14 ms·

Garfield's proof of the Pythagorean Theorem

190 points by benbreen 10mo ago

WCSTombs 10mo ago
There are thousands of different proofs of the Pythagorean theorem, and some of them are really cool. The purely trigonometric proof that was found by some high school students recently is a great one. However, I think the greatest proof of all is this little gem that has been attributed to Einstein [1]. Take any right triangle. You can divide it into two non-overlapping right triangles that are both similar to the original triangle by dropping a perpendicular from the right angle to the hypotenuse. To see that the triangles are similar, you just compare interior angles. (It's better to leave that as an exercise than to describe it in words, but in any case, this is a very commonly known construction.) The areas of the two small triangles add up to the area of the big triangle, but the two small triangles have the two legs of the big triangle as their respective hypotenuses. Because area scales as the square of the similarity ratio (which I think is intuitively obvious), it follows that the squares of the legs' lengths must add up to the square of the hypotenuse's length, QED. It's really a perfect proof: it's simple, intuitive, as direct as possible, and it's pretty much impossible to forget. [1] https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99s%20Boyhood%20Proof%20of%20the%20Pythagorean%20Theorem%20-%20The%20New%20Yorker.pdf https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99...
- bryanrasmussen 10mo ago
  unfortunately doesn't work for me because of difficulty visualizing things, so I suppose there are probably a good number of people with the same problem. So I guess for one particular subset of the population it is difficult, impossible to understand, and because it cannot be understood it will not be remembered. Not complaining just noting the amusing thing that different explanations may have all sorts of problems with it. Although if there was a video of it I guess I would understand it then. Not sure if everyone with visualization issues would though.
  - WCSTombs 10mo ago
    To be fair, I'm constrained by plain text on Hacker News. The argument I wrote down requires a diagram to be fully understood, so I described it in words expecting the reader to draw it themselves, or at least mentally visualize it (for those used to doing it). To be clear though, as far as I know every proof of the Pythagorean Theorem requires some sort of diagram, and the one I gave requires literally the least amount of drawing out of all the proofs (which is a bold claim, but call it a conjecture). That's why I felt comfortable writing out the proof just in words.
- zeroonetwothree 10mo ago
  This proof assumes that the area a triangle is some function k c^2 of the hypotenuse c where k is constant for similar triangles. This doesn’t seem super obvious to me, and it’s a bit more than just assuming area scales with the square of hypotenuse length, it indeed needs to be a constant fraction. To me that truth isn’t necessarily any less fundamental than the Pythagorean theorem itself. But to each their own. BTW Terrence Tao has a write up of this proof as well: https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/ https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem...
  - fiso64 10mo ago
    I don't get his "modern" proof. Specifically the step where he says "it's easy to see geometrically that these matrices differ by a rotation" seems to be doing a lot of heavy lifting. The first matrix transforms e1 to (a,-b), the second scales e1 to (c,0). If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.
    - degamad 10mo ago
      You're assuming that we know that the length of vector (a, -b) is a²+b². We don't know that. We start by assuming that the position vector (a, -b) has length c. This implies that we can rotate that vector until it becomes the position vector (c, 0). As you note, we can create the two vectors above from (1, 0) using linear transformation matrices [(a, b), (-b, a)] and [(c, 0), (0, c)] So we could create the position vector (c, 0) by starting at (1, 0), applying the linear transformation [(a, b), (-b, a)], then applying a rotation to bring it back to the e1 axis. Thus for some rotation matrix R, R × [(a, b), (-b, a)] = [(c, 0), (0, c)] The determinant of a rotation matrix is 1, so the determinant of the left side is 1×(a²+b²), while the determinant of the right side is c², which is how we end up with a²+b²=c². Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already.
      - 10mo ago
- zeroonetwothree 10mo ago
  I think proof #6 on this page is easier to follow and uses the same similar triangles. But then it’s just some basic algebra without assuming anything about areas of similar triangles :) https://www.cut-the-knot.org/pythagoras/index.shtml#6 https://www.cut-the-knot.org/pythagoras/index.shtml#6
  - WCSTombs 10mo ago
    That one's also very neat!
- dataflow 10mo ago
  > it follows that... "Now just draw the rest of the owl." Does it not feel like you skipped something here? The areas add up and area scales quadratically, therefore... Pythagorean Theorem? It definitely is not clear how this follows, even after the questionable assumption that it's obvious area scales quadratically.
  - codethief 10mo ago
    He didn't skip anything but he left the "obvious" details (for a mathematician) to the reader: Let C be the area of the big triangle, A and B be the areas of the two small triangles. By construction we know that C = A + B. Moreover, a, b, c are the hypotenuses of the triangles A, B and C. The area scaling quadratically with the similarity ratio means that A = (a/c)² C, and B = (b/c)² C. Now, plug this into A + B = C, cancel C, rearrange.
    - dataflow 10mo ago
      The math is obvious enough, I agree. But the description of the approach feels like it's lacking something - specifically, something along the lines of "now write down the scaling equations and simplify the area summation." I feel like it's not at all clear they're switching to an algebraic argument there.
      - WCSTombs 10mo ago
