Comment by zeroonetwothree

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__ 4 days 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.

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WCSTombs 4 days 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.

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.

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aaplok 4 days ago

You got a bunch of responses already, here is an intuitive reason.

In similar triangles all distances are scaled by a factor k, by definition. Then, intuitively the areas are scaled by a factor of k^2, since you obtain an area by multiplying two distances.

So the ratio of the area over the hypothenuse is scaled by a factor of k^2/k=k.

It is not hard to confirm the intuition that the areas are scaled by a factor of k^2, since it is precisely the product of the lengths of the two sides adjacent to the right angle.

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xeonmc 4 days ago

I think it would be clearer with an explicit function of a rectangle's area with respect to its diagonals:

                   A = x y
                   x = r cos(θ)
                   y = r sin(θ)
             => A(r) = r²cos(θ)sin(θ)
                A(c) = A(a) + A(b)
    = c²cos(θ)sin(θ) = a²cos(θ)sin(θ) + b²cos(θ)sin(θ)
               => c² = a² + b²

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fiso64 4 days 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.

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lupire 4 days 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!

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degamad 4 days 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.

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- lupire 4 days 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.
  
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  degamad 4 days ago
  
  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.
  
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layer8 4 days 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.

I would state it differently: Given the specific triangle, the ratio between the area of the square on the hypotenuse and the area of the triangle is some constant k that is invariant under scaling of the triangle.

This should be intuitively obvious: When you have a picture with some shapes in it, scaling the picture won’t change the relative proportions of the shapes in the picture. You don’t have to know the absolute dimensions of the picture to determine the area ratio of two shapes within the picture. The constant k above will be different for differently shaped triangles, but will be the same for triangles of the same shape (same angles).

So, for a given triangle T1 with hypotenuse c we have, for some k: area(T1) = k x c²

Now, we subdivide T1 into two smaller triangles T2 and T3 that have the property of both being a scaled version of T1, and their hypotenuses being the a and b of T1. Hence we have:

area(T2) = k x a²

area(T3) = k x b²

all with the same k, since they have the same shape (only differing by scaling).

Because we have

area(T1) = area(T2) + area(T3),

it follows that

k x c² = k x a² + k x b²

and since k ≠ 0, we get

c² = a² + b².

In pictures (as far as possible with ASCII art):

   _______
  |       |\   
  |   a²  | \  
  |       |T2\ 
  |       |.‘ 
   ¯¯¯¯¯¯¯

        +  

             .
           .T3\
          |¯¯¯¯|
          | b² |
          |____|
           

        =

                 / \
              /     \
           /         \
          |\    c²    \
          | \         /
          |T1\     /
          |   \ /
           ¯¯¯¯

(The three are supposed to be the same exact picture, just rotated and scaled, and T1 can be split into T2 and T3, hence c² must also be the sum of a² and b².)

The underlying “miracle” is that you can subdivide any right triangle into two smaller copies of itself. The Pythagorean theorem then follows immediately from that. This subdivision capability is something that might be amenable to some further underlying explanation.

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[removed] 4 days ago

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layer8 4 days ago

P.S.: T2/T3 are also mirrored, not just scaled, from T1, but obviously that doesn’t change anything about their area.

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zem 4 days 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

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thaumasiotes 4 days 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.

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