The inscribed square is completely determined by the right triangle so x2/x1=x1/a; x3/x2=x2/x1, ... so x_k/x_(k-1)=x1/a, geometric sequence with ratio (x1/a)^2, total area = x1^2/(1-x1^2/a^2) solve x1 and put it in. bx1/a+x1+ax1/b=c x1=abc/(a^2+ab+b^2) A = a^2b^2c^2/((a^2+ab+b^2+bc)(a^2+ab+b^2-bc))
I imagine in the exam/HW you'd have to substitute c = √(a²+b²). It doesn't simplify much, but it is reasonable imo. You'd get a²b²(a²+b²)/[(a²+ab+b²)² - b²(a²+b²)] If you wanted to keep c then I'd write a²+ab+b² as c²+ab, which does lend itself to simplification a²b²c²/[(c²-bc+ab)(c²+bc+ab)] The last way I can think of expressing this is in terms of the full area A=½ab and the side ratio r = b/a. Then a = √(2A/r), b = √(2Ar), c = √[2A(r+1/r)] Working through the symbol manipulation, we get the expression: A_squares = 2Ar(r²+1)/(2r³+2r²+2r+1) What I like about this is that you immediately see what portion of the triangle the squares take, and it clearly doesn't care about scaling, only the ratio. The ratio is also the cotangent of the angle on the left (tangent of the upper angle), maybe this gives something nice, I'm not sure. r²+1 is 1/cos² but the polynomial in the denominator is harder to see through
Here is how I solved it. If you take away the first square and look at the smaller right triangle that contains the remaining squares then you have a triangle that similar to the larger triangle. Let's say the common ratio of sides for these 2 triangles is r and let A be the total area of all the squares. If s is the side length of the first square then we have A=s^2+r^2A or A=s^2/(1-r^2) we can use similar triangles to get that s=abc/(c^2+ab) and r=bc/(c^2+ab) putting these into the equation above gives A=a^2b^2c^2/(c^4+2abc^2+a^2b^2-b^2c^2) which if you substitute c^2=a^2+b^2 in the denominator for the first two occurrences of c (not the b^2c^2 term) then you get the same solution that you got.
I also like to see the ratio of the areas for some example triangles. Suppose the area of the bigger triangle is 1, and the total area of all squares is A: if a/b=3/4, A=200/323 ≈0.62; if a=b, A=4/7 ≈0.57; if a/b=4/3, A=75/143 ≈0.52.
@daniel_77. We're in luck here; although my last name isn't Riemann, it is instead Reiman. I'm gonna have to work on it. I'm less of a zeta and more of a beta to be honest
We can describe your life as a dynamical system F : E -> E in a metric space E, which we call "earth." It follows that we must show your life L/in E is a repeller with quadratic divergence of the dynamical system F.
Recursive argument from similar triangles: label each square X1, X2...; label the triangle in which X1 was the "largest square" T1, T2...; When a largest square is taken from right triangle Tn, that square has a side parallel to the hypotenuse, and therefore also one side perpendicular to it; All triangles Tn have the same lower-left angle, and all have right angles; therefore they are all similar. We can conclude that the ratios Xn : Tn : T(n+1) must be constant across all n; Tn = T(n+1) + Xn + Sn, where Sn is some difference made up of the two triangles whose areas are covered by Tn, but not by T(n+1) or Xn. Let R = T(n+1)/T(n), a constant ratio of areas less than 1. we can form an algebraic expression for the ratio in multiple ways. 1. Tn = T(n+1) + Xn + Sn = T(n+1)/R; (1/R-1) T(n+1) = Xn + Sn; 2. Tn = (Xn+Sn) + R(Xn+Sn) + R^2(Xn+Sn) + ... = (Xn+Sn)/(1-R) these two expressions are actually equivalent. The desired sum is the infinite sum of Xn, which is X1 + R X1 + R^2 X1 ... = X1/(1-R); specifically: R = T2/T1, so we find that the area sum(Xn) = X1/(1 - T2/T1) = X1 T1 / (T1 - T2); arguing more direclty from geometric similarity: the ratio of the infinite sum Xn to the area T1 is equal to the ratio of X1 to S1, which means we can discard all of the infinite areas and just look at the quadrilateral made up of X1 and S1 (T1-T2). The real challenge is figuring out what x1 even is in the first place, I can't even get an expression.
