Proofs without words

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[+383] [79]
Mariano Suárez-Álvarez

[2009-12-14 06:04:14]

[
reference-request
big-list
examples
intuition
alternative-proof
]

[ https://mathoverflow.net/questions/8846/proofs-without-words ]

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
^{[1]}, 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
^{[2]} with determination until we joyously *see* it.)

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

[1] https://mathoverflow.net/faq#whatquestions

[2] https://en.wikipedia.org/wiki/Autostereogram

[2] https://en.wikipedia.org/wiki/Autostereogram

[+540]
[2009-12-14 06:05:09]
Mariano Suárez-Álvarez

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

(Adapted from an entry I saw at
Wolfram Demonstrations
^{[1]}, see also the
original faster animation
^{[2]})

This proof 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, DOI:
10.1080/07468342.1985.11972910
^{[3]},
JSTOR
^{[4]}).

[2] https://i.imgur.com/HCfGOYp.gif

[3] https://doi.org/10.1080/07468342.1985.11972910

[4] https://www.jstor.org/stable/2686996

(3)
Fantastic! I'm reminded of a question I wrote for a high-school math competition way back when, which hinged on a similar diagram that counts solutions to a + b + c = n in nonnegative integers. (Imagine a third, horizontal, line.) The problem was something like to count the solutions to a + b + c = 100 with a, b, c bounded above by some numbers close to 100; I still don't know an elegant way to solve the problem other than to see it from the triangle. - ** Harrison Brown**

This one is just awesome! - ** Jan Weidner**

(4)
This is a beautiful proof. I like it very much. But I wonder what status proofs such as this one have in the eyes of people who think that mathematics is about deducing theorems from axioms. If you try to reduce such things to axioms the beauty vanishes. - ** Johann Cigler**

(11)
I gave this result +1, despite the fact that I am a formalist and do not regard this single image as a proof, because I think it is a very good proof **virus** to which experienced mathematicians are susceptible. - ** Niel de Beaudrap**

(28)
@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-Álvarez**

(16)
@Johann: some days of the week, I am such a person; and from that point of view, the picture is a beautifully clear encoding of a certain bijection, and the formal construction of the bijection itself is a very beautiful proof. No beauty is destroyed!// I strongly believe that a proof with a clear intuition should *also* be clear as a formal proof. If not, either (usually) our formalism isn't as good as it could be, or (occasionally) our intuition really is overlooking some non-trivial subtleties. - ** Peter LeFanu Lumsdaine**

(1)
That's beautiful! - ** Jim Conant**

(77)
Am I the only one who doesn't understand this "proof" at all? - ** mathreader**

(56)
@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**

(6)
Steve: there are n blue dots. :) - ** Adam Hughes**

(1)
@Mariano: Then help me... what would be a nonmistaken idea of the mathematical activity? (so far I have no other way of seen what math is...) - ** David Fernandez-Breton**

Damn, the picture's link won't load. Can anyone find another? - ** Elizabeth S. Q. Goodman**

(8)
@Adam: Only if $n=7$ - ** kangdon**

(58)
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**

(1)
+1 Everybody has already noted this but I just have to add my vote that this proof is stunning. - ** benblumsmith**

I wish you had shared this question to Math. S.E. :-) - ** Mikasa**

(4)
i.imgur.com/HCfGOYp.gif An animated version, which makes it a bit more clear. - ** DoubleAW**

(2)
The article that @BarryCipra mentions is available on JSTOR at jstor.org/stable/10.2307/2686996. - ** LSpice**

(3)
The image link seems to be broken now. - ** shreevatsa**

(4)
This is still broken. Or perhaps this is without words *and* without image and thus the topvoted one ;-) - ** user9072**

(1)
I edited it to substitute the broken image link to one in the comments. It's not optimal, but until the original link is restored it is better than nothing. - ** José Figueroa-O'Farrill**

I really still didn't understand this one. But I found another visual explanation for it maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html - ** ihebiheb**

(5)
I preferred the original static image (+1), which made the idea clear. The chaotic animated image (now -1) is mainly irritating---I cannot have it on my screen for any length of time. Perhaps it would become acceptable if the animation was slowed by a factor of ten or so. - ** Joel David Hamkins**

(5)
I find the underlying argument extremely nice but the picture (especially the animated one) does not really work for me without any words. The animation makes me think that is has some meaning that the yellow dots are visited in a certain order. - ** Dirk**

@NieldeBeaudrap sorry if this is a simple question, but what do u mean by "proof virus"? Not sure if I understood ur comment. - ** Charlie Parker**

How this can be a prove? Since its only displayed for $n=7$. - ** Benjamin**

(1)
Gauss discovered a solution to this sum at the age of 9. His solution is an example where a proof with words is more intuitive than a proof without words. math.stackexchange.com/a/1017743/202346 - ** Bob Woodley**

Could someone please edit the image to slow down the animation by a factor of 10 or so? The current chaotic form is beyond irritating. - ** Joel David Hamkins**

(1)
With help from Jeremy Kun, I have posted a slower version of the animation (20% speed), which I think makes a much more pleasant experience. - ** Joel David Hamkins**

I cannot see why this is considered a "proof without words": It takes a lot of words to state what it's a proof of - and to give hints how it does prove it. (Only the experts see it at first sight - but aren't proofs without words targeted more to the **non**-experts?) - ** Hans-Peter Stricker**

@NieldeBeaudrap Interesting distinction you make. Why is a written 'proof' a real proof and not a proof virus? Or is every so-called proof just a proof virus? I could probably get behind that, though perhaps then we should be worried about mutant strains? - ** lukeuser**

@Dirk Good point about the arbitrary order. I got it fairly quickly so I think it very much just varies person to person (many that's arbitrary too?). The arbitrariness of the order could be corrected by making it interactive—the yellow circle and corresponding blue pair is highlighted by where the cursor is. This could surely be done easily as svg I'm thinking. - ** lukeuser**

1

[+239]
[2010-03-07 02:16:51]
Russell O'Connor

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

An animation of the above is:

(_{This work has been released into the public domain by its author, Trekky0623 at English Wikipedia. This applies worldwide.})

There does not seem to be any necessity for the particular 'path in the relevant configuration space' that was used by the author of the above animated gif. This may be seen as an argument *against* including an animation.

(99)
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**

(61)
@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**

(17)
@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**

(8)
+1 , Here is the wiki page for this. en.wikipedia.org/wiki/Missing_square_puzzle - ** PKumar**

(3)
It might be noted that the success of the illusion partly depends on the fact this uses Fibonacci numbers (it is a coincidence I guess that the next newest answer is also about Fibonacci numbers!). - ** Todd Trimble**

This argument against proofs by picture is itself a proof by picture. - ** pseudosudo**

(1)
And if you use the actual area of the figure, it proves that 31.5 and 32.5 are both also equal to 32. - ** Toby Bartels**

This is not a proof without words: It's a trick question invented by an "amateur magician" which exploits the fact that the combined geometry looks like a 13x5 triangle. For us it's an exercise in exposing an optical illusion. Why would we try to prove that a statement that is explicitly false (e.g. ** Honest Abe**

`32.5 = 31.5`

) is in fact true? - 2

[+198]
[2009-12-14 15:50:16]
Jason Dyer

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

(6)
I suppose this picture can also be adapted to obtain the stereographic projection proof that a sphere is a manifold? - ** Kevin H. Lin**

@Jason Dyer: What software did you use? Inkscape? - ** user577**

(1)
I usually use Inkscape for my vector-based needs, but this was just done with my Smartboard presentation software. - ** Jason Dyer**

(1)
This picture shows not only that they have same cardinality but that they are homeomorphic. - ** Stefan Witzel**

3

[+184]
[2010-03-06 23:18:28]
shreevatsa

It is known (see
this other answer
^{[1]}) that an 8x8 board in which squares at opposite corners have been removed cannot be tiled with dominoes, as the removed squares are of the same "colour". But what if two squares of *different* 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:

(Imagine A and B are the squares removed.) The image is from *Mathematical Gems I* by Ross Honsberger.

