@@calmsouls4502 "they" is a gender neutral personal pronoun in the English language, as well as being 3rd person plural. this means it can be used to describe any person (without the need to discriminate between genders) as well as a group of people
These were the kinds of videos I craved back when I was in middle and highschool, there just weren't enough good quality ones. I am certain you'll have a number of passionate young mathematicians learn a lot of important skills from these videos.
This whole time I was trying to think of what the “ideal” case would be like if you started with an infinitely large “circle” that was parallel to your line, so Csub1 was congruent to C, maximizing all CsubN
What's cool is that you can choose any initial position for the C0 and the infinte sequence of circles will all still converge to the same point, this is a pretty neat intuition for a how an infinite sequence is determined more by how it converges rather than what it converges on. Each set of circles is following the same general rule, but they all converge to an indistinguishable state in the limit; so, intuitively, it makes sense that you cannot infer the intial circle only by knowing that limit (or by only looking at that indistinguishable state) . Seems pretty obvious in this context but it can be confusing for limits of 1^n for example.
Very nice question. I had a little play around, i even summed the areas of the circles. lfhcvsqjyo (desmos code) if anyone wants to have a look for anyone too lazy, the total area is pi/24 * D * psi[3](sqrt(D/D_0)) where psi[3](x) is the 3rd derivative of the digamma function
When trying to solve the initial problem, I found that the locus of the secondary circles follows the equation of a parabola (x^2/4) This means that you can have a secondary circle of any size. If you go backwards, you *can* continue the sequence past the size of the original circle, however, this requires that the circles alternate sides of the base circle, which isn't as pretty.
I did something similar to this with stacking them vertically between two adjacent equally sized circles on a line. It ended up being an elegant proof of the sum of the inverse of triangular numbers being 2
The largest circle you could make depends on if you are counting up or counting down. The largest circle circle that you could draw with the two points of tangency would be a congruent circle. Then the question becomes how can you count up so that the final result is a congruent circle where you can draw a line between the two diameters that is parallel with the bottom line. If the second circle has to be smaer it can be reconceptualized as a limit
So the outro question boils down to replacing +k with -k Dk = ( sqrt(D D0) / [ sqrt(D) - k sqrt(D0) ] )^2 Now the impossibility of drawing Dk strictly speaking only happens when `sqrt(D) - k sqrt(D0) = 0`, in other words `D/D0 = k^2` is a perfect square. To construct these circles start with with D0 = D, you have two identical circles on line L touching each other. Now fit a smaller circle inbetween those two, you'll get D/D0 = 4. Keep going and these circles' diameters will always be a perfect square factor from D. The "impossible" circle would have to touch the two identical circles, thus it has to be a degenerate circle of infinite size, aka. a straight line parallel to L. If D/D0 is not a perfect square, the denominator `sqrt(D) - k sqrt(D0)` will get the closest to 0 on the positive side at `k = floor(sqrt(D/D0))`. Let's rewrite Dk = ( sqrt(D) / [sqrt(D/D0) - k] )^2 = D / [sqrt(D/D0) - floor(sqrt(D/D0))]^2 Note that the denominator is just the fractional part of sqrt(D/D0) squared which is less than 1, so Dk > D. After that C{k+1} would have to touch L on the left side of C and all the following circles will form squeezing circles on the left side of C.
considering D != D_0 (more generally, there exist no k such that D_k = D/4), max amount of circles we can construct: n = ceil(sqrt(D/D_0)) We can quick check the trivial case D_0 > D, and see that only 1 circle can be constructed, before we can no longer make new circles, without overlapping old ones.
i think it will work until one circle is as big as the original circle. as long as the last circle is smaller than the main circle you can add another. it may require it to be very large though
ok now assuming the big circle radius is 1 for simplicity, what do the sum of the area of the circles converge to as the amount of circles approach infinity, i know it will be less than 1 but whats the value
Hi! the sum of the areas of all of the circles would be pi/24 * D * phi[3](sqrt(D/D_0)) where phi[3](x) is the 3rd derivative of the digamma function, D is the diameter of the Big circle, and D_0 is the diameter of the first circle in the chain. lfhcvsqjyo is the desmos code if you want to have a look around
I have another challenge- find the value of D0 such that the circle C0+1 “squeezes” C with the same proportions as C/C0, allowing a fractal “anti-squeezing” series
Giving it a go, and I like your animation style. The circumference is irrational. If there is a rational L then you must be measuring the two inner axes of the circle, and the circles will fall inside so measuring the area is the sum of all the circles'. The line must be irrational to follow the described pattern, which means there is no pattern to the circles' diameters yet and they are under pressure. A pattern must be related to some external factor, and not yet disclosed.
As I see it from the thumbnail; I'd have to conclude it'd be in kinda a curved shape with a flat bottom so long as we are only looking at a flat mathematical plane. You should trust me. I'm a very good armchair scientist
Maybe it would depend on the makeup and density of the circle. But it definitely would depend on the shape you're trying to squeeze it into, would it not?
