov 10 '10 at 6:21
1  
By contrast, here you might observe that the difference between, say, how 2 circular wedges differ from their triangular counterparts in ratio, and how a wedge of twice the size differs from its triangular counterpart in ratio, does give on the order of geometric convergence. You can more or less just see that. –  AndrewLMarshall Nov 10 '10 at 6:21
4  
Wikipedia attributes this proof to Leonardo da Vinci. You can make establish rigorous convergence by using triangles that inscribe and circumscribe the wedges. –  S. Carnahan Nov 11 '10 at 3:04
1  
Hah, this is actually the proof appear in my primary school textbook. (I went to primary school in China, it was like 6th or 5th year) I'm amazed by this proof, but I'm not sure many kids can remember this though. –  temp May 30 '12 at 2:40

Another proof of the sum of the first $n$ squares, relying on the knowledge of the formula for the sum of the first $n$ numbers:

$$1^2 + 2^2 + \dots + n^2 = n(n+1)(2n+1)/6$$

This one has a similar flavor to the fabled proof by Gauss of the sum of the first $n$ numbers. It's a good follow up for students after Gauss's proof.

share|improve this answer
4  
+1 Superb. Is this original? If not, to whom is it attributed? –  I. J. Kennedy Nov 22 '13 at 2:05

This might be trivial but integration by parts has a nice proof without words:

enter image description here

(Got from: Roger B. Nelsen, Proof without Words: Integration by Parts, Mathematics Magazine, Vol. 64, No. 2 (Apr., 1991), p. 130; the original link is to http://www.math.ufl.edu/~mathguy/year/S10/int_by_parts.pdf)

share|improve this answer
1  
@Daniel, I've turned the PDF into a PNG, and inserted the relevant part. I did keep the URL to the PDF for reference. Thanks, by the way! –  Mariano Suárez-Alvarez Feb 7 '11 at 2:55
3  
The same picture also gives an interesting formula for the integral of an inverse function! –  Matt Noonan Jun 29 '11 at 0:57

I'm quite surprised no-one pointed out this one yet:

Theorem. The trefoil knot is knotted.

Proof.

3-colored trefoil know $\square$

Some comments: a 3-colouring of a knot diagram D is a choice of one of three colours for each arc D, such that at each crossing one sees either all three colours or one single colour. Every diagram admits at least three colourings, i.e. the constant ones. We'll call nontrivial every 3-colouring in which at least two colours (and therefore all three) actually show up. It's easy to see (one theorem, more pictures!) that Reidemeister moves preserve the property of having a nontrivial 3-colouring, and that the unknot doesn't have any nontrivial colouring.

The picture shows a (nontrivial) 3-colouring of the trefoil.

EDIT: I've made explicit what "nontrivial" meant ― see comments below. Since I'm here, let me also point out that the number of 3-colourings is independent of the diagram, and is itself a knot invariant. It also happens to be a power of 3, and is related to the fundamental group of the knot complement (see Justin Robert's Knot knotes if you're interested).

share|improve this answer
2  
The image isn't loading... –  Nate Eldredge Aug 16 '12 at 22:32

$$2 \pi > 6$$

hexagon inscribed in a circle

$ $ $ $ $ $ $ $

share|improve this answer
8  
And similarly one proves that $\pi < 4$ by inscribing a circle in a square. –  Michael Hardy Nov 16 '10 at 21:46
13  
At first I was thrown off by this, because I was looking at area and not circumference. The area of an inscribed regular 12-sided polygon in the unit circle is also 3. –  Todd Trimble Mar 12 '11 at 22:07

Means inequalities:

alt text

The image was sent to me by James M. Lawrence, grazie! See also page 53 of "Proofs without words: exercises in visual thinking, Volume 2" for a very different layout of the same 4 inequalities.

