Constructing a Square of Equal Area to a given Polygon

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About the video:
1. Pick any polygon
2. Split it up into triangles (While it is trivial to triangulate any convex polygon it is has been proven that any polygon, convex or concave, can be decomposed into triangles and there are many triangulation techniques).
3. Construct a rectangle of equal area for each triangle.
4. Construct a square of equal area for each rectangle.
5. Construct a larger square equal to the sum of each smaller squares via Pythagorean Theorem.

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Any further questions:

Programs used:
► Cinema 4D
► Processing

Music by

[ThinkTwiceLtu]

For anyone who is interested, I have also made a little interactive sketch with p5.js library where you can play around by squaring a rectangle. I am still learning so let me know if you encounter some bugs, also it will not work that well on your mobile phone so open it up on your PC.

[-----bkle]

If we take n-gon where n tends to infinity, so this problem can be named as..

[erfanshekarriz]

I love your videos they inspire me to do more Mathematics in my free time!!!

[Carl-Gauss]

There is also a 3D version of this problem: can you cut given polyhedron into finite number of pieces and make cube with equal volume of them? It even was in the list of Hilbert’s problems (№3). Interestingly, it turns out that unlike 2D version answer for this one is “no”.

[AneR]

2:53 bottom corner of pink parallelogram: "am I a joke to you?"

[redsalmon]

That’s so cool.
Especially how you can turn the squares into a larger one using the simple Pythagorean theorem.
Love your work.

[alexandersanchez]

You guys, this is technically different from the numberphile video: that one said you can cut up a polygon and re-assemble the pieces into a square of equal area; this one says that you can *construct* a square of equal area (in the sense of straight-edge/compass constructions). The upshot is that from each regular n-gon one can construct a square of equal area. This is kind of interesting, since in some certain sense, the circle is almost like a limit of regular n-gons. So, if we weren't being careful, we might begin to suspect that squaring the circle is possible; of course, that's nonsense because we couldn't do the first step (of triangulation) on a circle which was assumed at the outset. However, this does show that to square a shape, it is sufficient to triangulate it, which is cool in its own right.

[madhuragrawal]

It's exactly the equidecomposability problem, no? Maybe viewers would like the numberphile video on the dehn invariant

[HuslWusl]

That's such a beautiful visualization of pythagoras theorem a^2 + b^2 = c^2
I never realized the potential of it. I always thought like "yeah whatever I'll never use it for anything else other than finding c" but damn, you opened my eyes

[VibingMath]

Nice one! Enjoy this video with relaxing music

[FG]

There's also the mathematical way:
I assume we know the height and bottom length of the triangles as h1, h2, h3 and a, b, c
A1 = (h1*a)/2
A2 = (h2*b)/2
A3 = (h3*c)/2
And of course, the area of the polygon is Ap = A1+A2+A3 while the area of the square will be Ap = As = B*B, or B*(d+e+f) where d, e, f are the lengths of the individual rectangles.
Each rectangle must have the same area as the respective triangle, and the end result must be a square:
(1): A1 = (h1*a)/2 = B*d
(2): A2 = (h2*b)/2 = B*e
(3): A3 = (h3*c)/2 = B*f
(4): B = d+e+f
Rearranging the first three and substituting into the fourth:
B = = (h1*a+h2*b+h3*c)/(2*B)
Moving the B under the division to the left hand side means it's squared on the left, so the value of B is:
B = sqrt((h1*a+h2*b+h3*c)/2)
Where all variables are known.
Then from (1), (2) and (3):
d = (h1*a)/(2*B)
e = (h2*b)/(2*B)
f = (h3*c)/(2*B)
And yeah, then you're done really. Proof:
Ap = A1+A2+A3 = (h1*a+h2*b+h3*c)/2
As = B*(d+e+f) = B*((h1*a+h2*b+h3*c)/(2*B)) = (h1*a+h2*b+h3*c)/2
And as we can see: As = Ap

[DiegoMathemagician]

Before my hollidays I challenged myself to "square" a triangle, like Leonardo DaVinci did. Then I forgot about it and you just upload this video. Not sure if I should watch it :(( I want to figure it out by myself but I have a lot of work to do because I also want to find out so many things on my own. A little bit frustrating :/
Anyway, I love your content man! If you see this, Think Twice, i am legenddaryum from twitter (we chatted a while ago about geometry and stuff)

[mrkhunt.]

Beautiful Animation, Proving with just visuals plus leading us with the intuition. Just Brilliant!!

[ivarangquist]

2:30 Best Pythagorean theorem visualization I've ever seen!

[NotEnoughMs]

Why can't we just transform the polygon in triangles, get each triangle area, add the areas, take the square root and that would be the side of the square?

[oliverhoare]

I’ve never seen the skewing squares method shown here before, I’m glad I did because it looks beautiful.

[Ymitzna]

Your videos are just perfect! Short and sweet, showing how beautiful agh can be. Just incredible, always with simple and to the point explanations and fantastic music. Just incredible.

[mohammedal-haddad]

It is amazing how how simple geometry tools you are using to achieve something like this. Thank you very much.

[ich]

That was actually a pretty nice visualisation of the Pythagorean theorem in your video. 👍

[mathemaniac]

Although I have seen on Numberphile a similar problem, seeing the animation is a lot more satisfying.

By the way, which software do you use? I might put some animations in my videos using it.