        The thing is that in my head there is no algebraic argument: we go from (1) similarity ratios being A:B:C and (2) the first two areas adding up to the third area, straight to the conclusion of A^2 + B^2 = C^2. I think your point about a step being missing here is valid, but when I search my intuition, it's still not coming up as algebraic. I suspect this is the same for others like me who are inclined to think geometrically, but I'd like to hear their opinions. Here's an attempt at filling in the geometric intuition with something more concrete. You know how it's common to visualize the theorem with squares on the three sides of the triangle and saying that the two small squares add up to the big one? And then everyone stares at it and says "huh?" because that fact is far from obvious from that diagram. Here's the thing though, we're free to choose different area units if we want. So just choose units where our triangle itself with a given hypotenuse H has area H^2 units. Then we can give the argument above without any extra factors and cancellations. To fully justify the "choose any units," you do need to check that it's logically consistent, which you could say is more missing steps, but I think this idea is far more fundamental than the Pythagorean Theorem. Our use of squares to define the fundamental units of area really is a completely arbitrary choice. We call them "square units," which already biases us to think of area in a specific way, but there's absolutely no reason we can't use any other shape. Of course squares are convenient because you can stack them up neatly and count them, but that doesn't seem to be helpful at all in this context, so it's natural to choose something else.
- card_zero 10mo ago
  Einstein's proof relies on the fact that the theorem works with any shape, not just squares, such as pentagons: https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentagons.svg https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentag... Or any arbitrary vector graphics, like Einstein's face. So in the proof, the shape on the hypotenuse is the same as the original triangle, and on the other two sides there are two smaller versions of it, which when joined have the same area (and shape) as the big one. Fair enough. However, none of the hundreds or thousands of proofs explain it. They all prove it, like by saying "this goes here, that goes there, this is the same as that, therefore logically you're stupid," but it still seems like weird magic to me. Some explanation is missing.
  - chronial 10mo ago
    Draw a square around Einstein's face. Call the side length of the square a and the area of the square A. We have A=a^2. Einstein takes up some portion p < 1 of that area, so Einstein has area E = pA. Now we scale the whole thing by factor f. So the new square has side lengths fa, and thus area A' = (fa)^2 = f^2×a^2 = f^2×A. Since the relative portion the face takes up doesn't change with scaling, the face now has size pA' = p×f^2×A = f^2 × pA = f^2 E. Does that help or was that not the part you were missing?
    - card_zero 10mo ago
      No, that part is fine: I'm happy with the fact that it works with arbitrary shapes. What bothers me is that the area on the hypotenuse is equal to the sum of the areas on the other two sides, when the triangle has a right angle. This somewhat like saying that I'm troubled by the fact that 1+1=2, I know. But that's a potentially distracting sidetrack, let's not get into that one.
  - lupire 10mo ago
    What definition of area are you using in the first place, for non-swuare objects? Most people find area intuitive and informal, but if you describe area formally, it should be easy to use your definition to account for scaling.
- JanisErdmanis 10mo ago
  Indeed, a wonderful proof. It does, though, make one implicit assumption that if one stretches the fabric by the same amount, all holes in it stretch by the same amount. In particular, it assumes that triangle stretching is size-independent. Perhaps there are fabrics where that is not true...
  - lupire 10mo ago
    You mean non-Euclidean fabric? https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-geometry/ https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-ge... https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://en.wikipedia.org/wiki/Minkowski_distance https://en.wikipedia.org/wiki/Minkowski_distance
    - JanisErdmanis 10mo ago
      That is one possibility. Probably the only one, though.
  - layer8 10mo ago
    The Pythagorean theorem is only true in Euclidian space, where that “stretching” assumption is true. So you are right about there being assumptions, and indeed they are imposing limitations to the applicability of the theorem.
- lupire 10mo ago
  > The purely trigonometric proof that was found by some high school students recently is a great one. It was geometric, using trigonometric vocabulary.
  - SJC_Hacker 10mo ago
    Trig relies on some geometric assumptions. The definition of sin and cos is going to require a right triangle, for instance,
- SJC_Hacker 10mo ago
  > The purely trigonometric proof that was found by some high school students recently is a great one. I failed to understand what was so cool about that proof. It relied on concepts such as Cartesian coordinate systems, and the measure of an angle (not just a pure geometric concept), and even concepts like convergence of infinite sums, which weren't purely geometric. Geometry had been formalized in the 20th century and had moved past informal proofs
- globalnode 10mo ago
  That is interesting and made me think. Only after following some of the other subcomments did I manage to understand it. Personally, replacing the word similarity ratio with scale factor made all the difference. At first I thought it was a circular argument, relying on pythag to prove pythag but that scale factor is the key actually, and the fact that side lengths scale linearly but the area scales quadratically. It feels like a similar trick we see when adding logarithms gives us multiplication.
  - WCSTombs 10mo ago
    Yeah, the similarity ratio/scale factor and its connection to area is the key part of the proof. Sorry, that could have been clearer.
    - globalnode 10mo ago
      All good! thanks for the maths though, just when I think I understand it... I dont! Its one of those love hate relationships, I love it, it hates me.
- layer8 10mo ago
  Also, iterating that triangular subdivision gives rise to the Pythagorean tree fractal: https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal) https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal)
b800h 10mo ago
I was ready for it to involve lasagna.
- esher 10mo ago
  Same!
- Lio 10mo ago
  There's no feline involvement but there is a good proof of Pythagorean theorem in Neil Stephenson's Anathem[1] demonstrated by deviding cake. 1. https://en.wikipedia.org/wiki/Anathem https://en.wikipedia.org/wiki/Anathem
- hmokiguess 10mo ago
  Glad it wasn’t just me hahaha
vee-kay 10mo ago