(as said in the other comments) the largest square is always completely determined by the right triangle (it's the largest inscribtion s.t. one side of the square is parallel and in the hypotenuse) so you don't have to compare anything and just say xn/xn-1=xn+1/xn (all you've done is rescale it and reorient it once so automatically similar)
To get a feel for it, I calculated the ratio of the green area vs the area of the triangle for the 16 primitive Pythagorean triples < 100 and got an average of 35%, with minimum of 31% for (3, 4, 5), a maximum of 42.4% for (11, 60, 61) and standard deviation of 4.6%
As expected the series of squares tumbles into a geometric series. The same happens for an infinite series of circles trapped at one angle of the trangle. In the limit a=b=c. x(n)=a.(1/3)^n This yields for the tumbling squares: A=(a^2)/8, which is 1/(2.sqrt(3))=28.8% of the total triangle area.
And a good place to stop is also an excellent place to start as ... Consider going the other way. From smallest triangle to a larger triangle with following constraints: It is an equilateral triangle with seed triangle (a) side length = Planck's constant with NO inner square contained initially and square added with recursion when required and seed triangle (b) an equilateral triangle with inner square having square side length = Planck's constant and seed triangle (c) further limited to integer multiples of Planck's constant applied to both triangle and enclosed square (yes side length of triangle and side length of square must be integer multiples of Planck's constant. Increasing from x₀ to x₁ to xₙ should be easy in each case as they are equilateral triangles Hidden complexity: At what size do additional squares with side length multiples of Planck's constant and is that also recursively well behaved? I would do it myself but there must be at least one unoccupied post-grad student out there willing and daring to give it a go. Besides I do not wish to misquote Fermat at what he said just after 25 minutes to 5 in the afternoon. Note that all lengths must as in M-U-S-T be integer lengths of Planck's constant - yes I know it is joule seconds but the answer in total joule seconds might also be something or nothing? Added complexities rhomboid, rhombus, cube, and all other regular 'oids and regular 3d 'oids, sphere maybe even tori? Growth constraints all the same: must be filled with regular squares, cubes, ... all with size integer multiples of yeh - you got it Further complexities ... filled with regular copies of the same. eg a torus of tori *hmmm sounds like a Star Wars character "I am Atorus of Tori"
The only thing that is kind of missing here is a proof that the squares are uniquely determined. It seems clear in a hand wavy way that among all the inscribed rectangles in a given right triangle which have a side along the hypotenuse that only one of them is a square, but it also is something which might be worth proving since sometimes intuitions are wrong.
I feel like a lot of the symbolic manipulation being handwaved through could just be circumvented entirely by a triangle similarity argument. Argue the squares will be a GM and then find the ratio and start term as done in the first half of the vid
I have had to stop watching at 3:45 into video. Too much to do (sigh) But wondered if there are differences between taking square size to infinitesimally small or to Planks constant yiend different resuts? That is termination at infinity OR terminating at Planks constant
![F.HERO Ft. JSPKK x ลำไย ไหทองคำ x M-PEE - ไม่สนิทบิดหมด (Thai Riders Anthem) [Official MV]](http://i.ytimg.com/vi/kPJZ0UihBfk/mqdefault.jpg)
The inscribed square is completely determined by the right triangle
so x2/x1=x1/a; x3/x2=x2/x1, ...
so x_k/x_(k-1)=x1/a, geometric sequence with ratio (x1/a)^2, total area = x1^2/(1-x1^2/a^2)
solve x1 and put it in.
bx1/a+x1+ax1/b=c
x1=abc/(a^2+ab+b^2)
A = a^2b^2c^2/((a^2+ab+b^2+bc)(a^2+ab+b^2-bc))
I imagine in the exam/HW you'd have to substitute c = √(a²+b²). It doesn't simplify much, but it is reasonable imo. You'd get
a²b²(a²+b²)/[(a²+ab+b²)² - b²(a²+b²)]
If you wanted to keep c then I'd write a²+ab+b² as c²+ab, which does lend itself to simplification
a²b²c²/[(c²-bc+ab)(c²+bc+ab)]
The last way I can think of expressing this is in terms of the full area A=½ab and the side ratio r = b/a.