Another one exists involving the sum $$1^3 + 2^3 + \cdots + n^3:$$

alt text

The second image is due to Brian Sears

share|improve this answer
1  
I used the second proof (involving sum of cubes) in my class today after proving it by induction. A few were quite inspired by it! –  Somnath Basu Feb 24 '12 at 18:42
1  
2nd proof: It would be nicer if the small strips were above and to the left of the big square. –  Günter Rote Feb 25 '13 at 22:56

Duality between $\ell^1$ and $\ell^\infty$ norms.

p-norm 1 -> infinity

and the reverse animation

p-norm infinity -> 1

share|improve this answer
7  
I... don't quite get it. I think I need a few more words: What's the dot representing in each picture? –  Harrison Brown Dec 16 '09 at 15:01
9  
The red line in xy-space satisfies the given equation. The dot gives the (a,b) coordinates of the same line in ab-space. The xy- and ab-spaces are linearly dual to each other. The resulting black and red shapes represent the unit balls in respective norms. –  Igor Khavkine Dec 16 '09 at 15:34

I'm partial to the proof using Dandelin spheres that (certain) cross sections of cones are ellipses, where an ellipse is defined as the locus of points whose total distance to two foci is constant. It's particularly nice because it explains the foci geometrically, as well as the focus-directrix property with some more work.

Dandelin spheres touch the light blue plane that intersects the cone.

share|improve this answer
1  
Yes, this one is beautiful. –  Andres Caicedo May 15 '10 at 18:47
1  
@PatrickDaSilva: $PF1 = PP1$ because tangents to a circle/sphere have equal length. The total distance is thus equal to $PF1 + PF2 = PP1 + PP2 = P1P2$, which is constant. –  aorq Jul 29 '13 at 8:53

There's a picture proof in the Princeton Companion, or alternatively on p. 340 of Hatcher, of the fact that the higher homotopy groups are abelian. Actually, here's a screenshot of the one in Hatcher (hopefully fair-use!):

Hatcher p. 340

Here $f$ and $g$ are mappings (with basepoint) of $S^n$ into some space for $n > 1$; the picture shows a homotopy between $f + g$ and $g + f$.

The above diagrams show an application of the interchange law, a more general expression of the Eckmann-Hilton argument, for double categories or groupoids. Here is a more general picture

dbgxmod

which shows that the interchange law for a double groupoid implies the second rule $v^{-1}uv= u^{\delta v} $, where in the picture $a=\delta v$, for the crossed module associated to a double groupoid, taken from the book advertised here. There are many $2$-dimensional rewriting arguments which are essential to the results of this book.

share|improve this answer
5  
Page 340 of Hatcher's book: math.cornell.edu/~hatcher/AT/AT.pdf –  Dan Piponi Dec 14 '09 at 18:27
12  
This is sometimes called the Eckmann-Hilton argument: en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument –  Kevin H. Lin Dec 14 '09 at 20:46
3  
I've heard that term, but I've never quite understood how the diagram is supposed to prove the more general abstract nonsense theorem. But if you can explain it, that's what community wiki's for! :D –  Harrison Brown Dec 14 '09 at 20:53
4  
There are lots of places on the web where this is explained nicely: youtube.com/watch?v=Rjdo-RWQVIY , math.ucr.edu/home/baez/week258.html , ncatlab.org/nlab/show/Eckmann-Hilton+argument , etc.... –  Kevin H. Lin Dec 14 '09 at 23:50

The cardinality of the real number line is the same as a finite open interval of the real number line.

Proof without words

share|improve this answer
1  
I suppose this picture can also be adapted to obtain the stereographic projection proof that a sphere is a manifold? –  Kevin H. Lin Dec 14 '09 at 23:47
1  
I usually use Inkscape for my vector-based needs, but this was just done with my Smartboard presentation software. –  Jason Dyer Dec 15 '09 at 14:21

There are a couple of Fibonacci identities, I think. For example

$F_0^2+F_1^2+\cdots+F_n^2=F_{n}F_{n+1}$, with $F_0=1$.