AFAIK: Pythagoras never wrote about this Triangle Theorem. There's no proof that he ever even knew about it. But he had mandated to his Pythagorean school (students) that any discovery or invention they made would be attributed to him instead. The earliest known mention of Pythagoras's name in connection with the theorem occurred five centuries after his death, in the writings of Cicero and Plutarch. Interestingly: the Triangle Theorem was discovered, known and used by the ancient Indians and ancient Babylonians & Egyptians long before the ancient Greeks came to know about it. India's ancient temples are built using this theorem, India's mathematician Boudhyana (c. ~800 BCE) wrote about it in his Baudhayana Shulba (Shulva) Sutras around 800 BCE, the Egyptian pharoahs built the pyramids using this triangle theorem. Baudhāyana, (fl. c. 800 BCE) was the author of the Baudhayana sūtras, which cover dharma, daily ritual, mathematics, etc. He belongs to the Yajurveda school, and is older than the other sūtra author Āpastambha. He was the author of the earliest Sulba Sūtra—appendices to the Vedas giving rules for the construction of altars—called the Baudhāyana Śulbasûtra. These are notable from the point of view of mathematics, for containing several important mathematical results, including giving a value of pi to some degree of precision, and stating a version of what is now known as the Pythagorean theorem. Source: http://en.wikipedia.org/wiki/Baudhayana http://en.wikipedia.org/wiki/Baudhayana Baudhyana lived and wrote such incredible mathematical insights several centuries before Pythagoras. Note that Baudhayana Shulba Sutra not only gives a statement of the Triangle Theorem, it also gives proof of it. There is a difference between discovering Pythagorean triplets (ex 6:8:10) and proving the Pythagorean theorem (a2 + b2 = c2 ). Ancient Babylonians accomplished only the former, whereas ancient Indians accomplished both. Specifically, Baudhayana gives a geometrical proof of the triangle theorem for an isosceles right triangle. The four major Shulba Sutras, which are mathematically the most significant, are those attributed to Baudhayana, Manava, Apastamba and Katyayana. Refer to: Boyer, Carl B. (1991). A History of Mathematics (Second ed.), John Wiley & Sons. ISBN 0-471-54397-7. Boyer (1991), p. 207, says: "We find rules for the construction of right angles by means of triples of cords the lengths of which form Pythagorean triages, such as 3, 4, and 5, or 5, 12, and 13, or 8, 15, and 17, or 12, 35, and 37. However all of these triads are easily derived from the old Babylonian rule; hence, Mesopotamian influence in the Sulvasutras is not unlikely. Aspastamba knew that the square on the diagonal of a rectangle is equal to the sum of the squares on the two adjacent sides, but this form of the Pythagorean theorem also may have been derived from Mesopotamia. ... So conjectural are the origin and period of the Sulbasutras that we cannot tell whether or not the rules are related to early Egyptian surveying or to the later Greek problem of altar doubling. They are variously dated within an interval of almost a thousand years stretching from the eighth century B.C. to the second century of our era."
wunderlust 10mo ago
American presidents used to be smart.
- xeonmc 10mo ago
  they also use to be bullet magnets
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- deleted 10mo ago
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einpoklum 10mo ago
That looks like "half" of the proof using a square: https://www.onlinemathlearning.com/image-files/xpythagorean-theorem-proof.png.pagespeed.ic.cxJaOlvF7a.png https://www.onlinemathlearning.com/image-files/xpythagorean-... where you draw three extra triangles, not just one, and they surround a square of c x c. Think about it as making two copies of the trapezoid, one rotated on top of the other.
- sorokod 10mo ago
  I shared this one with my son, the step where the 2ab expressions cancel out gave him a little aha moment.
- procrastitron 10mo ago
  This was exactly what I thought of as well. Garfield’s version seems more complicated since you have to calculate the area of a trapezoid instead of the area of a square, but conceptually they are the same.
makmende 10mo ago
Netflix recently released a mini-series on Garfield's election and presidency: https://www.themoviedb.org/tv/245219-death-by-lightning https://www.themoviedb.org/tv/245219-death-by-lightning.
dataflow 10mo ago
Not to bash the former president, but I'm failing to see what's so clever or nice about the proof... could someone please explain if I'm missing something? If you're going solve it with algebra on top of the similar triangles and geometry anyway, why complicate it so much? Why not just drop the height h and be done with it? You have 2 a b = 2 c h, c1/a = h/b, c2/b = h/a, c = c1 + c2, so just solve for h and c1 and c2 and simplify. So why would you go through the trouble of introducing an extra point outside the diagram, drawing an extra triangle, proving that you get a trapezoid, assuming you know the formula for the area of a trapezoid, then solving the resulting equations...? Is there any advantage at all to doing this? It seems to make strictly more assumptions and be strictly more complicated, and it doesn't seem to be any easier to see, or to convey any sort of new intuition... does it?
- da_chicken 10mo ago
  I can't follow your reasoning at at. "Drop the height h" is completely ambiguous. And the nice thing about Garfield's proof is that all it requires that you know is the area of a right triangle and the basic Euclidean premises. You can easily get the area of a trapezoid from that.
  - dataflow 10mo ago
    > I can't follow your reasoning at at. "Drop the height h" is completely ambiguous. I'm referring to the classic proof where you drop the height perpendicularly to the hypotenuse from the opposite corner.
rcarmo 10mo ago
I must confess I clicked through hoping to see a comic of Garfield the cat using pizza slices to approximate right triangles.
- emigre 10mo ago
  Me too...
- RobotToaster 10mo ago
  I hate triangular Mondays
- medwards666 10mo ago
  This was also similar to my first thought. Damn. I want pizza now as well...
  - monocularvision 10mo ago
    I believe you mean lasagna.
    - deleted 10mo ago
      [deleted]
    - parpfish 10mo ago
      pizza is portable lasagna.
nullbyte808 10mo ago
He was also the 20th president of the USA.
- kingofmen 10mo ago
  "Better known for other work".
charlieyu1 10mo ago
This is actually one of the most well-known proof
BitsAndObjects 10mo ago
Imagine having a president with the intellectual ability to create a novel mathematical proof, and the humility to publish it without claiming to be the greatest mathematician of all time…
- adventured 10mo ago
  If we had a president that could create novel mathematic proofs, I'd be happy to tolerate the arrogance at this point. At least there'd be some substance there.
  - YesThatTom2 10mo ago
    I hope you would but… The only intellectual that served as president in the last 100 years was Obama and we(1) shit all over him every day for “atrocities” such as wearing a brown suit. (1) not me and probably not you but it was a national talking point for weeks
    - isoprophlex 10mo ago
      I thought that was quite a fashionable suit, and wish I could pull off wearing something like that
    - BigTTYGothGF 10mo ago
      If you bump that up to 110 years you bring in Woodrow Wilson, one of the all-time worst presidents.