(21)
What I like about this example is that there seems to be no straightforward proof *without* the picture; the crux of the proof's idea is specifically this picture. - ** shreevatsa**

(6)
Very nice. I guess the crux of the proof is that, when $mn$ is even, $P_m\times P_n$ is a Hamiltonian bipartite graph? - ** bof**

Well, I'd call that a generalization, not the crux of the proof. :-) Staying concrete, for the question about the specific case of $m=n=8$, the crux of this proof (that this graph *is* Hamiltonian) is this picture. Similar pictures can be draw whenever $mn$ is even, sure. - ** shreevatsa**

A complete result (guessed not shown) is for m or n odd : Any mxn board with 1 square removed has a neighborhood graph that has an hamiltonian cycle. - ** Jérôme JEAN-CHARLES**

I call this the "hungry snake" proof (imagine a game of Snake and you have to get the high score. The snake would have to assume some pose like this one). The exact same idea can be used to prove a lot of boards have the same property (for general n-dimensions). - ** MaudPieTheRocktorate**

4

[+177]
[2010-01-16 21:42:19]
serargus

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 putting 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

(10)
fantastic ! - ** Martin Brandenburg**

Really exceptional! - ** Koundinya Vajjha**

(1)
I think that there *is* a nice pictorial proof for this fact, but I don't think this is it. It's a proof for a specific $n$. To make it a general proof, the inductive step needs to be illustrated. - ** Max**

(30)
@Max: The inductive step is easy to figure out, since the rectangle above contains the rectangles from previous steps. - ** Daniel Litt**

5

[+145]
[2009-12-14 06:29:44]
Mike

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)*

(8)
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**

(62)
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**

(4)
@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**

(3)
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**

(1)
@Michael Lugo: The continuous version of this proof is "elementary geometry": the volume of a pyramid is one third of its height times the area of its ground surface! - ** nsrt**

For what it's worth, I disagree with @Marc van Leeuwen; I think that the geometric correctness of this image is sufficiently self-evident as to serve as a proof. - ** BD107**

6

[+128]
[2010-05-20 01:27:35]
Vaughn Climenhaga

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*:

*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.

(8)
Why would you resist? - ** Mariano Suárez-Álvarez**

(6)
+1 for the "Conjecture & Proof" shout-out. Best, course, ever! - ** Kevin O'Bryant**

(1)
Igen, nagyon jó. - ** Douglas Zare**

Beautiful! Yes! (I had to add "Yes!" to make my comment long enough). - ** Włodzimierz Holsztyński**

köszönöm, Google Translate. - ** Todd Trimble**

(1)
I really like this image and proof. The same idea works for other regular polygons, and I made images for pentagons, heptagons and so on, which you can find on my blog post at jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice. (The code accept $n$ as input, and makes the image for an $n$-gon.) - ** Joel David Hamkins**

(3)
And here is my post on the hexagonal lattice: the only regular polygons to be found are triangles and hexagons. jdh.hamkins.org/no-regular-polygons-in-the-hexagonal-lattice - ** Joel David Hamkins**

7

[+120]
[2011-02-07 01:27:59]
Daniel Parry

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

(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 https://www.maa.org/sites/default/files/Roger_B04151._Nelsen.pdf).

(3)
@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-Álvarez**

(10)
The same picture also gives an interesting formula for the integral of an inverse function! - ** Matt Noonan**

(6)
I guess this proof works only when $f$ and $g$ are both increasing? - ** Greg Martin**

@GregMartin : The way this drawn assumes that $f(b)>f(a)$ and $g(b)>g(a)$, but that's all (and you can draw similar pictures for the other possibilities). If they're not increasing the whole way, then there will be stuff outside the shaded regions, but it will count both positively and negatively and so cancel. Thus, the shaded regions' areas do equal the stated integrals. - ** Toby Bartels**

8

[+115]
[2014-04-19 14:12:18]
Zurab Silagadze

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).

(11)
This proof is actually known to Archimedes and used in his *Quadrature of the Parabola* en.wikipedia.org/wiki/1/4_%2B_1/16_%2B_1/64_%2B_1/… - ** Machinato**

Visual recursion. Awesome. - ** Manuel Bärenz**

Pierre Arnoux made a nice video about the geometric series, which has a version (with colours and animation) of this picture: youtu.be/6KQiTJLBwEw The video has words though! - ** Samuel Lelièvre**

9

[+113]
[2009-12-14 16:08:51]
Harrison Brown

There's a picture proof in the *Princeton Companion*, or alternatively on p. 340 of
Hatcher
^{[1]}, of the fact that the higher homotopy groups are abelian. Actually, here's a screenshot of the one in Hatcher (hopefully fair-use!):

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

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
^{[2]}. There are many $2$-dimensional rewriting arguments which are essential to the results of this book.

[2] http://pages.bangor.ac.uk/%7Emas010/nonab-a-t.html

(6)
Page 340 of Hatcher's book: math.cornell.edu/~hatcher/AT/AT.pdf - ** Dan Piponi**

(14)
This is sometimes called the Eckmann-Hilton argument: en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument - ** Kevin H. Lin**

(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**

(5)
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**

The "more general picture" seems to be broken. - ** Gerry Myerson**

10

[+105]
[2009-12-22 16:54:49]
aorq

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.

(1)
Yes, this one is beautiful. - ** Andrés E. Caicedo**

Indeed. And there are also similar visual proofs for the hyperbola and the parabola - ** godelian**

(2)
How does the picture explain the invariance of the total distance to two foci? I don't see it ; I haven't done geometry in a while though, I'm guessing it's some triviality... refresh my memory please? :) - ** Patrick Da Silva**

(8)
@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**

@aorq : Riiiight. Thanks for the clarification! It is indeed very visual. - ** Patrick Da Silva**

(1)
I've learned this and related proofs from Hilbert and Cohn-Vossen (but these proofs still originated mostly with Dadelin). - ** Włodzimierz Holsztyński**

(2)
I was confused by the perspective. In case anyone is having the same problem: the perspective is from a point that is (below the apex $S$ of the cone but) *above* the base of the cone (circle $k2$) and the bottom half-sphere ($G2$). I mistakenly thought we were looking up through $k2$ into the inside of the cone -- I think this is because in my browser, at least, the the circle $k2$ gets *thicker* when it passes behind the ellipse and should if anything get *thinner*. It's generally a nice drawing, though, and a nice proof. - ** Tim Campion**

There are more gems in Dandelin's figure but I need words : Let circles $k_1,\; k_2$ lie in planes $J_1.\; J_2 $ respectively. Let the ellipse be in plane $J_E$. Then $J_1\cap J_E ,\; J_2\cap J_E$ are the directrices.. Let line $l_P$ be the tangent to the ellipse at $P,$ with $l_P\subset J_E.$ Let $l_P$ meet $J_1,\;J_2$ at $Q_1\;,Q_2$ respectively. For $ i=1,2 , $ triangles $Q_iPP_i ,\; Q_iPF_i$ are congruent (as $Q_iP_i.\; Q_iF_i$ are tangent to sphere $G_i$), So angles $Q_iPP_i = Q_iPF_i$ . But $Q_1PQ_2$ and $P_1PP_2$ are lines. So angles $Q_2PF_2=Q_2PP_2=Q_1PP_1=Q_1PF_1.$ - ** DanielWainfleet**

