Garfield's Proof of the Pythagorean Theorem

122 pointsposted 11 hours ago
by benbreen

61 Comments

zahlman

21 minutes 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/Fil...) in half and describing a different way to construct it.

WCSTombs

10 hours 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%99...

SJC_Hacker

an hour 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

zeroonetwothree

9 hours 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...

Sesse__

3 hours 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.

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.

fiso64

6 hours 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

5 hours 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.

lupire

4 hours ago

> 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.

lupire

5 hours ago

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

5 hours ago

> 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

4 hours ago

> 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.

dataflow

8 hours 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

6 hours 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

2 hours 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.

bryanrasmussen

9 hours 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.

JanisErdmanis

6 hours 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...

card_zero

7 hours 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_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

6 hours 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

6 hours 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

4 hours 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.

card_zero

3 hours ago

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.

lupire

4 hours ago

> The purely trigonometric proof that was found by some high school students recently is a great one.

It was geometric, using trigonometric vocabulary.

rcarmo

7 hours ago

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

medwards666

4 hours ago

This was also similar to my first thought.

Damn. I want pizza now as well...

surprisetalk

18 minutes ago

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

russfink

an hour 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.

wunderlust

7 hours ago

American presidents used to be smart.

xeonmc

7 hours ago

they also use to be bullet magnets

EdNutting

5 hours 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…

icpmoles

26 minutes ago

Peru's experience with a President with a math degree didn't go that well

adventured

5 hours 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

4 hours 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

4 hours ago

I thought that was quite a fashionable suit, and wish I could pull off wearing something like that

BigTTYGothGF

4 hours ago

If you bump that up to 110 years you bring in Woodrow Wilson, one of the all-time worst presidents.

rflrob

3 hours 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_preside...

b800h

9 hours ago

I was ready for it to involve lasagna.

hmokiguess

4 hours ago

Glad it wasn’t just me hahaha

dataflow

7 hours 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

2 hours 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

2 hours 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.

vee-kay

8 hours 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

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."

bombcar

3 hours ago

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

charlieyu1

6 hours ago

This is actually one of the most well-known proof

einpoklum

7 hours ago

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

7 hours ago

I shared this one with my son, the step where the 2ab expressions cancel out gave him a little aha moment.

NoNameHaveI

3 hours ago

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

nullbyte808

6 hours ago

He was also the 20th president of the USA.