      - rflrob 10mo ago
        I’m curious how you came to that conclusion. While he’s certainly not in the pantheon of best presidents, he ends up around the 75th percentile in rankings by historians. Even subtracting a few spots, he’s nowhere near close to “one of the all-time worst“. Or are you faulting him for not resigning when incapacitated by a stroke? https://en.wikipedia.org/wiki/Historical_rankings_of_presidents_of_the_United_States https://en.wikipedia.org/wiki/Historical_rankings_of_preside...
frostyel 10mo ago
It's as good as Piers Morgans legendary pythagorean theorem: https://www.youtube.com/watch?v=QZWS2g-fEAU https://www.youtube.com/watch?v=QZWS2g-fEAU
bombcar 10mo ago
There’s a sci-fi/time travel thriller here where he was assassinated because of his mathematical prowess.
NoNameHaveI 10mo ago
Fun fact: Garfield LOVES lasagna, and hates Mondays. Oh. Wait!
russfink 10mo ago
May be a repeat here, but best proof I saw was inscribe a square with sides of length c inside another square, but rotated such that the interior square’s corners intersect the outer square’s edges. The intersecting points divide the outer square’s edge making lengths a and b. This produces an inner square’s edge with sides length c and four equal right triangles of sides a, b, and c. Note that the area of the outer square equals the sum of the inner square plus the area of the four triangles. Solve this equality.
zahlman 10mo ago
We can imagine another copy of the trapezoid, rotated 180 degrees and situated on top; the pair of them create a square with side lengths of a + b. This cancels all the 1/2s out of Garfield's equations, and also makes the result more geometrically obvious: the entire square (a + b)^2 = a^2 + 2ab + b^2 is the inscribed square c^2 plus four copies of the original triangle 4 * ab/2 = 2ab. This then becomes a restatement of another classic proof (the simple algebraic proof given near the top of the main Wikipedia page for the theorem). So we can imagine Garfield discovering this approach by cutting that diagram (https://en.wikipedia.org/wiki/Pythagorean_theorem#/media/File:Animated_gif_version_of_SVG_of_rearrangement_proof_of_Pythagorean_theorem.gif https://en.wikipedia.org/wiki/Pythagorean_theorem#/media/Fil...) in half and describing a different way to construct it.
- deleted 10mo ago
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- torginus 10mo ago
  Yeah this is the high-school proof, just cut in half.
  - amelius 10mo ago
    Maybe the high school proof was inspired by Garfield?
    - torginus 10mo ago
      Afaik the ancient proof (the square inside a bigger square rotated by 45 degrees) comes from ancient Greece, so I guess Garfield can add 'time travel' to his list of discoveries.
surprisetalk 10mo ago
They assassinated him because he uncovered too much forbidden knowledge about the triangles
- SJC_Hacker 10mo ago
  The Triluminati
kingofmen 10mo ago
I'm having some trouble with this part of the explanation: > From the figure, one can easily see that the triangles ABC and BDE are congruent. I must confess I do not easily see this. It's been a long time since I did any geometry, could someone help me out? I'm probably forgetting some trivial fact about triangles.
- istjohn 10mo ago
  It wasn't obvious to me either. But we know the angles ABC, ABD, and DBE equal 180 degrees, as do the interior angles of triangle ABC. From that we can deduce that angle BAC = angle DBE, from which it follows that angle ABC = angle BDE.
- desertrider12 10mo ago
  We know angle EBD equals BAC, since the sum of triangle ABC's interior angles is 180 degrees and the sum of the 3 angles at B are also 180 degrees. We also know angle DEB is 90 degrees since DE was constructed to be perpendicular to CB. Finally, D was placed at a distance c from B. The two triangles have the same angles and the same side lengths opposite the right angles, so they must be congruent.
- da_chicken 10mo ago
  So, the line BE is just the line CB extended. It's the same line. And we know that the angles of a triangle add up to 180. And we know that the line BD is defined as perpendicular to AB. That means the angle ABC and angle DBE must add up to 90. But that's also true of the angles ABC and angle CAB. That means that angle DBE and angle CAB must be the same. Both triangles ABC and BDE are both right triangles, so that means angles ABC and BDE are the same. So they're similar triangles: They have all the same angles. Additionally, the point D is just at a point so that the length of line segment BD and the length of line segment AB are both the same: c. Since we know that the hypotenuse of triangle ABC is c, and the hypotenuse of triangle BDE is also c, and we know they're both similar triangles, then these triangles must be congruent as well.
  - kingofmen 10mo ago
waldrews 10mo ago
Garfield was in many ways the most personally appealing and brilliant of the American presidents, rising from poverty and obscurity by being absurdly talented across many fields and eloquent. He was assassinated early and barely got to serve. The story of his life, the shooting, and the subsequent medical drama (featuring even a cameo by Alexander Graham Bell improvising a diagnostic device) are so epic you have to wonder if time travelers are messing with us. His legacy was the nonpartisan professional civil service, a key part of his agenda that his successor felt obligated to carry out, an accomplishment that recently came under particularly heavy attack. Netflix just came out with the miniseries about him, 'Death by Lightning,' based on the book 'Destiny of the Republic.' His earlier life is featured prominently in '1861: The Civil War Awakening' by Adam Goodheart. There are a few great C-SPAN/Book TV videos by some of the authors that tell the story concisely and convey why some of us are so fascinated by that history.
- Apocryphon 10mo ago
  The Netflix miniseries was very funny and engaging, but I can’t help but to think that it over-valorizes Garfield as some sort of fallen benevolent sage-king in the same way Oliver Stone’s JFK and other Camelot hagiographies do. The man was not above the corrupt politics of the day, at least earlier in his career, after all.
amelius 10mo ago
I sometimes wonder what mathematics and physics would have looked like if the Pythagorean theorem was a really ugly formula, or something you couldn't write in closed form.
nvlled 10mo ago
A side thought, how would a natural number sequence of hypotenuses that satisfies the equation a^2 + b^2 = c^2 look like? Or what interesting properties would it have if any.
- Joker_vD 10mo ago
  You mean, what does sequence A004431 [0] look like? Well, like itself. Some of its properties are also listed at that link. [0] https://oeis.org/A004431 https://oeis.org/A004431
nottorp 10mo ago
Oh. It’s not that cat.
tug2024 10mo ago
[dead]