11

[+99]
[2010-09-14 07:44:10]
muad

$ $ $ $ $ $ $ $

(12)
And similarly one proves that $\pi < 4$ by inscribing a circle in a square. - ** Michael Hardy**

(27)
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**

jstor.org/stable/10.5951/mathteacher.105.8.0632 - ** Benjamin Dickman**

Helpful to remember that pi is ratio of circumference of the circle to diameter. - ** Talespin_Kit**

(7)
Another case for $\tau$. - ** Daniel R. Collins**

12

[+92]
[2010-05-15 22:32:52]
Dave Pritchard

Means inequalities:

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 $$1^3 + 2^3 + \cdots + n^3:$$

The second image is due to
Brian Sears
^{[1]} (
Wayback Machine
^{[2]})

[2] http://web.archive.org/web/20180305035702/http://users.tru.eastlink.ca/%7Ebrsears/math/oldprob.htm#s32

(2)
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**

(2)
2nd proof: It would be nicer if the small strips were above and to the left of the big square. - ** Günter Rote**

(3)
The first proof could use some words. How is HM constructed? What is the small circle for? How does one prove that those segments have the claimed lengths? - ** Federico Poloni**

(1)
For completeness, since a link is still missing: Mariano Suárez-Álvarez has given a beautiful improved version of the second image here: math.stackexchange.com/q/61483 - ** Peter Heinig**

(1)
I didn't hear about "Means inequalities" until today and I don't know what this is useful for, but I have re-created the first image in GeoGebra, so you can drag around the point E to get a better feel for the line lengths: geogebra.org/classic/ndrfsstq - ** Fabian Röling**

13

[+84]
[2011-07-07 23:25:37]
JeremyKun

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.

There's probably a nice three-dimensional rendition of this that doesn't require writing down all those numbers. - ** Michael Lugo**

(11)
+1 Superb. Is this original? If not, to whom is it attributed? - ** I. J. Kennedy**

(2)
@MichaelLugo the 3D rendition is elsewhere on this page mathoverflow.net/q/8851 (you commented on it), but I do prefer this version. - ** adl**

14

[+82]
[2009-12-14 15:35:33]
Steve Flammia

Wikipedia has a few nice proofs of the pythagorean theorem. Elementary, but elegant.

(1)
oops! didn't see the word "non-trivial" in there... - ** Steve Flammia**

(26)
Pythagoras' theorem is trivial? I had no idea … Seriously, I don't necessarily think that the existence of a very simple proof implies triviality. Such proofs are, after all, not so easily discovered. Anyway, this is my favourite proof of the theorem. - ** Harald Hanche-Olsen**

(4)
The 20th President of the US, James Garfield, independently discovered the proof obtained by halving the right-hand diagram along a diagonal of the square of side length c. It requires you to write down an equation, though. That's my favorite proof, but mostly because of the corollary that B. Obama isn't the first geeky POTUS. - ** Harrison Brown**

(14)
@HB: Um, Thomas Jefferson? - ** Pete L. Clark**

(1)
I was told about an Indian mathematician who included the diagram on the right in his book about geometry. His proof of the Pythagorean theorem then consisted of just one word: "See?" - ** Paul Siegel**

Paul Siegel: That sounds very familiar to Bhaskara's "Behold!", which was a slightly different proof that the one displayed here. - ** muad**

(17)
A typical fake proof --- a simple statement as Pythagorean theorem is proved using much more advanced theorem on existence of area... - ** Anton Petrunin**

(31)
A typical fake refutation. You don't need to define Lebesgue measure to do manipulations in geometry. All operations can be defined geometrically if I associate a number X with the segment of length X, and define $X \mapsto X^2$ as a function, mapping a segment to a square with such side. In fact, even many of infinite summations can be done geometrically, using the obvious topology and metric on shapes. Thanks to this formalistic tradition it took 100 years of pain to get from non-trivial Lebesgue construction to much more natural motivic integration. - ** Anton Fetisov**

(1)
There's another really elegant proof of the Pythagorean theorem that involves basically drawing a single line. See Frank Wilczek who attributes it to Einstein, "A Polished Jewel": frankwilczek.com/2013/pythagorasTwo(Einstein)WithFigure.pdf - ** Dave Pritchard**

@DavePritchard That just blew my mind, thanks! - ** lukeuser**

15

[+77]
[2009-12-16 11:32:32]
Igor Khavkine

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

and the reverse animation

Mariano, thanks for fixing up my post! - ** Igor Khavkine**

(26)
I... don't quite get it. I think I need a few more words: What's the dot representing in each picture? - ** Harrison Brown**

(14)
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**

16

[+76]
[2010-03-25 19:00:35]
muad

If we have 3 circles on the plane with tangent lines, we can notice they have colinear intersection!

To prove it, we can visualize the same configuration in 3D, the balls lay on a surface and rather than tangent lines we take cones: The colinearity comes from the fact that if we lay a plane ontop of this configuration it will intersect the table in a line!

This is from 'curious and interesting geometry' and the proof is attributed to John Edson Sweet. I really like this proof because it gives a vivid example of the general idea that sometimes, to solve a problem in the most simple way you need to view it as a part of some bigger whole.

(28)
You need to draw a 3D picture of this to get rid of the words! - ** Ian Agol**

Don't we need the cones to all have the same slope? - ** benblumsmith**

(2)
In this pretty solution there is another pretty geometric problem: Given three spheres there is a plane which is tangent to all three. - ** Rogelio Fernández-Alonso**

(5)
Where is the picture? - ** Patrick Da Silva**

(1)
This was one of my favorite proofs in this list... it's a shame that imageshack took this picture off to promote their site. - ** KalEl**

I warmly recommend the (briefly cited) very interesting book "The Penguin Dictionnary of Curious and Interesting Geometry", by David Wells that can be accessed at (archive.org/details/…) - ** Jean Marie Becker**

17

[+72]
[2011-04-10 16:57:19]
Marco Golla

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

**Theorem**. The trefoil knot is knotted.

*Proof*.

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
^{[2]} if you're interested).

[2] http://math.ucsd.edu/~justin/Papers/knotes.pdf

This is wonderful. - ** Kevin H. Lin**

What does "nontrivial" mean? - ** Tom Goodwillie**

@Tom Goodwillie: I've edited, and added some remarks. Thank you. - ** Marco Golla**

18

[+70]
[2009-12-15 02:32:12]
Darsh Ranjan

Here's a proof of the inequality of the arithmetic and geometric means in the form $$\frac{x_1^n}{n} + \cdots + \frac{x_n^n}{n} \geq x_1\cdots x_n.$$

Proof for $n=3$:

The "figure" for general $n$ is similar, with $n$ right pyramids, one with an $(n-1)$-cube of side length $x_k$ as its base and height $x_k$ for each $k=1,\ldots,n$.

(I made this in
Inkscape
^{[1]}, a wonderful free-software vector drawing application. For the inequality and associated labels, I used the
textext
^{[2]} extension.)

[2] http://pav.iki.fi/software/textext/

(3)
And what exactly is a proof about this? - ** darij grinberg**

(9)
The box has volume xyz and is contained in the union of the three square pyramids, which respectively have volumes x^3/3, y^3/3, and z^3/3. Thus xyz <= x^3/3 + y^3/3 + z^3/3. - ** Darsh Ranjan**

19

[+60]
[2010-09-27 15:36:25]
AndrewLMarshall

from Steven Strogatz's column:
http://opinionator.blogs.nytimes.com/2010/04/04/take-it-to-the-limit/ (
Wayback Machine
^{[1]})

(20)
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**

(3)
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**

(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**

(6)
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**

(2)
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**

20

[+57]
[2009-12-15 01:31:55]
Steven Gubkin

Here is the very first piece of original mathematics I ever did, in high school:

The derivative of sine is cosine.

