the famous Calculus proof of the Pythagorean theorem.

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[luizmenezes]

I had a terrible maths teacher back in highschool, that whenever I solved a problem using a non-standard approach, she said that I was killing an ant with an RPG shot.

I hated that witch.

That being said, using diff eqs to prove Pythagoras is killing an ant with an RPG.

[Phi]

I always thought it was weird when mathematicians called mathematical proofs "beautiful" but now I get it.

[tolberthobson]

VERY Interesting proof. I've never actually seen an analytical proof of the Pythagorean theorem I was just always told draw 2 squares from the legs of the right triangle and observe their areas add to the area of a square drawn from the hypotenuse of the right triangle. So very cool to see this

[erwinmulder]

My favorite i s the geometrical one where you have a big square that is cut twice perpendicular to the sides, so that you end up with four parts where two of them are squares (A and B) and two are identical rectangles.The rectangles are then cut diagonally to create four identical right angled triangles. If you put the right angles of those triangles in the corners of the original square, they will leave a square shaped hole 'C' in the center with the sides made up by the hypotenuses of the triangles. Since the four triangles plus the newly formed square hole fill up the same square as the one when they were combined with squares A and B, the total area of C has to be the same as A + B.

[digxx]

I think the step DE -> y(x+dx) - y(x) needs a bit more love. For the proof to work you would have to show that DE = y(x+dx) - y(x) + O(dx^2).

[TedHopp]

My favorite proofs are based on similar triangles. Let the triangle be ABC with opposing side lengths a, b, c, with c the hypotenuse. Drop a perpendicular from the right angle vertex C to the hypotenuse AB. Label the point of intersection D. Then triangles ABC, ACD, and CBD are all similar.

From here, we can construct two proofs. The first uses the fact that areas of similar triangles are proportional to the squares of corresponding sides. Call the area of ACD x and the area of CBD y. Then the area of ABC is x+y. The areas are in proportion x:y:x+y and the hypotenuses are in proportion a^2:b^2:c^2, which is the Pythagorean theorem.

The second proof uses the property that corresponding sides of similar triangles are in equal ratio. From △ABC ~ △ACD, we have that AD/AC = AC/AB, or AC^2 = AB × AD. From △ABC ~ △CBD, we have that BD/BC = BC/AB, or BC^2 = AB × BD. Adding these two equations, we have that AC^2 + BC^2 = AB × (AD + BD). But since AD + BD = AB, we end up with AC^2 + BC^2 = AB^2.

[Miyamoto_]

I like how instead of simplifying tricky things, he complicates the most generic and simplest of things 😂

[talinuva]

Starting with a triangle ABC which has a right angle at B, construct the altitude from B to a point D on AC.
Given that both new triangles ADB and CDB share an angle with ABC, in addition to a right angle, we have that all three triangles are similar.
By similarity, |AD|/|AB|=|AB|/|AC| and |CD|/|BC|=|BC|/|AC|.
Clearing denominators gives us |AD|*|AC|=|AB|² and |CD|*|AC|=|BC|²
Adding the two results in (|AD|+|DC|)*|AC|=|AB|²+|BC|² which, since D is between A and C, gives the Pythagorean Theorem

[butchkow]

There was a book written in the 40's and published by I believe the NCTM that had a lot of proofs of this theorem.

[multipontushd]

I like this channel - it reminds me of my two years of university math courses actually almost three (majored in CS). Also would like to say that there are some 80s vibes over his videos dunno why...

[shaiavraham]

My favorite proof uses dimensional analysis, that is “checking the units” (I am a physics and math student):

We have a right triangle with sides a and b, and hypotenuse c, and an acute angle theta. We know that a right triangle is completely and uniquely specified by its hypotenuse c and its acute angle theta. Therefore, the area a right triangle is a function of the hypotenuse and the acute angle: A(c, theta).

c has dimension of length, [L], theta is dimensionless, and area has dimension of length squared, [L^2]. Therefore, by dimensional analysis, the area has to be proportional to c^2, and it’s dependence on theta is arbitrary: A(c, theta) = f(theta) * c^2. The important thing is that it’s the same function f(theta) for every right triangle.