190 points by benbreen 10mo ago

There are thousands of different proofs of the Pythagorean theorem, and some of them are really cool. The purely trigonometric proof that was found by some high school students recently is a great one. However, I think the greatest proof of all is this little gem that has been attributed to Einstein [1]. Take any right triangle. You can divide it into two non-overlapping right triangles that are both similar to the original triangle by dropping a perpendicular from the right angle to the hypotenuse. To see that the triangles are similar, you just compare interior angles. (It's better to leave that as an exercise than to describe it in words, but in any case, this is a very commonly known construction.) The areas of the two small triangles add up to the area of the big triangle, but the two small triangles have the two legs of the big triangle as their respective hypotenuses. Because area scales as the square of the similarity ratio (which I think is intuitively obvious), it follows that the squares of the legs' lengths must add up to the square of the hypotenuse's length, QED. It's really a perfect proof: it's simple, intuitive, as direct as possible, and it's pretty much impossible to forget. [1] https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99s%20Boyhood%20Proof%20of%20the%20Pythagorean%20Theorem%20-%20The%20New%20Yorker.pdf https://paradise.caltech.edu/ist4/lectures/Einstein%E2%80%99...

bryanrasmussen 10mo ago
unfortunately doesn't work for me because of difficulty visualizing things, so I suppose there are probably a good number of people with the same problem. So I guess for one particular subset of the population it is difficult, impossible to understand, and because it cannot be understood it will not be remembered. Not complaining just noting the amusing thing that different explanations may have all sorts of problems with it. Although if there was a video of it I guess I would understand it then. Not sure if everyone with visualization issues would though.
- WCSTombs 10mo ago
  To be fair, I'm constrained by plain text on Hacker News. The argument I wrote down requires a diagram to be fully understood, so I described it in words expecting the reader to draw it themselves, or at least mentally visualize it (for those used to doing it). To be clear though, as far as I know every proof of the Pythagorean Theorem requires some sort of diagram, and the one I gave requires literally the least amount of drawing out of all the proofs (which is a bold claim, but call it a conjecture). That's why I felt comfortable writing out the proof just in words.
zeroonetwothree 10mo ago
This proof assumes that the area a triangle is some function k c^2 of the hypotenuse c where k is constant for similar triangles. This doesn’t seem super obvious to me, and it’s a bit more than just assuming area scales with the square of hypotenuse length, it indeed needs to be a constant fraction. To me that truth isn’t necessarily any less fundamental than the Pythagorean theorem itself. But to each their own. BTW Terrence Tao has a write up of this proof as well: https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/ https://terrytao.wordpress.com/2007/09/14/pythagoras-theorem...
- fiso64 10mo ago
  I don't get his "modern" proof. Specifically the step where he says "it's easy to see geometrically that these matrices differ by a rotation" seems to be doing a lot of heavy lifting. The first matrix transforms e1 to (a,-b), the second scales e1 to (c,0). If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.
  - degamad 10mo ago
    You're assuming that we know that the length of vector (a, -b) is a²+b². We don't know that. We start by assuming that the position vector (a, -b) has length c. This implies that we can rotate that vector until it becomes the position vector (c, 0). As you note, we can create the two vectors above from (1, 0) using linear transformation matrices [(a, b), (-b, a)] and [(c, 0), (0, c)] So we could create the position vector (c, 0) by starting at (1, 0), applying the linear transformation [(a, b), (-b, a)], then applying a rotation to bring it back to the e1 axis. Thus for some rotation matrix R, R × [(a, b), (-b, a)] = [(c, 0), (0, c)] The determinant of a rotation matrix is 1, so the determinant of the left side is 1×(a²+b²), while the determinant of the right side is c², which is how we end up with a²+b²=c². Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already.
    - 10mo ago
zeroonetwothree 10mo ago
I think proof #6 on this page is easier to follow and uses the same similar triangles. But then it’s just some basic algebra without assuming anything about areas of similar triangles :) https://www.cut-the-knot.org/pythagoras/index.shtml#6 https://www.cut-the-knot.org/pythagoras/index.shtml#6
- WCSTombs 10mo ago
  That one's also very neat!
dataflow 10mo ago
> it follows that... "Now just draw the rest of the owl." Does it not feel like you skipped something here? The areas add up and area scales quadratically, therefore... Pythagorean Theorem? It definitely is not clear how this follows, even after the questionable assumption that it's obvious area scales quadratically.
- codethief 10mo ago
  He didn't skip anything but he left the "obvious" details (for a mathematician) to the reader: Let C be the area of the big triangle, A and B be the areas of the two small triangles. By construction we know that C = A + B. Moreover, a, b, c are the hypotenuses of the triangles A, B and C. The area scaling quadratically with the similarity ratio means that A = (a/c)² C, and B = (b/c)² C. Now, plug this into A + B = C, cancel C, rearrange.
  - dataflow 10mo ago
    The math is obvious enough, I agree. But the description of the approach feels like it's lacking something - specifically, something along the lines of "now write down the scaling equations and simplify the area summation." I feel like it's not at all clear they're switching to an algebraic argument there.
    - WCSTombs 10mo ago