By puting together squares of side $F_n$, one at a time, you get a rectangle of dimension $F_nF_{n+1}$: The two squares of side 1, then the square of side 2, then the square of side 3 and so on.

Here is an image I found online

fibonacci_rectangle

share|improve this answer
7  
fantastic ! –  Martin Brandenburg Apr 17 '10 at 23:30
2  
@Max: The inductive step is easy to figure out, since the rectangle above contains the rectangles from previous steps. –  Daniel Litt Mar 16 '11 at 20:01

Because I think proof by picture is potentially dangerous, I'll present a link to the standard proof that 32.5 = 31.5:

"proof" that 32.5 = 31.5

share|improve this answer
54  
I think it is just as easy to introduce some kind of logical gap in a written proof as in a graphical one. –  Steven Gubkin Mar 7 '10 at 23:41
38  
@Steven: I think there is some truth to your claim, but I don't agree fully. First, we may notice that most proofs rely much more on writing than on pictures, and so mathematicians have developed a better radar for "written gaps". Second, there is a very strong sense in which written proofs may be formalized and checked by computer. Picture proofs, unless they share quite a bit of the "discrete" character of written proofs, usually are not amenable to such treatment. (And the notions of discreteness I can think of pretty much ensure that the picture proof could be turned into words.) –  Pietro KC May 15 '10 at 20:22
24  
+1 for "the standard proof that $32.5 = 31.5$." Made me laugh. :) –  Quadrescence Oct 15 '10 at 20:10
10  
@Pietro: “there is a very strong sense in which written proofs may be formalised”? Formalisation is a highly non-trivial task, and typically depends on quite a lot of mathematical background. What affects the difficulty is not whether the proof is written or graphical, but whether it’s detailed or highly abstracted. Formalising a good proof-by-picture is no harder than formalising a high-level written proof. Insofar as there’s a difference, I’d say it’s just that written proofs can be made detailed enough that formalising them is straightforward, whereas picture proofs perhaps can’t. –  Peter LeFanu Lumsdaine Nov 29 '10 at 1:04

Sphere eversion


And here's a two-dimensional rendering of the sphere eversion: enter image description here

share|improve this answer
6  
As pretty as it is, that is nowhere understandable as a proof. More as an illustration. –  Willie Wong Mar 11 '10 at 16:44
12  
@Willie: Suppose someone wrote down the equations/formulas for the sphere eversion in that video. It seems to me that checking that the formulas indeed give a sphere eversion would be a rather difficult and tedious task, whereas a video animation is, although not a rigorous proof, much more immediately convincing. –  Kevin H. Lin Apr 6 '10 at 16:30
2  
I just watched the video, which was excellent, but it had a lot of words in it. –  Patricia Hersh Aug 19 '12 at 0:23

I am surprised that no one had cited the "proof" that the rationals are countable yet. See, for example, this picture

enter image description here

share|improve this answer
3  
I think that the fact that the rationals are countable qualifies as non-trivial, when put in historical perspective –  Geoff Robinson Mar 19 at 19:02

This other answer shows that an 8x8 board with opposite squares removed cannot be tiled with dominoes, as they are of the same "colour". But what if two squares of opposite colours are removed? Ralph E. Gomory showed that it is always possible, no matter where the two removed squares are, and this is his proof.

Image of board

(Imagine A and B are the squares removed.) The image is from Honsberger's Mathematical Gems I.

share|improve this answer

Algebraic manipulations in monoidal categories can also be performed in a graphical calculus. And the best part is that this is completely rigorous: a statement holds in the graphical language if and only if it holds (in the algebraic formulation). See for example Peter Selinger's "A survey of graphical languages for monoidal categories". There are many instances, for example in knot theory studied via braided categories. The following specific example comes from Joachim Kock's book "Frobenius Algebras and 2D Topological Quantum Field Theories", and proves that the comultiplication of a Frobenius algebra is cocommutative if and only if the multiplication is commutative.

enter image description here

share|improve this answer
3  
this proof makes me wonder what is a 'picture' and what is –  Jeremy Lane Apr 19 at 14:43

Also elementary, but here is a proof that

$C_n = \binom{2n}{n} - \binom{2n}{n+1} = \frac{\binom{2n}{n}}{n+1},$

where $C_n$ is the $n$th Catalan number.

http://utdallas.edu/~hagge/images/Catalan.pdf

Sorry for the link; new users may not use image tags.