(6)
It looks like your image is no longer available... - ** I. J. Kennedy**

(3)
Leibniz actually did this drawing. It's very nice because you can teach it to undergrads. You can do the same with any of the trig functions and their inverses. For tangent, you can extend the hypotenuse of the above triangle until it intersects the line tangent at the point $1$ (assuming this is the unit circle in the complex plane). Then you get a triangle with base $1$, height tangent, and hypotenuse secant. For cosecant and cotangent, you draw a tangent line from the point i. Then through similar triangles you can differentiate all these functions and their inverses. - ** Phil Isett**

(1)
Yup, I have drawn pictures for all of those as well, but this always seemed like the simplest one. Never understood why this isn't in calculus books. - ** Steven Gubkin**

(3)
@StevenGubkin If you still have it, could you put the picture back in? No one can see it. - ** Todd Trimble**

(1)
This post is very much related to this. - ** Simply Beautiful Art**

21

[+52]
[2010-02-22 15:25:59]
CuriousUser

The sequence of pictures

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

(3)
Thomas Harriot first proved this formula in 1603, apparently by a similar argument, though I have not seen his picture(s). - ** John Stillwell**

(6)
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**

Incidentally, I tried to find a similar proof for the area formula for hyperbolic triangles. Unfortunately, that did not work due to non-compactness of hyperbolic space. If anyone knows whether such a proof exists, I'd be happy to see it. - ** Igor Khavkine**

(3)
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**

This same proof also appears at the very opening of this paper: arxiv.org/abs/1301.0352 - ** Yaakov Baruch**

22

[+43]
[2011-03-15 22:02:59]
Chris Heunen

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
^{[1]}". 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"
^{[2]}, and proves that the comultiplication of a Frobenius algebra is cocommutative if and only if the multiplication is commutative.

[2] https://books.google.com/books?id=6dZZW08Z04MC

The link to Selinger's paper wasn't working - ** Yemon Choi**

There are many proofs of similar flavor about 4-manifolds using the Kirby calculus. - ** Matt Brin**

Picture is dead - ** BlueRaja**

I'm voting this up because I like string diagrams, even though you don't mention them specifically. - ** Ryan Reich**

(9)
this proof makes me wonder what is a 'picture' and what is - ** Jeremy**

23

[+42]
[2009-12-15 01:52:27]
Ian Agol

Sphere eversion
^{[1]}

And here's a two-dimensional rendering of the sphere eversion: [1] https://www.youtube.com/watch?v=BVVfs4zKrgk

(15)
As pretty as it is, that is nowhere understandable as a proof. More as an illustration. - ** Willie Wong**

(16)
@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**

(8)
I just watched the video, which was excellent, but it had a lot of words in it. - ** Patricia Hersh**

(2)
The original picture has better image quality: andreghenriques.com/PDF/Eversion.pdf - ** André Henriques**

24

[+42]
[2016-12-29 21:16:17]
Robin Saunders

Late to the party, but David Lehavi and Bob Palais both mentioned the proof that $\pi_1(SO(3))$ has an element of order 2. In fact it is the only nontrivial element, and so the double cover of $SO(3)$ is simply connected.

Here's an animated illustration of that fact, courtesy of Wikipedia (
here
^{[1]}):

25

[+36]
[2010-09-14 08:24:03]
Alexis Monnerot-Dumaine

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** :

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

(40)
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**

(17)
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**

(11)
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**

(53)
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**

(5)
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**

(3)
@Michael Hardy, I think it has been done: youtube.com/watch?v=4RQmLNa5ZNo&feature=related - ** Gerry Myerson**

(6)
Contrary to gowers, I don't think it's clear that we are defining a sequence of continuous functions which converges uniformly. We aren't defining a sequence of functions at all, only a sequence of images of the interval under a function. If we choose the "wrong" sequence of parameterisations of this sequence of curves then we do not get a limit function. The sequence of parameterisations (which is not illustrated) is crucial to proving the existence of the limit object: without some indication of which parameterisations we must choose there is no proof. - ** Ian Morris**

well, it is clear that if the curves are parametrized with constant velocity they converge uniformly - ** Pietro Majer**

26

[+36]
[2015-10-28 06:51:26]
L.Z. Wong

$$\arctan \frac{1}{3} + \arctan \frac{1}{2} = \arctan 1$$

It's easy to generalize this to

$$ \arctan \frac{1}{n} + \arctan \frac{n-1}{n+1} = \arctan 1, \text{ for } n \in \mathbb{N}$$

which can further be generalized to

$$ \arctan \frac{a}{b} + \arctan \frac{b-a}{b+a} = \arctan 1, \text{ for } a,b \in \mathbb{N}, a \leq b $$

Edit: A similar result relating Fibonacci numbers to arctangents can be found
here
^{[1]} and
here
^{[2]}.

[2] https://www.maa.org/sites/default/files/Fibonacci_Numbers55908.pdf

(3)
It needed quite a long time for me to understand this. But, well, then it is amazing! - ** Gottfried Helms**

(2)
Very nice. It might be a bit more clear if the right triangle for arctan(1/2) and arctan(1/3) were colored in the first picture, and the triangle for arctan(1) was colored in the second picture. - ** Steven Gubkin**

27

[+33]
[2016-12-04 05:40:44]
Steven Stadnicki

This proof-without-words of the Pythagorean Theorem is far from a new one, but it's the first one I've ever seen 'in the wild' (this photo was snapped after finishing dinner at a Mongolian Grill restaurant):

(27)
I imagine people from the other tables, watching somebody while taking a picture of an *empty* plate! - ** Pietro Majer**

Wait a minute -- since the triangles are tilted up toward the viewer, isn't this rather a *disproof* of the Pythagorean theorem? - ** Tim Campion**

28

[+29]
[2009-12-14 06:13:11]
Mike

As you probably already know — there are lots of these in
Proofs without Words
^{[1]} (and II) by Roger Nelson.

Yup. I was wondering if there are examples proving non elementary results (my example is not exactly non-elementary, I know...) - ** Mariano Suárez-Álvarez**

29

[+29]
[2018-09-05 20:57:14]
Hans-Peter Stricker

I wonder why
Apostol's proof of the irrationality of
$\sqrt{2}$
^{[1]} - which is as visual as a proof can be (in my opinion) - has not been mentioned: One can literally *see* at a glance that it proves what it's supposed to prove: the **impossibility of a isosceles right triangle with integer side length** (by infinite descent):

Note that it's not a proof completely without words. It helps a lot to read the comments of the author:

Each line segment in the diagram has integer length, and the three segments with double tick marks have equal lengths. (Two of them are tangents to the circle from the same point.) Therefore the smaller isosceles right triangle with hypotenuse on the horizontal base also has integer sides.

But through own thinking one could come up with this by oneself (having in mind what's to be proved).

