[HN Gopher] Garfield's Proof of the Pythagorean Theorem
___________________________________________________________________
Garfield's Proof of the Pythagorean Theorem
Author : benbreen
Score : 160 points
Date : 2025-11-29 06:37 UTC (16 hours ago)
(HTM) web link (en.wikipedia.org)
(TXT) w3m dump (en.wikipedia.org)
| WCSTombs wrote:
| 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%99...
| bryanrasmussen wrote:
| 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 wrote:
| 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 wrote:
| 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...
| fiso64 wrote:
| 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 wrote:
| 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 x [(a, b), (-b, a)] = [(c, 0), (0, c)]
|
| The determinant of a rotation matrix is 1, so the
| determinant of the left side is 1x(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.
| lupire wrote:
| > 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
|
| You can define the determinant that way. Now the question
| is why the cross multiplication formula for determinant
| accurately computes the area.
|
| You can prove that via decomposition into right triangles
| https://youtu.be/_OiMiQGKvvc?si=TyEge1_0W4rb648b
|
| Or you can go in reverse from the coordinate formula, to
| prove that the area is correctly predicted by the
| determinant.
| degamad wrote:
| Yep - I'm just not sure if any of those proofs implicitly
| assume Pythagoras, and haven't thought through them
| properly.
|
| I was initially going to say we know that det R = 1 by
| using the trigonometric identity cos2x+sin2x=1, but then
| found out that all the proofs of it seem to assume
| Pythagoras, and in fact, the identity is called the
| Pythagorean trigonometric identity.
| lupire wrote:
| Draw it on graph paper:
|
| Set B as the origin, and let BC (the 'a' side), be on the
| the positive side of the x-axis. Let AC be on the positive
| side of the y-axis.
|
| The left matrix is a clockwise rotation and scaling. This
| is clearly seen if you draw the transformation applied to
| the two axis basis vectors. (The scaling factor isn't
| obvious yet.)
|
| Then the left matrix varies (1,0) to the side AB, which has
| magnitude c. Z and carries (0,1) to an perpendicular line
| of the (importantly) same magnitude,
|
| So it's a rotation and a scaling by c.
|
| The right matrix obviously is a scaling by c.
|
| > 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.
|
| Yes, that's the point!
| zem wrote:
| > where k is constant for similar triangles.
|
| you can see that by simply scaling the figure of (the
| triangle + square on its hypotenuse) as a whole; whatever
| size the triangle is the ratio of the two pieces doesn't
| change
| thaumasiotes wrote:
| > 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.
|
| The second half of your sentence is not correct; if area
| scales with the square of any one-dimensional measurement
| (including hypotenuse length, because the hypotenuse is one-
| dimensional), that is sufficient to prove the theorem.
|
| The statement you're looking for is: "triangle A is similar
| to triangle C with a length ratio of a/c, therefore the area
| of triangle A is equal to the area of triangle C multiplied
| by the square of that ratio".
|
| It is in fact necessary that the area will scale with the
| square of hypotenuse length, because the hypotenuse is one-
| dimensional and area is two-dimensional. If you decided to
| measure the area of the circle that runs through the three
| corners of the triangle, the triangle's area would scale
| linearly with that.
|
| It isn't clear to me what scenario you're thinking might mess
| with the proof.
|
| > 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.
|
| So, for similar shapes, you can set your own measurements.
|
| 1. Say I have two triangles X and Y and they're similar. I
| take a straightedge, mark off the length of the longest side
| (x) of triangle X, and say "this length is 1". Then I
| calculate the area of triangle X. It will be something. Call
| it k.
|
| 2. Now I take a _second_ straightedge, mark off the length of
| the longest side (y) of triangle Y, and I label that length
| "1". I can calculate the area of triangle Y and, by
| definition, it must be k. But it is equal to k using a scale
| that differs from the scale I used to measure triangle X.
|
| 3. We can ask what the area of triangle Y would be if I
| measured it using the ruler marked in "x"es instead of the
| one marked in "y"s. This is easier if we have the same area
| in a shape that's easier to measure. So construct a square,
| using the "y" ruler, with area equal to k.
|
| 4. Now measure that square with the "x" ruler. The side
| length, measured in y units, is [?]k. Measured in our new x
| units, it's (y/x)[?]k. When we square that, we find that the
| x-normed area is equal to... k(y/x)2.