      The thing is that in my head there is no algebraic argument: we go from (1) similarity ratios being A:B:C and (2) the first two areas adding up to the third area, straight to the conclusion of A^2 + B^2 = C^2. I think your point about a step being missing here is valid, but when I search my intuition, it's still not coming up as algebraic. I suspect this is the same for others like me who are inclined to think geometrically, but I'd like to hear their opinions. Here's an attempt at filling in the geometric intuition with something more concrete. You know how it's common to visualize the theorem with squares on the three sides of the triangle and saying that the two small squares add up to the big one? And then everyone stares at it and says "huh?" because that fact is far from obvious from that diagram. Here's the thing though, we're free to choose different area units if we want. So just choose units where our triangle itself with a given hypotenuse H has area H^2 units. Then we can give the argument above without any extra factors and cancellations. To fully justify the "choose any units," you do need to check that it's logically consistent, which you could say is more missing steps, but I think this idea is far more fundamental than the Pythagorean Theorem. Our use of squares to define the fundamental units of area really is a completely arbitrary choice. We call them "square units," which already biases us to think of area in a specific way, but there's absolutely no reason we can't use any other shape. Of course squares are convenient because you can stack them up neatly and count them, but that doesn't seem to be helpful at all in this context, so it's natural to choose something else.
card_zero 10mo ago
Einstein's proof relies on the fact that the theorem works with any shape, not just squares, such as pentagons: https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentagons.svg https://commons.wikimedia.org/wiki/File:Pythagoras_by_pentag... Or any arbitrary vector graphics, like Einstein's face. So in the proof, the shape on the hypotenuse is the same as the original triangle, and on the other two sides there are two smaller versions of it, which when joined have the same area (and shape) as the big one. Fair enough. However, none of the hundreds or thousands of proofs explain it. They all prove it, like by saying "this goes here, that goes there, this is the same as that, therefore logically you're stupid," but it still seems like weird magic to me. Some explanation is missing.
- chronial 10mo ago
  Draw a square around Einstein's face. Call the side length of the square a and the area of the square A. We have A=a^2. Einstein takes up some portion p < 1 of that area, so Einstein has area E = pA. Now we scale the whole thing by factor f. So the new square has side lengths fa, and thus area A' = (fa)^2 = f^2×a^2 = f^2×A. Since the relative portion the face takes up doesn't change with scaling, the face now has size pA' = p×f^2×A = f^2 × pA = f^2 E. Does that help or was that not the part you were missing?
  - card_zero 10mo ago
    No, that part is fine: I'm happy with the fact that it works with arbitrary shapes. What bothers me is that the area on the hypotenuse is equal to the sum of the areas on the other two sides, when the triangle has a right angle. This somewhat like saying that I'm troubled by the fact that 1+1=2, I know. But that's a potentially distracting sidetrack, let's not get into that one.
- lupire 10mo ago
  What definition of area are you using in the first place, for non-swuare objects? Most people find area intuitive and informal, but if you describe area formally, it should be easy to use your definition to account for scaling.
JanisErdmanis 10mo ago
Indeed, a wonderful proof. It does, though, make one implicit assumption that if one stretches the fabric by the same amount, all holes in it stretch by the same amount. In particular, it assumes that triangle stretching is size-independent. Perhaps there are fabrics where that is not true...
- lupire 10mo ago
  You mean non-Euclidean fabric? https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-geometry/ https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-ge... https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/ https://en.wikipedia.org/wiki/Minkowski_distance https://en.wikipedia.org/wiki/Minkowski_distance
  - JanisErdmanis 10mo ago
    That is one possibility. Probably the only one, though.
- layer8 10mo ago
  The Pythagorean theorem is only true in Euclidian space, where that “stretching” assumption is true. So you are right about there being assumptions, and indeed they are imposing limitations to the applicability of the theorem.
lupire 10mo ago
> The purely trigonometric proof that was found by some high school students recently is a great one. It was geometric, using trigonometric vocabulary.
- SJC_Hacker 10mo ago
  Trig relies on some geometric assumptions. The definition of sin and cos is going to require a right triangle, for instance,
SJC_Hacker 10mo ago
> The purely trigonometric proof that was found by some high school students recently is a great one. I failed to understand what was so cool about that proof. It relied on concepts such as Cartesian coordinate systems, and the measure of an angle (not just a pure geometric concept), and even concepts like convergence of infinite sums, which weren't purely geometric. Geometry had been formalized in the 20th century and had moved past informal proofs
globalnode 10mo ago
That is interesting and made me think. Only after following some of the other subcomments did I manage to understand it. Personally, replacing the word similarity ratio with scale factor made all the difference. At first I thought it was a circular argument, relying on pythag to prove pythag but that scale factor is the key actually, and the fact that side lengths scale linearly but the area scales quadratically. It feels like a similar trick we see when adding logarithms gives us multiplication.
- WCSTombs 10mo ago
  Yeah, the similarity ratio/scale factor and its connection to area is the key part of the proof. Sorry, that could have been clearer.
  - globalnode 10mo ago
    All good! thanks for the maths though, just when I think I understand it... I dont! Its one of those love hate relationships, I love it, it hates me.
layer8 10mo ago
Also, iterating that triangular subdivision gives rise to the Pythagorean tree fractal: https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal) https://en.wikipedia.org/wiki/Pythagoras_tree_(fractal)