[1] http://hipatia.dma.ulpgc.es/profesores/personal/aph/ficheros/resolver/ficheros/crp/2root_2000.pdf
I think you mean to refer to the impossibility of an integer isosceles *right* triangle, since clearly one can have isosceles triangles with integer sides, such as an equilateral unit triangle. - ** Joel David Hamkins**

(1)
@JoelDavidHamkins: Happy to hear from you again after all those years;-) I made the correction. Thanks for the hint. - ** Hans-Peter Stricker**

Oh yes, I've enjoyed many of your questions on MO over the years. - ** Joel David Hamkins**

So happy to hear that! But your enjoyment could not have been greater than mine was about your answers. - ** Hans-Peter Stricker**

It might be worth noting that one doesn't need to introduce the circle here; drawing the line between the top vertex of the triangle and the interior point on the bottom edge shows that the kite-shaped quadrilateral is bilaterally symmetric (the two triangles it's split into are congruent by side-angle-side equivalence) - ** Steven Stadnicki**

@StevenStadnicki You don't *need* the arc (since the tics are used to show equidistance), but surely the arc is by far the easiest way to construct the vertex at the correct position? You do after all need it to be possible to find such a point—I realise this could be taken as given, but that would make the proof slightly harder to follow in my opinion. - ** lukeuser**

30

[+28]
[2014-03-19 12:11:07]
Campello

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

Maybe it doesn't fit into the "non-trivial" category? - ** Campello**

(8)
I think that the fact that the rationals are countable qualifies as non-trivial, when put in historical perspective - ** Geoff Robinson**

31

[+25]
[2009-12-14 07:53:45]
Alon Amit

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 hexagon 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.

(3)
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**

That rhombus tilings are equinumerous to plane partitions which fit in a box. - ** Mariano Suárez-Álvarez**

(3)
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**

(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**

Sorry - I didn't want to spoil the fun right away. The statement being proven is the one indicated by Darsh Ranjan: however you tile a hexagon with rhombi, there's an equal number of tiles in each of the three orientations. The picture-proof asks you to believe that all such tilings can be regarded as the facets of a cubical arrangement, and the orientations correspond to the viewing angle. As far as I know, the colors are random and are just a distraction. - ** Alon Amit**

There should be 3 colors, for the 3 orientations. (I think it even says this in the book.) I think the gray is a mistake introduced by the cover designer - unless there is some hidden meaning in the arrangement of gray rhombi? - ** Emil**

@Alon: could you repost the image using a more permanent upload site? - ** Mariano Suárez-Álvarez**

By the way - I highly recommend this book. Practically every problem in it is both challenging and interesting. - ** Yaakov Baruch**

Is there a name for this theorem? I vaguely recall someone referring to it by the name of some French cookie which happens to have a rhomboid shape and come in a hexagonal tin, but I can't find a reference online - ** Xodarap**

(1)
@Xodarap sorry, doesn’t ring a bell. - ** Alon Amit**

@Xoderap this french rhomboid cookie is called a "calisson d'Aix" (en Provence) - ** Jean Marie Becker**

32

[+25]
[2010-06-25 02:45:08]
BlueRaja

Conway and Soifer tried to set a record for least number of words in a mathematical paper. I've reproduced it here in its entirety.

**Can n ^{2} + 1 unit equilateral triangles cover an equilateral triangle of side > n, say n + ε?**

n^{2} + 2 can:

(7)
"This paper is worth publication, provided the authors add some words of explanation on their construction." - ** Pietro Majer**

33

[+24]
[2009-12-14 08:05:58]
Alon Amit

In an attempt to push the bar towards the non-trivial, I'll mention the proof that the boundary complex of every polytope is shellable. The proof is virtually word-free but requires an actual movie rather than a still image: imagine yourself in a spaceship, taking off in a straight line from one of the facets, away from the polytope. Every once in a while a new facet is visible to you; under assumptions of general position, this provides a shelling of the complex (obviously, you need to fly off to projective infinity and come back on the other side).

This was assumed by Euler but first proved only in 1970 by Brugesser and Mani, who said that the idea came to him in a dream. More details
here
^{[1]} (search for "shellability") or
here
^{[2]}.

[2] https://gilkalai.wordpress.com/2008/09/18/annotating-kimmo-erikssons-poem/

(32)
Why are there so many words and so few pictures in this answer? - ** David Eppstein**

(17)
Because I couldn't a way to draw this, let alone animate, in a reasonable time. I trust that the description is helpful in imagining what the actual wordless proof is. - ** Alon Amit**

(13)
I want a video! - ** Emil**

@AlonAmit The difference between what we want and what you gave us is explained in this video. - ** Simply Beautiful Art**

34

[+23]
[2011-06-28 14:42:08]
leonbloy

(I'd post this as a comment to Mariano Suárez-Alvarez, but I've not enough rep). From a
ME thread
^{[1]}.

$$\sum_{k=1}^n (-1)^{n-k} k^2 = {n+1 \choose 2} = \sum_{k=1}^n \; k = \frac{(n+1) \; n}{2}$$

[1] https://math.stackexchange.com/questions/44759/combinatorial-proof-that-binomial-coefficients-are-given-by-alternating-sums-of-s/44782#4478235

[+22]
[2011-09-16 03:05:50]
sclv

A line that bisects the right angle in a right triangle also bisects a square erected on the hypotenuse:

source: https://www.futilitycloset.com/2011/09/12/half-and-half/

36

[+22]
[2012-05-29 22:27:02]
O.R.

I just saw this proof, which is of course not mine.

Proof with words: The [area of a] circle
^{[1]}

Similar in concept to the above video:

[1] https://youtu.be/whYqhpc6S6g37

[+20]
[2010-05-15 16:40:20]
Vaughn Climenhaga

This should really be a comment on Marco Radeschi's
answer
^{[1]} 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.

(That picture is slightly modified from p. 221 of
this book
^{[2]}, which has the whole proof in more detail.)

[2] https://books.google.com/books?id=XFkc0Yn-TE8C&printsec=frontcover&dq=Lectures+on+Surfaces&ei=bMvuS86VK4-2zQTRufDxCg&cd=1#v=onepage&q&f=false

38

[+19]
[2014-10-07 19:29:11]
Thierry de la Rue

The area under a cycloid is three times the area of the generating circle.

(1)
How do you know each of the red bars of the football shape has exactly the width as the corresponding chord on the circle at the same height? - ** Jonah**

(2)
@Jonah: Consider a circle moving rightwards, starting at the left corner of the football. If we put a dot on the bottom of this circle moving clockwise at the same speed as the circle, we can think of the dot as painted on the circle, and the circle as rolling on the bottom line. But if the dot instead moves counterclockwise, we can imagine the dot on a wheel rolling on the *top* line. This produces the cycloids in the diagram. But for a fixed position of the circle, the dots will be at the same altitude, and so their distance will be the width of the circle at that altitude. - ** RavenclawPrefect**

39

[+18]
[2009-12-19 17:45:58]
Gil Kalai

Q: Can you tile
^{[1]} with
^{[2]}?

$ $ $ $

[1] http://www.cl.cam.ac.uk/~mj201/images/mutilated-chess-board2.gif[2] http://www.stewart.hinsley.me.uk/Fractals/Scripts/Polyplet.php?r0=3

I don't think this is clear enough to be self-contained, although I have something in mind to fix it. Do you mind if I try? - ** Jason Dyer**

I have edited and put in my modification of the image. - ** Jason Dyer**

can someone explain this in words? - ** Turbo**

(10)
It's easier if you *do* use words. If you take away opposite squares, you have more of one color than another... - ** Todd Trimble**

@Turbo, When you tile with dominoes, each domino covers two adjacent squares. Because *every* two adjacent squares contain 1 black and 1 white, the original question is equivalent to the question: Can you tile with black-and-white dominoes where each domino's colors match the colors of the two squares it covers. (So we have replaced the original question with a more constrained question, but know that the answers must be identical). Yet when tiling with black-and-white dominos that match what they cover, it's clear you can only cover boards with an equal number of black and white squares. - ** Jonah**

40

[+15]
[2010-08-07 03:37:33]
Thierry Zell

In the movie category, I'm surprised that no-one has yet posted a link to
Moebius Transformations Revealed
^{[1]}.