|
| This is why it's obvious that k must be constant for similar
| triangles. k is just a name for the scale-free representation
| of whatever it is that you're measuring. It has to be
| constant because, when you change the scaling that you use to
| label a shape, the shape itself doesn't change. And that's
| what similarity means.
| Sesse__ wrote:
| > 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.
|
| It is elementary to show that the area of a triangle is base
| * height / 2. (It follows from the fact that you can make a
| rectangle out of it using two identical sub-triangles. I
| assume you're willing to concede that the area of a rectangle
| is base * height.) If you scale your triangle by c, both base
| and height will be multiplied by c, and 2 will not.
| WCSTombs wrote:
| > _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._
|
| Area in what units? "Square" units? But we're free to choose
| any unit we want, so I choose units where the triangle itself
| with hypotenuse H has area H^2 units. To justify that, I
| think the only thing we need is the fact that area scales as
| the square of length. (There's that word "square" again,
| which implies a specific shape that is actually completely
| arbitrary when talking about area. Perhaps it's better to say
| that "area scales as length times length.")
|
| > _To me that truth isn't necessarily any less fundamental
| than the Pythagorean theorem itself._
|
| I think the Pythagorean Theorem is surprisingly _non_
| -fundamental, in that you can get surprisingly far without
| it. It's surprising because we usually learn about it so
| early.
| zeroonetwothree wrote:
| 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
| WCSTombs wrote:
| That one's also very neat!
| dataflow wrote:
| > 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 wrote:
| 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 wrote:
| 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 wrote:
| 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.
| codethief wrote:
| > So just choose units where our triangle itself with a
| given hypotenuse H has area H^2 units.
|
| This is not at all trivial. You're claiming you can
| choose units in such a way (reusing my notation from
| before) that _simultaneously_
|
| A = a2, B = b2, C = c2.
|
| Intuitively, you can do that precisely because the
| triangles are similar and area is quadratic in the
| similarity ratio. But there is definitely some algebra
| behind that.
| jancsika wrote:
| Mathematicians explain things the way I imagine musicians
| would if the ancient Greeks had insisted on making all
| musical instruments in a range audible only to dogs.
|
| I'd be like, "How do I actually _hear_ the difference
| between a major and minor sixth? " And the musician would
| be like, "Just play them into the cryptophone and note
| the difference in the way your dog raises its eyebrows."
|
| The very few remaining musicians in this hellscape would
| be the ones who are unwittingly transposing _everything_
| to the human range in their sleep, then spending the day
| teaching from the Second Edition of the Principles of
| Harmonic Dog Whistling for all us schmucks.
|
| Luckily we don't live in that musical universe. But
| mathwise, something like that seems to be the case.
| WCSTombs wrote:
| Look, I think it's pretty hard for most of us to read
| long math arguments in plain text, so I wrote in the
| simplest language I could, leaving the simple details for
| the reader to fill in.
|
| I will add that in the vast majority of mathematical
| literature, both in pedagogy and in research, the active
| participation of the reader is assumed: the reader is
| expected to verify the argument for themselves, and that
| often includes filling in the details of some simple
| arguments. That's exactly why math literature uses the
| plural first-person "we," because it's supposed to be as
| if the writer and reader are developing the argument
| together.
|
| In contrast, listening to music can be purely passive
| (but doesn't need to be).
| card_zero wrote:
| 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_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 wrote:
| 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^2xa^2 = f^2xA. Since the relative
| portion the face takes up doesn't change with scaling, the
| face now has size pA' = pxf^2xA = f^2 x pA = f^2 E.
|
| Does that help or was that not the part you were missing?
| card_zero wrote:
| 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 wrote:
| 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.
| card_zero wrote:
| I was saying two separate things. Thing 1, the non-square
| shapes are relevant to Einstein's nice proof. Thing 2,
| considering squares now if you like, pythagoras's theorem
| has a magical quality which proofs can't dispel.
|
| If you travel some distance, square it, travel some other
| distance perpendicularly, square that too, and add the
| results, you get the square of the straight distance from
| start to finish. Every proof just seems like a
| reformulation of this freaky fact.
| JanisErdmanis wrote:
| 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 wrote:
| You mean non-Euclidean fabric?
|
| https://www.johndcook.com/blog/2022/09/08/trig-hyperbolic-
| ge...
