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Date : 1732226181419 - unknown on Apple WebKit
Mystery text : UFFUSHhpSGVvaFkgaSAgbG92ICB1IGV1LXByb3h5LnBva2V0dWJlLmZ1bg==
143 : true
177,686 Views • Apr 8, 2024 • Click to toggle off description
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This animation is based on a proof that appears in the following sources:
Mathematical Morsels by Ross Honsberger (MAA, 1978)
Proofs without Words II by Roger B. Nelsen (MAA, 2000) (bookstore.ams.org/clrm-14/)
And can be traced to an article by an author known as Nev R. Mind.
To learn more about animating with manim, check out:
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#manim #math #mathvideo #mathshorts #geometry #square #animation #theorem #pww #proofwithoutwords #visualproof #proof #area #onefifth
Views : 177,686
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Uploaded At Apr 8, 2024 ^^
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RYD date created : 2024-07-03T18:06:27.116325Z
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192 Comments
Top Comments of this video!! :3
I solved it a different way. It's a little hard to explain in a comment, but basically I looked at the right triangles with hypotenuse 1 and ½ respectively, and noticed that since they shared an angle, they were similar. Thus all their sides were in ratio 2:1, so if we call the little side length x, we see that 5x² = 1, so x = 1/sqrt(5). Now since x is half the length of the long side, the side of the square is the other half, so it must also be 1/sqrt(5), and so the square's area must be ⅕.
145 |
Almost thought it was 1/4
341 |
The general expression for any Area inner square, for any incline, is:
A2 =
(1/cos(atan(a))
- a*sin(atan(a))
- a*cos(atan(a)))^2 ;
a: length side 1 is reduced by;
atan(a) : angle of lowest diagonal to square 1 lowest side;
1/5 only correct when alpha=atan(a)=45°;
when a=.5 then A2=.2; but only for this angle
34 |
Thats amazing man u open my mind to think like this thanks
4 |
I did it in a complex way. I did sqrt(1^2+0.5^2) which gave me sqrt(5)/2 to get the length of one of the lines that connect from the corner to the midpoint. I thought that if the lines got lengthened by a factor of root 5 over 2 then it would shorten the distance between a parallel pair of those lines by the same factor, luckily I was right. I divided 0.5 by root 5 over 2 and got root 5 over 5 for the length of the squares side and then squared that and got 1/5.
4 |
How do we know that we can fold the triangles on that way?
72 |
Pythagore : hypotenuse h= sqrt(5)/2
Thales : Square side s= h2/5 ( the 4 inner identical triangles )
Area a=(h×2/5)^2 = 5/25=1/5
1 |
No. Because with any θ(0<θ<π/2, θ!=π/4) given, we can draw a line that makes the angle θ with a side of the square and do it four times it forms a square in the middle. We need more information about the lines before we know the area of the square. Sofar we only know it's smaller than 1 and larger than 0
1 |
I used the cartesian system, developing eqns for 2 positive slope lines (Y=1/2X; Y=1/2X+1/2) and the first negative slope line (Y=-2X+1). With these, I found the coordinates where they intersect [(0.4,0.2) (0.2,0.6)]. Subtracting the 2 Xs and 2 Ys gave the 2 short sides of a right triangle (0.2, 0.4). Enter Mr Pythagoras, giving the square of the hypoteneuse, which is also the area of the square (0.2)
1 |
I actually had this problem for school. In geometry, we had something called the Myriad, which was a collection of difficult geometry problems that were worth more or less depending on how difficult they were, and you had to get a certain number of points before the end of the year
|
I paused it and it took me about 5-10 minutes to figure it out. 1/5. I got this because the length of the lines disecting the areas is sqrt(5)/2 because it is the hypotenuse of a 1/2 by 1 triangle, then I knew the area of the triangle, if looking at it where the hypotenuse on bottom as the base, it would be 1/2 b*h, so I found the height with b = sqrt(5)/2, but we also know that the area is 1/2 (1/2) * 1 = 1/4 because that is the area given by the smaller sides of the right triangle, so h = (1/4)/(sqrt(5)/4) = 1/sqrt(5) = sqrt(5)/5, but the smallest triangle that has the side given by that length of sqrt(5)/5 is similar to the bigger triangle and the bigger triangle has one side length of size 1, where the smaller triangle has side length 1/2, so because they are similar and the ratio is 2, we know that the place where they are connected is exactly the same length of the side as the smaller triangle, so it is sqrt(5)/5 on the side length of the square, and then squaring that number we get 5/25 = 1/5.
EDIT:
Looking at his solution it looks much more obvious, but it’s not rigorous at all, and is just for visual intuition, he used pretty much the same principle, but didn’t give any indication of why it’s true. Really good for getting people interested, but not good overall because it convinces people they know the math behind it, which creates people who think they know a lot more than they do. Good intuition though
5 |
bottom and top triangle are collectively half the square. then if you rearrange the parallelogram in the middle you get that the square is 2/5ths of that and 0.5*(2/5) = 1/5 so the square is 1/5th
1 |
I did it mathematically, although I couldn't remember off-hand whether sine was opposite over hypotenuse or adjacent over hypotenuse so I had to do a little extra work to cover for that. So I used the arctangent of 2 and 0.5 to get the values for the angles that weren't right angles. Then I did the sine of both of those angles and used the value that was greater than one, as that would be the length of the hypotenuse. Since this was the length of the pink lines and I now had two angles to work with, using one of the smaller triangles and knowing its hypotenuse was 0.5, I was able to use sine and multiplying by the hypotenuse to determine the length of the other two sides. After subtracting those values from the full length of the pink line I was able to determine the length of the sides of the blue square. Then I just squared it and I got 0.2.
|
From the framing of the question, the only way we would know the hypotenuse of the triangles are length 0.5 are by reading your mind
|
I did the visualization "wrong", still got 1/5 though
6 |
It's plausible that the trapezoid plus the triangle makes a rectangle, since the hypotenuse and the slopey trapezoid side have the same length, and the angles are complementary.
It's a bit hand wavy to assert that those rectangles are actually squares.
|
Nice math movie, like it very much❤
1 |
I stopped at "are those all triangles which sides we know ? Great we know the area of the square" without doing any calculation 😂
like my old math teacher used to say "there's calculations remaining but morally we solved the exercise"
your solution is far more elegant tho I'll admi
8 |
do y=mx+b for 3 lines and then solve.
this is how i cheesed every single geometry problem
so
a) y=x/2
b) y=-2x+1
c) y=-2x+2
the intersection between a) and b) is at (2/5, 1/5)
the intersection between a) and c) is at (4/5, 2/5)
therefore, the change in y is 1/5 and change in x is 2/5
using the pythagoreas theorem,
side length = sqrt((2/5)^2+(1/5)^2)
which ends up being sqrt(5/25)
then, to calculate the area, we square it
(sqrt(5/25))^2 and it cancels out
5/25=1/5
therefore the square has an area of 1/5
|
@acrien
7 months ago
Mathematically, you'd have to prove that when you rotate the other pieces, they fit perfectly to become the same size as the center piece. You can't say visually it looks like it and make it a proof.
822 |