(2)
But what does that movie *prove*? - ** Mariano Suárez-Álvarez**

(4)
@Mariano: it doesn't prove anything, but then again neither do *any* proofs without words. They merely give us insight into the proof, and in that respect, any movie has even more potential than a simple image. I think we will soon see very innovative approaches in movie-proofs. - ** Thierry Zell**

(1)
Very beautiful. I suppose it proves the usefulness of abstraction in obtaining unity - ** William**

41

[+14]
[2015-08-17 12:23:46]
Bhaskar Vashishth

From the book "*Proofs without words*", there are ton of others too but this one I had trouble proving in UG, so like it most.

I like using $\int_{-1}^01+r+r^2+\dots+r^{n-1}dr=\int_{-1}^0\frac{1-r^N}{1-r}dr$, but this way is very much visual. - ** Simply Beautiful Art**

42

[+13]
[2017-10-20 08:59:55]
Turbo

Given three mutually tangent circles in the plane, there exist exactly two circles tangent to all three.

43

[+12]
[2010-01-15 16:31:52]
vonjd

Have a look at this document from an MIT-instructor (Sanjoy Mahajan): http://mit.edu/18.098/book/extract2009-01-21.pdf

(This is a draft of Chapter 4 of:
Sanjoy Mahajan,
*
Street-Fighting Mathematics
*
, MIT Press 2010
^{[1]}.)

44

[+11]
[2009-12-14 07:05:44]
David Lehavi

The first homotopy group of SO_3 has an order 2 element (that's a classic).

The surface area of a quarter of the unit sphere is Pi via Gauss-Bonnet (My source is Ariel Shaqed - it should have been a classic, but no one I asked seems to knew it). The sphere is what you reach with a straight hand while standing still. Hold a Pencil in your hand, that's your tangent vector. Now parallel transport the pencil on a quarter sphere: it points in the opposite direction. QED

(3)
For SO(3) has order 2 element: gregegan.customer.netspace.net.au/APPLETS/21/21.html - ** Dan Piponi**

Indeed, but it's not the one I wanted.... - ** David Lehavi**

(7)
Place a glass on the open palm of your hand. You can, with a bit of practice, rotate the glass twice (but not once) around the vertical axis without spilling any liquid from it, and return to your original position. Each part of your body goes through a loop in SO_3. Moving from the shoulder via the arm to the glass, you get a homotopy essentially proving the theorem. I have seen dancers from somewhere in south-east Asia incorporating this move into their dance. - ** Harald Hanche-Olsen**

David, it's not the one I wanted either :-) I wanted to find a video of Feynman himself doing the plate trick but there doesn't seem to be one online. - ** Dan Piponi**

(23)
Why are there so many words and so few pictures in this answer? - ** David Eppstein**

(4)
@David: well, you can think if this answer (or of Harald's comment, which gets my emphatic upvote) as a script for the choreography which, when acted out, is a proof without words :P - ** Mariano Suárez-Álvarez**

(2)
It doesn't feature Feynman, but here's a video of a human doing the plate trick (just after 1 minute in): youtube.com/watch?v=CYBqIRM8GiY - ** Harrison Brown**

@Herald: thanks - thats the one I wanted. @David E. becasue for both of them you need a movie (and I didn't mennage to verbaly describe the SO_3 proof which Herald did). - ** David Lehavi**

isnt this also showing about $\eta^2$? or rather $lim\pi_{n+1}(S^n)$? - ** Sean Tilson**

45

[+11]
[2011-11-08 09:33:26]
Martin Brandenburg

Proof of the associativity law $f * (g * h) = (f * g) * h$ in the fundamental groupoid of a topological space:

You can find more of these diagrams in J. P. May's *A Concise course in algebraic topology*.

46

[+11]
[2012-08-16 22:01:48]
Jon Cohen

The pathspace of any topological space is contractible.

Pf (as given in my homotopy theory class): slurp spaghetti.

47

[+10]
[2012-08-20 22:39:22]
Marc Chamberland

I like the tiling proof of the **Pythagorean Theorem**. The left image is credited to Al-Nayrizi and Thābit ibn Qurra (9th century) and the right by Henry Perigal (19th century).

I'm having trouble seeing a triangle (of the appropriate dimensions) in the Perigal tiling. - ** Gerry Myerson**

Gerry, slide the red square to the left by half the side length of a white square. The segment connecting the two lower red corners is the hypotnuse, and the legs have lengths which are the widths of the two tiling squares. Yes, I know that sounds confusing. - ** Marc Chamberland**

Thanks, Marc, not confusing at all. But I think if you have to add that to see that there's a triangle there, it's not really a proof without words. Well, at least, for me it's not a proof without words. - ** Gerry Myerson**

48

[+10]
[2016-12-04 13:40:14]
Pietro Majer

I was somehow challenged by the idea: do more abstract topics allow proofs without words? I came out with this example. Of course, it is disputable if it is really "without words", since some words of explanation should be given (here they are: vertical segments represent Banach spaces and subspaces; connecting segments between two of them represent a linear operator. Italic letters $a,b,c,d,e$ are the dimensions of the corresponding linear subspaces).

49

[+9]
[2010-03-06 21:55:58]
Tobias Hagge

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
^{[1]} (
Wayback Machine
^{[2]})

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

Here's the image:

[1] http://utdallas.edu/%7Ehagge/images/Catalan.pdf[2] http://web.archive.org/web/20160322132017/http://utdallas.edu/%7Ehagge/images/Catalan.pdf

(4)
Do you have an explanation for the picture? I looked at it, and looked at it, and don't get it. - ** Willie Wong**

(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**

50

[+9]
[2011-08-16 06:45:45]
Ron Maimon

(2)
I suppose I should link to the physics stackexchange question that this came from: physics.stackexchange.com/questions/12435/… - ** Ron Maimon**

51

[+9]
[2016-12-04 06:28:08]
მამუკა ჯიბლაძე

Posted this quite a while ago on math.SE as an answer to https://math.stackexchange.com/q/733754/214353; I think it has a place here too.

To be rigorous, it maybe only proves that $f(z):=\displaystyle\lim_{n\to\infty}\left(1+\frac zn\right)^n$ is $1$ at integer multiples of $2\pi i$ (perhaps also that $f$ is $2\pi i$-periodic), but does not give any insight into why $f(z)=\text{something}^z$; still...

I've seen it much better rendered once somewhere on the web but could not find it again, so tried to reconstruct it myself

52

[+9]
[2017-08-03 13:39:57]
assaferan

I've tried to read carefully through all the posts here, and I think no one has mentioned the excellent book by R.B. Nelsen which contains many more such examples. I give here an explicit citation:

Nelsen, Roger B. Proofs without words: Exercises in visual thinking. No. 1. MAA, 1993.

Here is an example which I have recently used:

(4)
Nelson's book was already mentioned eight years ago (but without any illustrations): mathoverflow.net/a/8849 (and was the *second* answer to question). - ** jeq**

53

[+9]
[2020-09-01 08:08:46]
C.F.G

Gluing two Mobius strips along their edges is a Klein bottle.
^{[1]}

Beautiful animation! Is this rendered or hand-drawn? The colors/shader are fantastic... - ** Steven Stadnicki**

54

[+8]
[2011-07-06 00:39:00]
JeremyKun

Can you tile an 8x8 chessboard with one corner cut off with dominoes of dimension 3x1?

This is a simple way to show that choosing a useful coloring can make a proof trivial.

This proof was also a result of the Conjecture and Proof class in the Budapest Semesters in Mathematics. It was one of the first problems encountered there, hence not *that* hard :)

(1)
See also: **MESE 220**. - ** Benjamin Dickman**

55

[+8]
[2014-09-11 05:37:33]
Zurab Silagadze

Here
^{[1]} 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)

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

[1] http://www.maa.org/sites/default/files/269122948517.pdf56

[+8]
[2015-01-23 00:51:10]
Loreno Heer

$S^2 \vee S^1 \vee S^1$ is homotopy equivalent to the Klein bottle with self-intersection.