b800h 10mo ago

I was ready for it to involve lasagna.

esher 10mo ago
Same!
Lio 10mo ago
There's no feline involvement but there is a good proof of Pythagorean theorem in Neil Stephenson's Anathem[1] demonstrated by deviding cake. 1. https://en.wikipedia.org/wiki/Anathem https://en.wikipedia.org/wiki/Anathem
hmokiguess 10mo ago
Glad it wasn’t just me hahaha

vee-kay 10mo ago

AFAIK: Pythagoras never wrote about this Triangle Theorem. There's no proof that he ever even knew about it. But he had mandated to his Pythagorean school (students) that any discovery or invention they made would be attributed to him instead. The earliest known mention of Pythagoras's name in connection with the theorem occurred five centuries after his death, in the writings of Cicero and Plutarch. Interestingly: the Triangle Theorem was discovered, known and used by the ancient Indians and ancient Babylonians & Egyptians long before the ancient Greeks came to know about it. India's ancient temples are built using this theorem, India's mathematician Boudhyana (c. ~800 BCE) wrote about it in his Baudhayana Shulba (Shulva) Sutras around 800 BCE, the Egyptian pharoahs built the pyramids using this triangle theorem. Baudhāyana, (fl. c. 800 BCE) was the author of the Baudhayana sūtras, which cover dharma, daily ritual, mathematics, etc. He belongs to the Yajurveda school, and is older than the other sūtra author Āpastambha. He was the author of the earliest Sulba Sūtra—appendices to the Vedas giving rules for the construction of altars—called the Baudhāyana Śulbasûtra. These are notable from the point of view of mathematics, for containing several important mathematical results, including giving a value of pi to some degree of precision, and stating a version of what is now known as the Pythagorean theorem. Source: http://en.wikipedia.org/wiki/Baudhayana http://en.wikipedia.org/wiki/Baudhayana Baudhyana lived and wrote such incredible mathematical insights several centuries before Pythagoras. Note that Baudhayana Shulba Sutra not only gives a statement of the Triangle Theorem, it also gives proof of it. There is a difference between discovering Pythagorean triplets (ex 6:8:10) and proving the Pythagorean theorem (a2 + b2 = c2 ). Ancient Babylonians accomplished only the former, whereas ancient Indians accomplished both. Specifically, Baudhayana gives a geometrical proof of the triangle theorem for an isosceles right triangle. The four major Shulba Sutras, which are mathematically the most significant, are those attributed to Baudhayana, Manava, Apastamba and Katyayana. Refer to: Boyer, Carl B. (1991). A History of Mathematics (Second ed.), John Wiley & Sons. ISBN 0-471-54397-7. Boyer (1991), p. 207, says: "We find rules for the construction of right angles by means of triples of cords the lengths of which form Pythagorean triages, such as 3, 4, and 5, or 5, 12, and 13, or 8, 15, and 17, or 12, 35, and 37. However all of these triads are easily derived from the old Babylonian rule; hence, Mesopotamian influence in the Sulvasutras is not unlikely. Aspastamba knew that the square on the diagonal of a rectangle is equal to the sum of the squares on the two adjacent sides, but this form of the Pythagorean theorem also may have been derived from Mesopotamia. ... So conjectural are the origin and period of the Sulbasutras that we cannot tell whether or not the rules are related to early Egyptian surveying or to the later Greek problem of altar doubling. They are variously dated within an interval of almost a thousand years stretching from the eighth century B.C. to the second century of our era."

wunderlust 10mo ago

American presidents used to be smart.

xeonmc 10mo ago
they also use to be bullet magnets
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einpoklum 10mo ago

That looks like "half" of the proof using a square: https://www.onlinemathlearning.com/image-files/xpythagorean-theorem-proof.png.pagespeed.ic.cxJaOlvF7a.png https://www.onlinemathlearning.com/image-files/xpythagorean-... where you draw three extra triangles, not just one, and they surround a square of c x c. Think about it as making two copies of the trapezoid, one rotated on top of the other.

sorokod 10mo ago
I shared this one with my son, the step where the 2ab expressions cancel out gave him a little aha moment.
procrastitron 10mo ago
This was exactly what I thought of as well. Garfield’s version seems more complicated since you have to calculate the area of a trapezoid instead of the area of a square, but conceptually they are the same.

makmende 10mo ago

Netflix recently released a mini-series on Garfield's election and presidency: https://www.themoviedb.org/tv/245219-death-by-lightning https://www.themoviedb.org/tv/245219-death-by-lightning.

dataflow 10mo ago

Not to bash the former president, but I'm failing to see what's so clever or nice about the proof... could someone please explain if I'm missing something? If you're going solve it with algebra on top of the similar triangles and geometry anyway, why complicate it so much? Why not just drop the height h and be done with it? You have 2 a b = 2 c h, c1/a = h/b, c2/b = h/a, c = c1 + c2, so just solve for h and c1 and c2 and simplify. So why would you go through the trouble of introducing an extra point outside the diagram, drawing an extra triangle, proving that you get a trapezoid, assuming you know the formula for the area of a trapezoid, then solving the resulting equations...? Is there any advantage at all to doing this? It seems to make strictly more assumptions and be strictly more complicated, and it doesn't seem to be any easier to see, or to convey any sort of new intuition... does it?

da_chicken 10mo ago
I can't follow your reasoning at at. "Drop the height h" is completely ambiguous. And the nice thing about Garfield's proof is that all it requires that you know is the area of a right triangle and the basic Euclidean premises. You can easily get the area of a trapezoid from that.
- dataflow 10mo ago
  > I can't follow your reasoning at at. "Drop the height h" is completely ambiguous. I'm referring to the classic proof where you drop the height perpendicularly to the hypotenuse from the opposite corner.

rcarmo 10mo ago

I must confess I clicked through hoping to see a comic of Garfield the cat using pizza slices to approximate right triangles.

emigre 10mo ago
Me too...
RobotToaster 10mo ago
I hate triangular Mondays
medwards666 10mo ago
This was also similar to my first thought. Damn. I want pizza now as well...
- monocularvision 10mo ago
  I believe you mean lasagna.
  - deleted 10mo ago
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  - parpfish 10mo ago
    pizza is portable lasagna.

nullbyte808 10mo ago

He was also the 20th president of the USA.

kingofmen 10mo ago
"Better known for other work".

charlieyu1 10mo ago

This is actually one of the most well-known proof

BitsAndObjects 10mo ago

Imagine having a president with the intellectual ability to create a novel mathematical proof, and the humility to publish it without claiming to be the greatest mathematician of all time…

adventured 10mo ago
If we had a president that could create novel mathematic proofs, I'd be happy to tolerate the arrogance at this point. At least there'd be some substance there.
- YesThatTom2 10mo ago
  I hope you would but… The only intellectual that served as president in the last 100 years was Obama and we(1) shit all over him every day for “atrocities” such as wearing a brown suit. (1) not me and probably not you but it was a national talking point for weeks
  - isoprophlex 10mo ago
    I thought that was quite a fashionable suit, and wish I could pull off wearing something like that
  - BigTTYGothGF 10mo ago
    If you bump that up to 110 years you bring in Woodrow Wilson, one of the all-time worst presidents.
    - rflrob 10mo ago
      I’m curious how you came to that conclusion. While he’s certainly not in the pantheon of best presidents, he ends up around the 75th percentile in rankings by historians. Even subtracting a few spots, he’s nowhere near close to “one of the all-time worst“. Or are you faulting him for not resigning when incapacitated by a stroke? https://en.wikipedia.org/wiki/Historical_rankings_of_presidents_of_the_United_States https://en.wikipedia.org/wiki/Historical_rankings_of_preside...

frostyel 10mo ago

It's as good as Piers Morgans legendary pythagorean theorem: https://www.youtube.com/watch?v=QZWS2g-fEAU https://www.youtube.com/watch?v=QZWS2g-fEAU

bombcar 10mo ago

There’s a sci-fi/time travel thriller here where he was assassinated because of his mathematical prowess.