Now from the vertex of the right angle, draw a perpendicular to the hypotenuse, creating three similar right triangles with the angle theta (two small triangles and the original triangle). The area of the original triangle is A(c, theta). For the two smaller triangles, note that their hypotenuses are a and b, and therefore their areas are A(a, theta) and A(b, theta). The sum of these areas equals the area of the original triangle, which gives us the equation:

A(a, theta) + A(b, theta) = A(c, theta)
f(theta) * a^2 + f(theta) * b^2 = f(theta) * c^2
a^2 + b^2 = c^2

which is the Pythagorean theorem.

I really like this proof since it’s extremely simple, elegant and can be taught to any kid who knows that length is measured in meters and area is measured in meter squared. This proof essentially says that the fact that area is measured as squared length forces the Pythagorean theorem to be true.

[the_informative_edge]

Sir your every segment is mind blowing..

[alikaperdue]

Given triangle of sides a:b:c, where a<=b<=c.
Scale the triangle by each of it's own sides to create three new triangles A, B, C:
A = aa:ab:ac
B = ba:bb:bc
C = ca:cb:cc
Join triangle A and B along common length ab, so that the right angles are touching to make a straight line.
The new triangle will have sides ac:bc:aa+bb
This triangle is congruent to C: ca:bc:cc
Since the triangles are equal, the sides a^2+b^2 = c^2

[DeJay]

The Theory of Relativity competing with the Pythagorean theorem for the most proofs of the same idea.
WE GET IT. IT'S TRUE!

[Anonymous-zphb]

Is this argument circular? (pun very much intended)

(d/dt) (sin(t)^2 + cos(t)^2)
= 2sin(t)cos(t) - 2sin(t)cos(t)
= 0
So for all t, sin(t)^2 + cos(t)^2 = sin(0)^2 + cos(0)^2 = 1
and therefore, Pythagoras.

[herghamoo]

I think my favorite proof of the Pythagorean theorem is that it follows trivially from Ptolemy's theorem (in my opinion one of the coolest results from ancient mathematics).

[PedroAssumpcao-ondb]

I liked how the proof manages to use only bits of calculus that don`t themselves require the pythagorean theorem. Very neat

[paulg]

I had a few ancient proofs but for the students I teach.. this is one for the ages !

[JamesLewis]

When I clicked on this video, I expected the more recent proof found by a couple of college students a couple of years ago that relies on infinite series and works only for non-isosceles right triangles:

Let the length of the larger leg be a and of the smaller leg be b; then on the leg of length b, construct a similar right triangle in which the longer leg has length b, and extend the hypotenuses of both triangles until they intersect (this is why the legs cannot have the same length), and keep drawing line segments perpendicular to the previous one, marking off similar right triangles infinitely often.

Then, looking at the next right triangle, its longer leg has length b=a*(b/a) and its shorter leg has length b²/a=b*(b/a); from this, the nth triangle after the first has legs of length a*(b/a)^n and b*(b/a)^n and area ½ab*(b/a)^(2n), from which the whole collection of right triangles has area ½ab/(1−(b/a)²)=½ba³/(a²−b²).

If the hypotenuse of the original right triangle has length c, then the hypotenuse of the next one has length cb/a and the extension of that original hypotenuse, as a leg of the larger triangle, has length c/(1−(b/a)²)=ca²/(a²−b²), while the extension of that next hypotenuse has length abc/(a²−b²); the triangle formed by dropping a line segment from the vertex between the sides of lengths c and a perpendicular to the other "extended hypotenuse" is also similar to the original right triangle…and I'm honestly not sure about how the proof goes from there, but you're supposed to add the area of this to the area of the whole collection and equate this to the area of a larger right triangle that contains all of the others, and after a bunch of cancellations, the end result should be c²=a²+b², but I kept making some sort of mistake while finishing it up.

[alexbennie]

Dude! The quality of your chalk!

That's no cheap chalk! 8:41, I've never used red chalk that made such vivid colour.

Also

Very cool application of Geometry and Limits. First time seeing it used like this!

Could anyone point me in a direction of where an accessible entry hole is for this rabbit hole?