Here's the image:

alt text

share|improve this answer
4  
Do you have an explanation for the picture? I looked at it, and looked at it, and don't get it. –  Willie Wong Mar 11 '10 at 16:38
2  
Sorry for not noticing your question (much) earlier. The differences between adjacent terms in Pascal's triangle form another triangle which obeys the same generation rules. In my picture of that triangle, the yellow squares count some of the downward paths on a square grid which has been rotated $45^\circ$, namely those that never fall to the left of the top square. One definition of $C_n$ is that it is the number of such paths which terminate at the bottom corner of an $n \times n$ grid. –  Tobias Hagge Oct 26 '10 at 5:46

This should really be a comment on Marco Radeschi's answer from Feb 22 involving the area formula for spherical triangles, but since I'm new here I don't have the reputation to leave comments yet.

In reply to Igor's comment (on Marco's answer) wondering about an analogous proof for the area formula of hyperbolic triangles: there is one along similar lines, and you're rescued from non-compactness by the fact that asymptotic triangles have finite area. In particular, the proof in the spherical case relies on the fact that the area of a double wedge with angle $\alpha$ is proportional to $\alpha$; in the hyperbolic case, you need to replace the double wedge with a doubly asymptotic triangle (one vertex in the hyperbolic plane and two vertices on the ideal boundary) and show that if the angle at the finite vertex is $\alpha$, then the area is proportional to $\pi - \alpha$. That follows from similar arguments to those in the spherical case (show that the area function depends affinely on $\alpha$ and use what you know about the cases $\alpha=0,\pi$).

Once you have that, then everything follows from the picture below, since you know the area of the triply asymptotic triangle and of the three (yellow, red, blue) doubly asymptotic triangles.

alt text

(That picture is slightly modified from p. 221 of this book, which has the whole proof in more detail.)

share|improve this answer

It's a long list of wonderful answers already, but I can't resist...

Question: Is it possible to find six points on a square lattice that form the vertices of a regular hexagon?

Proof without words:

alt text

Hint: A square lattice is invariant under rotation by π/2 around any lattice point. Use reductio ad absurdum.

Credit: I learned that proof from György Elekes during the Conjecture and Proof course in the Budapest Semesters in Mathematics, after constructing a proof of my own that used entirely too many words and made very laboured use of the fact that $\sqrt{3}$ is irrational. The picture here is my own creation (using Asymptote).

Follow-up: Can you find four points on a hexagonal lattice that form the vertices of a square? The proof is similar but not immediate.

share|improve this answer
4  
Why would you resist? –  Mariano Suárez-Alvarez May 20 '10 at 17:41
4  
+1 for the "Conjecture & Proof" shout-out. Best, course, ever! –  Kevin O'Bryant Nov 10 '10 at 23:18
1  
Igen, nagyon jó. –  Douglas Zare Feb 7 '11 at 5:08

This is a "proof without words" by an equation, not a picture.

Three complex numbers $a,b,c$ in the complex plane form the vertices of an equilateral triangle if and only if $~a^2 + b^2 + c^2 = ab + bc + ca$:

$$ $$

$$ \hspace{-3in} 2 |a^2 + b^2 + c^2 - ab - bc - ca|^2 $$ $$ = ( |a-b|^2 - |b-c|^2)^2 + ( |b-c|^2 - |c-a|^2)^2 + ( |c-a|^2 - |a-b|^2)^2 . $$

share|improve this answer

protected by Scott Morrison Oct 11 '13 at 0:51

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.