NoNameHaveI 10mo ago

Fun fact: Garfield LOVES lasagna, and hates Mondays. Oh. Wait!

russfink 10mo ago

May be a repeat here, but best proof I saw was inscribe a square with sides of length c inside another square, but rotated such that the interior square’s corners intersect the outer square’s edges. The intersecting points divide the outer square’s edge making lengths a and b. This produces an inner square’s edge with sides length c and four equal right triangles of sides a, b, and c. Note that the area of the outer square equals the sum of the inner square plus the area of the four triangles. Solve this equality.

zahlman 10mo ago

We can imagine another copy of the trapezoid, rotated 180 degrees and situated on top; the pair of them create a square with side lengths of a + b. This cancels all the 1/2s out of Garfield's equations, and also makes the result more geometrically obvious: the entire square (a + b)^2 = a^2 + 2ab + b^2 is the inscribed square c^2 plus four copies of the original triangle 4 * ab/2 = 2ab. This then becomes a restatement of another classic proof (the simple algebraic proof given near the top of the main Wikipedia page for the theorem). So we can imagine Garfield discovering this approach by cutting that diagram (https://en.wikipedia.org/wiki/Pythagorean_theorem#/media/File:Animated_gif_version_of_SVG_of_rearrangement_proof_of_Pythagorean_theorem.gif https://en.wikipedia.org/wiki/Pythagorean_theorem#/media/Fil...) in half and describing a different way to construct it.

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torginus 10mo ago
Yeah this is the high-school proof, just cut in half.
- amelius 10mo ago
  Maybe the high school proof was inspired by Garfield?
  - torginus 10mo ago
    Afaik the ancient proof (the square inside a bigger square rotated by 45 degrees) comes from ancient Greece, so I guess Garfield can add 'time travel' to his list of discoveries.

surprisetalk 10mo ago

They assassinated him because he uncovered too much forbidden knowledge about the triangles

SJC_Hacker 10mo ago
The Triluminati

kingofmen 10mo ago

I'm having some trouble with this part of the explanation: > From the figure, one can easily see that the triangles ABC and BDE are congruent. I must confess I do not easily see this. It's been a long time since I did any geometry, could someone help me out? I'm probably forgetting some trivial fact about triangles.

istjohn 10mo ago
It wasn't obvious to me either. But we know the angles ABC, ABD, and DBE equal 180 degrees, as do the interior angles of triangle ABC. From that we can deduce that angle BAC = angle DBE, from which it follows that angle ABC = angle BDE.
desertrider12 10mo ago
We know angle EBD equals BAC, since the sum of triangle ABC's interior angles is 180 degrees and the sum of the 3 angles at B are also 180 degrees. We also know angle DEB is 90 degrees since DE was constructed to be perpendicular to CB. Finally, D was placed at a distance c from B. The two triangles have the same angles and the same side lengths opposite the right angles, so they must be congruent.
da_chicken 10mo ago
So, the line BE is just the line CB extended. It's the same line. And we know that the angles of a triangle add up to 180. And we know that the line BD is defined as perpendicular to AB. That means the angle ABC and angle DBE must add up to 90. But that's also true of the angles ABC and angle CAB. That means that angle DBE and angle CAB must be the same. Both triangles ABC and BDE are both right triangles, so that means angles ABC and BDE are the same. So they're similar triangles: They have all the same angles. Additionally, the point D is just at a point so that the length of line segment BD and the length of line segment AB are both the same: c. Since we know that the hypotenuse of triangle ABC is c, and the hypotenuse of triangle BDE is also c, and we know they're both similar triangles, then these triangles must be congruent as well.
- kingofmen 10mo ago

waldrews 10mo ago

Garfield was in many ways the most personally appealing and brilliant of the American presidents, rising from poverty and obscurity by being absurdly talented across many fields and eloquent. He was assassinated early and barely got to serve. The story of his life, the shooting, and the subsequent medical drama (featuring even a cameo by Alexander Graham Bell improvising a diagnostic device) are so epic you have to wonder if time travelers are messing with us. His legacy was the nonpartisan professional civil service, a key part of his agenda that his successor felt obligated to carry out, an accomplishment that recently came under particularly heavy attack. Netflix just came out with the miniseries about him, 'Death by Lightning,' based on the book 'Destiny of the Republic.' His earlier life is featured prominently in '1861: The Civil War Awakening' by Adam Goodheart. There are a few great C-SPAN/Book TV videos by some of the authors that tell the story concisely and convey why some of us are so fascinated by that history.

Apocryphon 10mo ago
The Netflix miniseries was very funny and engaging, but I can’t help but to think that it over-valorizes Garfield as some sort of fallen benevolent sage-king in the same way Oliver Stone’s JFK and other Camelot hagiographies do. The man was not above the corrupt politics of the day, at least earlier in his career, after all.

amelius 10mo ago

I sometimes wonder what mathematics and physics would have looked like if the Pythagorean theorem was a really ugly formula, or something you couldn't write in closed form.

nvlled 10mo ago

A side thought, how would a natural number sequence of hypotenuses that satisfies the equation a^2 + b^2 = c^2 look like? Or what interesting properties would it have if any.

Joker_vD 10mo ago
You mean, what does sequence A004431 [0] look like? Well, like itself. Some of its properties are also listed at that link. [0] https://oeis.org/A004431 https://oeis.org/A004431

nottorp 10mo ago

Oh. It’s not that cat.

tug2024 10mo ago

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