Then a = √(2A/r), b = √(2Ar), c = √[2A(r+1/r)]
Working through the symbol manipulation, we get the expression:
A_squares = 2Ar(r²+1)/(2r³+2r²+2r+1)
What I like about this is that you immediately see what portion of the triangle the squares take, and it clearly doesn't care about scaling, only the ratio. The ratio is also the cotangent of the angle on the left (tangent of the upper angle), maybe this gives something nice, I'm not sure. r²+1 is 1/cos² but the polynomial in the denominator is harder to see through
Here is how I solved it.
If you take away the first square and look at the smaller right triangle that contains the remaining squares then you have a triangle that similar to the larger triangle.
Let's say the common ratio of sides for these 2 triangles is r and let A be the total area of all the squares. If s is the side length of the first square then we have
A=s^2+r^2A
or
A=s^2/(1-r^2)
we can use similar triangles to get that
s=abc/(c^2+ab)
and
r=bc/(c^2+ab)
putting these into the equation above gives
A=a^2b^2c^2/(c^4+2abc^2+a^2b^2-b^2c^2)
which if you substitute c^2=a^2+b^2 in the denominator for the first two occurrences of c (not the b^2c^2 term) then you get the same solution that you got.
This is exactly how I did it as well. It just felt simpler and easier. The GP is very obvious
I also like to see the ratio of the areas for some example triangles. Suppose the area of the bigger triangle is 1, and the total area of all squares is A:
if a/b=3/4, A=200/323 ≈0.62;
if a=b, A=4/7 ≈0.57;
if a/b=4/3, A=75/143 ≈0.52.
Next, please calculate how fast my life is tumbling into absurdity.
Let's work on it back & forth as a group; I'll start us off..
Abstract-
It will require the proof of the rienman Hypotheses
@daniel_77. We're in luck here; although my last name isn't Riemann, it is instead Reiman. I'm gonna have to work on it. I'm less of a zeta and more of a beta to be honest
We can describe your life as a dynamical system F : E -> E in a metric space E, which we call "earth." It follows that we must show your life L/in E is a repeller with quadratic divergence of the dynamical system F.
that's an easy one. about c, give or take a factor of 2
Recursive argument from similar triangles:
label each square X1, X2...;
label the triangle in which X1 was the "largest square" T1, T2...;
When a largest square is taken from right triangle Tn, that square has a side parallel to the hypotenuse, and therefore also one side perpendicular to it;
All triangles Tn have the same lower-left angle, and all have right angles; therefore they are all similar.
We can conclude that the ratios Xn : Tn : T(n+1) must be constant across all n;
Tn = T(n+1) + Xn + Sn, where Sn is some difference made up of the two triangles whose areas are covered by Tn, but not by T(n+1) or Xn. Let R = T(n+1)/T(n), a constant ratio of areas less than 1.
we can form an algebraic expression for the ratio in multiple ways.
1. Tn = T(n+1) + Xn + Sn = T(n+1)/R; (1/R-1) T(n+1) = Xn + Sn;
2. Tn = (Xn+Sn) + R(Xn+Sn) + R^2(Xn+Sn) + ... = (Xn+Sn)/(1-R)
these two expressions are actually equivalent. The desired sum is the infinite sum of Xn, which is X1 + R X1 + R^2 X1 ... = X1/(1-R); specifically: R = T2/T1, so we find that the area sum(Xn) = X1/(1 - T2/T1) = X1 T1 / (T1 - T2);
arguing more direclty from geometric similarity: the ratio of the infinite sum Xn to the area T1 is equal to the ratio of X1 to S1, which means we can discard all of the infinite areas and just look at the quadrilateral made up of X1 and S1 (T1-T2).
The real challenge is figuring out what x1 even is in the first place, I can't even get an expression.