(1)
What a masterpiece :)! - ** justadzr**

57

[+8]
[2015-11-11 17:40:41]
Guo Qi

^{[1]}.

This is an example I did when I was in high school.

Let it be a unit disc, consider the length of horizontal line, we know Yellow=$2\cos \frac{3}{7}\pi$, Yellow+Green=$-2\cos \frac{5}{7}\pi$, Red+Green=$2\cos \frac{1}{7}\pi$.

Then 1=Red=Red+Green-(Green+Yellow)+Yellow=$2(\cos \frac{3}{7}\pi+\cos \frac{5}{7}\pi+\cos \frac{1}{7}\pi).$

(1)
I don't understand it. Seems like it *does* require some words... - ** Johannes Hahn**

Sorry about that. Actually, I don't really know how to explain it well. - ** Guo Qi**

58

[+7]
[2010-03-05 16:59:16]
Sunni

There is a proof of
Erdös-Mordell Inequality
^{[1]} 'without words' which an impressive one.
Please follow the link
http://forumgeom.fau.edu/FG2007volume7/FG200711.pdf

How is that "without words"? :-/ - ** Andrea Ferretti**

(1)
If you observe carefully on the graphs, you don't even need to write a word. - ** Sunni**

59

[+7]
[2010-03-07 11:01:10]
Kumar

The idea is to prove things in ways that are obvious to different parts of your brain, right? Anyone found any "auditory proofs"? Some candidates -

Nyquist sampling theorem?

sin[a] + sin[b] = 2sin[(a+b)/2]cos[(a-b)/2]. If you use at and bt instead of a and b, you can translate that to show how the addition of two sine tones close in frequency can also be perceived as a modulation or "vibrato" around the centre frequency. The factor of 2 might be hard, though you can add a gain instead of 2 and show that the difference is silence when the gain is 2 :)

Sampling in frequency domain (comb filter) is periodicity in time domain?

Here are some "audio illusions" though, for your amusement - http://www.youtube.com/watch?v=e6JSTkwXg90

(1)
Are there more details on 1? 2 and 3 don't seem like proofs so much as examples, though maybe you are just putting them forth as challenges. - ** j.c.**

The link to the video seems to be dead. (OTOH, it was probably not that relevant for the topic here.) - ** Martin Sleziak**

60

[+7]
[2010-07-14 22:15:56]
Daniel Miller

There is a beautiful proof of the fact that a checkerboard with sides $2^{n}$, and one square removed can be tiled with $L$-shaped pieces formed by three squares. Given that a checkerboard of sides $2^{n-1}$ can be so tiled, then a square checkerboard of sides $2^{n}$ can be tiled by filling in the quarter in which the removed piece lies, and then placing an extra $L$-shaped tile with one square in each of the remaining three quarters.

I first learned this in Dan Velleman's book "How to prove it." I'm not sure if he originated it or not. - ** Jim Conant**

61

[+7]
[2015-01-23 01:10:06]
Loreno Heer

Proof of the lantern relation (taken from the book: A Primer on Mapping Class Groups by Farb, B. and Margalit, D.)

62

[+7]
[2017-10-23 13:19:20]
Peter McNamara

This image is a bijection between the puzzle rule and the semistandard tableau rule for Littlewood-Richardson coefficients. It is taken from
this paper of Ravi Vakil
^{[1]}, where it is attributed to Terry Tao.

The picture generalises to a bijection between rules in K-theoretic Schubert calculus, but I haven't seen a picture and don't currently have the patience to create one.

[1] https://arxiv.org/abs/math/0302294
too complicated.... - ** Turbo**

63

[+7]
[2017-11-29 17:48:55]
Vaughn Climenhaga

A crucial step in proving the triangle inequality for the $L^p$ norms is to prove **Young's inequality**: $ab \leq \frac{a^p}p + \frac{b^q}q$ when $a,b>0$ and $p,q > 1$ are conjugate ($\frac 1p + \frac 1q=1$). This can be proved by comparing the areas in the following picture:

*An easy calculus exercise gives the blue area as $\frac{a^p}p$ and the green area as $\frac{b^q}q$, hence their sum dominates the area $ab$ of the rectangle outlined in red.*

I learned this proof from Cohn's measure theory book, where it appears as an exercise (in Section 3.3).

You miss your graph.... - ** Hu xiyu**

That's one way of establishing $ab \leq \frac{a^p}{p} + \frac{b^q}{q}$, but I find this easier to grok via convexity of the exponential function. - ** Todd Trimble**

64

[+6]
[2010-03-25 18:11:49]
Chris Conway

This is apparently not was intended, but I think it qualifies. From
*
Principia Mathematica
*
^{[1]}:
the proof of 1+1=2
^{[2]}.

[2] https://en.wikipedia.org/wiki/File:Principia_Mathematica_54-43.png

Which image? The second link is dead and the first has no proof-like images. - ** Simply Beautiful Art**

(1)
en.wikipedia.org/wiki/File:Principia_Mathematica_54-43.png - ** Chris Conway**

I think you are referring to proof without English words... But I believe these statements are equivalent to English sentences. But still, it's a pleasure to read this. - ** justadzr**

65

[+6]
[2010-07-11 15:11:05]
O.R.

Let $0\leq x,y,z,t\leq1$ Prove that $x(1-y)+t(1-x)+z(1-t)+y(1-z)\leq 2$.

Draw a 1x1 square and mark in consecutive sides disjoint segments starting at the vertexes of lengths $x,y,z,t$. Joining the consecutive end points of the intervals that are not vertexes of the square form four triangles, the area of the triangles is the left hand side divided by 2, the area of the square is the right hand side divided by 2.

(3)
This proof without words has an awful lot of them! - ** I. J. Kennedy**

Indeed, there is a proof with only eleven words: rearrangement and arithmetic-geometric inequalities. (Details are left to the reader.) - ** dvitek**

I don't think drvitek's proof makes sense. - ** darij grinberg**

(1)
+1 It is quite a nice idea, if you actually do the drawing. - ** user22882**

66

[+6]
[2018-07-19 17:07:42]
Đào Thanh Oai

(1)
There's a similar figure for $(a+b)^2 = (a-b)^2 + 4ab$; but unfortunately there appears to be no similarly simple figure for $(a+b+c)^3 = (a+b-c)^3 + (a+c-b)^3 + (b+c-a)^3 + 24abc$, although there are some nice ideas for figures with some overlapping regions: mathoverflow.net/q/306394/88133. - ** Zach Teitler**

it would be simpler and more clear with all squares sharing a diagonal - ** Pietro Majer**

67

[+5]
[2009-12-15 00:19:29]
Aaron Mazel-Gee

Rich Schwartz had on his site a great paper consisting of only a picture which proved that every right triangle admits a periodic billiard path. Unfortunately, he's since deleted it, so I can't post it here. (It shouldn't take too long for anyone interested to re-construct the proof, though.)

(3)
I am guessing he did it by assembling four of the said right-triangles into a parallelogram. There is a path that bounces directly between the two longer sides. Mod out by the symmetry and you get a periodic path in the triangle. - ** Willie Wong**

68

[+4]
[2011-06-28 16:46:28]
Jesko Hüttenhain

Interesting how everyone understands *"proof without words"* as *"proof made of pictures"*. I read the title and immediately thought that every proof can be written without words, using just first order logic. I stopped there and thought that this is just another language, using different words - and I came to the conclusion that there can not be a mathematical proof without *"words"*, because you have to get some information across! Sure, you can use different languages than English. But in the end, this boils down to the question, **what is a word**?