|
| https://math.hmc.edu/funfacts/spherical-pythagorean-theorem/
|
| https://en.wikipedia.org/wiki/Minkowski_distance
| JanisErdmanis wrote:
| That is one possibility. Probably the only one, though.
| lupire wrote:
| > 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 wrote:
| > 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 wrote:
| 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.
| b800h wrote:
| I was ready for it to involve lasagna.
| esher wrote:
| Same!
| Lio wrote:
| 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
| hmokiguess wrote:
| Glad it wasn't just me hahaha
| vee-kay wrote:
| 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.
|
| Baudhayana, (fl. c. 800 BCE) was the author of the Baudhayana
| sutras, which cover dharma, daily ritual, mathematics, etc. He
| belongs to the Yajurveda school, and is older than the other
| sutra author Apastambha. He was the author of the earliest Sulba
| Sutra--appendices to the Vedas giving rules for the construction
| of altars--called the Baudhayana Sulbasutra. 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
|
| 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 wrote:
| American presidents used to be smart.
| xeonmc wrote:
| they also use to be bullet magnets
| einpoklum wrote:
| That looks like "half" of the proof using a square:
|
| 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 wrote:
| I shared this one with my son, the step where the 2ab
| expressions cancel out gave him a little aha moment.
| procrastitron wrote:
| 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 wrote:
| Netflix recently released a mini-series on Garfield's election
| and presidency: https://www.themoviedb.org/tv/245219-death-by-
| lightning.
| dataflow wrote:
| 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 wrote:
| 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 wrote:
| > 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 wrote:
| I must confess I clicked through hoping to see a comic of
| Garfield the cat using pizza slices to approximate right
| triangles.
| emigre wrote:
| Me too...
| RobotToaster wrote:
| I hate triangular Mondays
| medwards666 wrote:
| This was also similar to my first thought.
|
| Damn. I want pizza now as well...
| monocularvision wrote:
| I believe you mean lasagna.
| parpfish wrote:
| pizza is portable lasagna.
| nullbyte808 wrote:
| He was also the 20th president of the USA.
| kingofmen wrote:
| "Better known for other work".
| charlieyu1 wrote:
| This is actually one of the most well-known proof
| EdNutting wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| I thought that was quite a fashionable suit, and wish I
| could pull off wearing something like that
| BigTTYGothGF wrote:
| If you bump that up to 110 years you bring in Woodrow
| Wilson, one of the all-time worst presidents.
| rflrob wrote:
| 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_pres
| ide...
| BigTTYGothGF wrote:
| WWI plus all the racism, including re-segregation (https:
| //en.wikipedia.org/wiki/Woodrow_Wilson_and_race#Segreg...
| )
| zorked wrote:
| Never heard about the brown suit. But I heard about the
| drone strikes.
| Waterluvian wrote:
| I don't have to imagine.
| https://www.proquest.com/docview/301464456/
| stephenlf wrote:
| Nice
| xpe wrote:
| For those who can't be bothered to go a'link-clicking, the
| above refers Mark Carney, Prime Minister of Canada, which
| means he ministers to prime numbers living at the north pole.
| ThrowawayTestr wrote:
| That's a Prime Minister, not a president
| icpmoles wrote:
| Peru's experience with a President with a math degree didn't go
| that well
| frostyel wrote:
| It's as good as Piers Morgans legendary pythagorean theorem:
|
| https://www.youtube.com/watch?v=QZWS2g-fEAU
| bombcar wrote:
| There's a sci-fi/time travel thriller here where he was
| assassinated because of his mathematical prowess.
| NoNameHaveI wrote:
| Fun fact: Garfield LOVES lasagna, and hates Mondays. Oh. Wait!
| russfink wrote:
| 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 wrote:
| 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/Fil...)
| in half and describing a different way to construct it.
| torginus wrote:
| Yeah this is the high-school proof, just cut in half.
| amelius wrote:
| Maybe the high school proof was inspired by Garfield?
| surprisetalk wrote:
| They assassinated him because he uncovered too much forbidden
| knowledge about the triangles
| kingofmen wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| 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 wrote:
| Thank you! Rephrasing for my own understanding: The point of
| attack that I was missing was that angles BDE and ABC are
| equal, and now we have two equal angles (which immediately
| gives us the third) and one equal side, so we're good to go.
| waldrews wrote:
| 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 wrote:
| 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 wrote:
| 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.
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