(as said in the other comments) the largest square is always completely determined by the right triangle (it's the largest inscribtion s.t. one side of the square is parallel and in the hypotenuse) so you don't have to compare anything and just say xn/xn-1=xn+1/xn (all you've done is rescale it and reorient it once so automatically similar)
actually don't think you have to say "and in" (I juked myself and removed mid writing this) but I'll keep it in as a redundancy
To get a feel for it, I calculated the ratio of the green area vs the area of the triangle for the 16 primitive Pythagorean triples < 100 and got an average of 35%, with minimum of 31% for (3, 4, 5), a maximum of 42.4%
for (11, 60, 61) and standard deviation of 4.6%
As expected the series of squares tumbles into a geometric series. The same happens for an infinite series of circles trapped at one angle of the trangle. In the limit a=b=c. x(n)=a.(1/3)^n This yields for the tumbling squares: A=(a^2)/8, which is 1/(2.sqrt(3))=28.8% of the total triangle area.
This channel never disappoints in providing nice problems to dive into! Thank you
3:22 putting y at the bottom is unnerving but z on the hypotenuse doesn't matter. ☺
I was half expecting phi to show up...
When you're simplifying the denominator, why not replace (a² + b²) with c², since we've got a right triangle?
And a good place to stop is also an excellent place to start as ...
Consider going the other way. From smallest triangle to a larger triangle with following constraints:
It is an equilateral triangle with
seed triangle (a) side length = Planck's constant with NO inner square contained initially and square added with recursion when required and
seed triangle (b) an equilateral triangle with inner square having square side length = Planck's constant and
seed triangle (c) further limited to integer multiples of Planck's constant applied to both triangle and enclosed square (yes side length of triangle and side length of square must be integer multiples of Planck's constant.
Increasing from x₀ to x₁ to xₙ should be easy in each case as they are equilateral triangles
Hidden complexity: At what size do additional squares with side length multiples of Planck's constant and is that also recursively well behaved?
I would do it myself but there must be at least one unoccupied post-grad student out there willing and daring to give it a go.
Besides I do not wish to misquote Fermat at what he said just after 25 minutes to 5 in the afternoon.
Note that all lengths must as in M-U-S-T be integer lengths of Planck's constant - yes I know it is joule seconds but the answer in total joule seconds might also be something or nothing?
Added complexities rhomboid, rhombus, cube, and all other regular 'oids and regular 3d 'oids, sphere maybe even tori?
Growth constraints all the same: must be filled with regular squares, cubes, ... all with size integer multiples of yeh - you got it
Further complexities ... filled with regular copies of the same. eg a torus of tori *hmmm sounds like a Star Wars character "I am Atorus of Tori"
The only thing that is kind of missing here is a proof that the squares are uniquely determined. It seems clear in a hand wavy way that among all the inscribed rectangles in a given right triangle which have a side along the hypotenuse that only one of them is a square, but it also is something which might be worth proving since sometimes intuitions are wrong.
As the height of the inscribed rectangle increases, its width decreases, so there cannot be more than one square
I feel like a lot of the symbolic manipulation being handwaved through could just be circumvented entirely by a triangle similarity argument. Argue the squares will be a GM and then find the ratio and start term as done in the first half of the vid
It's not mentioned early that the first square is as big as it can possibly be, but it would be a similar problem regardless I think.
It really triggered me that abc are labelled clockwise...
I'm triggered by the fact that he never used a²+b²=c², so a²+ab+b² = c²+ab, which is shorter.
Are you right-handed?
The geometric ratio IS less than 1:
bc/(a^2+ab+b^2) < 1 iff
bc < a^2+ab+b^2 iff
bc < c^2+ab iff
0 < c^2-bc+ab iff
0 < c(c-b)+ab.
13:00 it's ironic since the word algebra come from the Arabic words for book of shifting and changing
Ok. Mislio sam na prosječnu širinu i prosječnu dužinu, pa onda može biti i trikvadroduetar. Nije bitno kojeg je oblika.
I have had to stop watching at 3:45 into video. Too much to do (sigh)
But wondered if there are differences between taking square size to infinitesimally small or to Planks constant yiend different resuts?
That is termination at infinity OR terminating at Planks constant
Iba a decir Magenta😅😅😅. Instead of Mayenta.
PLEASE VOTE IF YOURE OVER 18 AND IN THE US. I DONT CARE WHO YOU VOTE FOR, JUST VOTE!!!
Math is more important.
l@maxhagenauer24 lol, I'm queen and neurodivetgent, so the election is important to me! (I'm too young to vote myself)
@@sebas31415 Vote for a math teacher to be president when you can.
@@maxhagenauer24 so vote for noone?
Vote for Michael Penn