BTW: Unmentioned so far are category-theoretical proofs, which can sometimes be expressed very comprehensively as a sequence of diagrams. I am too lazy to look up a good example, because I already explained that I don't believe in the question.

(5)
Formulas (first order or what not) are just words written in abbreviated form. That really does not count... - ** Mariano Suárez-Álvarez**

(5)
So are pictures. That's my point. - ** Jesko Hüttenhain**

(5)
I doubt there is any sense in which one can formalize the notion, but I think it is pretty clear that a proof written in the first order calculus, or any other calculus, is simply *not* a "proof without words". You do not believe in the question, you say, but I honestly cannot understand what that can possibly mean: there is certainly *something* that gets the name proof-without-words (there is even a section in the MAA Monthly dedicated exclusively to this, and it has run for decades!) and most people ---while probably not being able to explain exactly what they are--- recognize them. - ** Mariano Suárez-Álvarez**

On the contrary, I would say that any picture that is a rigorous proof must first be formalised in some sense, and will then most probably be in words of some form. - ** Manuel Bärenz**

69

[+4]
[2015-05-14 08:45:49]
VividD

A 3D proof of a Fibonacci identity, that even includes a video:

Link does not work. - ** ThiKu**

You are right, I'll delete the link until I hopefully find working one, thanks. @ThiKu - ** VividD**

(3)
This isn't a Fibonacci identity per se; for *all* $a$ and $b$, $(a+b)^3 = a^3+b^3+3ab(a+b)$. - ** Steven Stadnicki**

70

[+3]
[2011-02-07 04:01:50]
Bob Palais

Here are some dynamic versions:

http://www.math.utah.edu/~palais/sums.html (two of the summation formulas mentioned above)

Several belt, plate, and tangle trick animations:

http://www.math.utah.edu/~palais/links.html

A visual derivation of complex multiplication:

http://www.math.utah.edu/~palais/newrot.swf

Pythagoras in the Isosceles case, based on the Yale tablet:

http://www.math.utah.edu/~palais/PythagorasIsosceles.html

and the general case:

http://www.math.utah.edu/~palais/Pythagoras.html

There is an animation of the Dandelin Spheres construction depicted above in 3d-XplorMath in anaglyph 3d. 3d-xplormath.org Not really a proof, but visual descriptions of the relationship of cosine and sine curves and uniform circular motion: math.utah.edu/~palais/cose.html math.utah.edu/~palais/sine.html and epicycloids: math.utah.edu/~palais/daledots.swf - ** Bob Palais**

(2)
All the links in this post appear to be broken. - ** I. J. Kennedy**

71

[+3]
[2011-04-10 15:54:03]
user11235

I suggest the videos of Viennot explaining the bijections between different families of objects counted by Catalan numbers:

http://www.xavierviennot.org/contscience/videos.html

72

[+3]
[2011-07-08 21:49:24]
Phil Isett

Here's a proof of the area of a circle (or sector) which is different from the one posted previously.

**EDIT:** I was unable to embed the file, which is in pdf form. Here is a link:

http://wildpositron.files.wordpress.com/2011/04/sectorarea2.pdf

I discussed what goes into making the proof complete to show that the map preserves area on my blog here (it requires just another picture or two, but it's essentially still only a geometric argument):

http://wildpositron.wordpress.com/2011/04/05/calculating-the-area-of-a-sector/

73

[+3]
[2011-09-16 03:19:31]
euklid345

This is not entirely without words, but Byrne's edition of Euclid's elements has cut down the number of words to a bare minimum.

http://www.math.ubc.ca/~cass/Euclid/byrne.html

(1)
This is not quite in the spirit of the question... - ** Mariano Suárez-Álvarez**

(3)
+1: Thanks for this wonderful and beautiful link (be it in the spirit of the question or not). - ** Hans-Peter Stricker**

74

[+2]
[2010-12-28 02:42:47]
user11863

The composition of two continuous mappings is continuous.

(1)
Can you explain that? - ** M. Winter**

75

[+2]
[2011-10-01 23:31:18]
isomorphismes

From Wikipedia: here is a "proof without words" of the Yoneda Lemma.

(10)
This answer has already been proposed, and after some discussion it was more or less agreed that this is *not* a proof-without-words in standard sense of the term. - ** Mariano Suárez-Álvarez**

It's not a **proof without words** but a visualization (i.e. **statement without words**) of the Yoneda Lemma. (But it depends: for the experts it may count as a proof.) - ** Hans-Peter Stricker**

76

[+2]
[2012-02-15 00:00:13]
Roberto Mizzoni

For $0 \lt k \lt n$,

$$\binom{n}{k} = \frac{n}{n-k}\binom{n-1}{k}$$

How k-subsets of [n], marked dark green in the rows, come from k-subsets of [n-1] after n-fold duplication and rearrangement:

Exactly $n-k$ times:

By induction, a base case, and taking $k=n$ and $k=0$ for granted: $$\binom{n}{k} = \frac{n}{(n-k)} \frac{(n-1)!}{(n-1-k)!\ k!} = \frac{n!}{(n-k)!\ k!}$$

(1)
For me the standard argument is just as visual, and clearer: I have n!/(n-k)! ordered lists of length k with no repetitions from an alphabet of n letters. If I group together all words using the same letters, there are k! members of each group, hence n!/(n-k)!(k!) groups. Each group corresponds to an unordered list. - ** Steven Gubkin**

I find it just as visual, and very clean, but even harder to depict. - ** Roberto Mizzoni**

77

[-3]
[2014-09-11 07:24:46]
Mostafa Mirabi

A nice proof for trigonometric equation:

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

(15)
This is an exhibit of the fact, but it isn't really a *proof* - it doesn't explain why those two functions sum to 1, just shows (arguably, just claims) that they do. You could replace the curve with any function $f$ with $f(\pi/2)=1$. - ** Steven Stadnicki**

(2)
@StevenStadnicki In fact, I'm pretty sure that the function in the picture is not $\sin x$, the inflection point has a sharper third derivative than it should (although I'm sure this is just a limitation of the means by which the picture was drawn). There is certainly nothing geometrical constraining the shape of the diagram. - ** Mario Carneiro**

(10)
However, this could be a nice proof of $\int_{0}^{\pi/2}\sin^2 x\,\mathrm{d}x = \int_{0}^{\pi/2}\cos^2 x\,\mathrm{d}x = \frac{1}{2}\left(\frac{\pi}{2}\cdot 1\right)$ - ** Machinato**

(2)
Well this might be actually interpreted as a visualization of the fact that squaring a sine/cosine curve produces another (shifted up) sine/cosine curve, viz. $\cos^2(x)=\frac12(1+\cos(2x))$ and $\sin^2(x)=\frac12(1-\cos(2x))$; in this way it comes closer to be a proof of something. - ** მამუკა ჯიბლაძე**

78

[-6]
[2012-08-21 01:28:28]
Marc Chamberland

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 $~a^2 + b^2 + c^2 = ab + bc + ca$:

$$ $$

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

79

Carter Tazio SchonwaldSuvritSelene RoutleySelene RoutleySuvritGil KalaiDavid RobertsMariano Suárez-ÁlvarezDavid RobertsJoel David HamkinsRoberto Mizzonidohave an astonishing example.) -Hans-Peter Strickermatovitchuser56097goblin GONEMariano Suárez-Álvarezgoblin GONEgoblin GONEMariano Suárez-Álvarezmath137Martin SleziakDanielWainfleetDan DascalescuDaniel CharryEliethesaiyanFederico Poloniuser141368user141368user141368msh210