This can be solved using Descartes' theorem (generalized to generalized circles) which states that the curvatures of 4 touching generalized circles in the plane satisfy (k_1+k_2+k_3+k_4)^2 = 2(k_1^2+k_2^2+k_3^2+k_4^2) In this case, we have (using radii instead of diameters): k_1 = 1/R k_2 = 1/R_0 k_3 = 0 (this is the line) So we get (1/R + 1/R_0 + 0 + k_4)^2 = 2(1/R^2 + 1/R_0^2 + 0^2 + k_4^2) And after some computation k_4^2 - 2(1/R + 1/R_0)k_4 + (1/R - 1/R_0)^2 = 0 So k_4 = 1/R + 1/R_0 ± √( (1/R + 1/R_0)^2 - (1/R - 1/R_0)^2 ) = 1/R + 1/R_0 ± 2√( 1/(R R_0) ) = (1/√R ± 1/√R_0)^2 = ((√R ± √R_0)/√(R R_0))^2 We're getting two solutions here, but we want the smaller circle so we want k_4 to be bigger than k_1 and k_2, so lets take the positive solution k_4 = ((√R + √R_0)/√(R R_0))^2 So R_1 = (√(R R_0)/(√R + √R_0))^2 And if we want the previous circle, it would simply be R_{-1} = (√(R R_0)/(√R - √R_0))^2 (note that if R=R_0, then this is not well defined, instead if we look at curvatures we can see that we get that the previous "circle" is a line parallel to the original one) I learned this while working on a visualization of an Apollonian gasket ( basically do the above computation over an over and over and over and over and over, but not really since once you have 4 circles you can use some linear transformation to get the fifth, and in fact you can calculate the centers in essentially the same way! )
This one was pulled from the book "Parabolic Problems: 60 Years of Mathematical Puzzles in Parabola" by David Angell and Thomas Britz. There are so many great problems from Parabola!
Soooo, I solved it, but on radiuses(so had to half on start and double the result) and forgot that R = sqrt(R)^2 so I ended up with the ()^2 in the denominator and R above it, also took me a longer way because of bad ideas I got and mistakes It was fun
Because for any given circle C_k along the chain, you can easily prove that there exists a circle C_k+1. By induction this means theres infinite circles
This puzzle reminds me of a one in VIsual Complex Analysis by Tristan Needham. I wonder if inverting this problem with respect to the bigger circle makes it easier Edit: I think the problem becomes harder if you invert it in a circle :)
5:36 "How many circles will we be able to add before the next circle can no longer touch both the line L and the circle C?" So I haven't done the algebra but I'm pretty sure that never happens. To see this, invert about any circle whose center is the point of tangency between L and C. Then the image of L is L, the image of C is a line parallel to L, and the image of the C_n's is a sequence of mutually tangent circles sandwiched between the two parallels. Seen from this perspective, it's obvious that no matter which direction you squeeze the circles into, they'll both result in infinite chains.
Neither of you read the construction I described, so here's the id of a desmos graph demonstrating it: djipnet4am (youtube won't let me comment the link for some reason). What you're both missing is that after the circles get big enough, they wrap around to the other side of the line.
@@boderaner No matter how big you make the diameter of the next it won't simultaneously touch the floor, the previous circle, and the C circle. Even if this isn't immediately obvious to you, consider the geometry of having the previous circle as big as the C circle. How can it simultaneously touch the floor, the previous circle and the C circle? It's just impossible. If you think it is possible, draw it in paper. At most you will be able to make one circle as big as C (probably not even that), but I guarantee more than that won't fit.
@jurd-e3d + in a word. Not rotation x. + is axySyMMetry, death,potENtial, AChirality, r-nose-ance, scalar, twist, helice,propeller,dielectric, lon-digit-unal sound(closed system),aDDition (MIXing MM I CS, STr8 line really, because Archimedes screw is upDown, unlike str8 line Aries mushroom of rising empty Shu, eddy or vortex or spinor, 2 opposite swirls whirLings giving 2 atoms or 1 electron. Pi. Plus, plasma, syMMetry diMMing(saDDLe miDDle ) contains itself and its opposite. X is bobbin, chiral life, rotation, heat, transverse sound,magnetism,light illumination, phi. More on twist in tilings. Also quanta magazine on how two spheres are connected. Sphere is a sphere in any dimension. Mixing ...ST-andi-ng STationary STiLL time vibration..anda or ence is ellipse or cenote. EN or e9 is hole pore..evaPoration or voLaTile, eversion. Tion is ohi light 89. 1+1=4 means 1 straight back and forth. Atom + atom = 4 atoms of voID or vcUUm. Light + light = darkness ShaDow. HeliCity= knots + links Euler eq connecting exponents and angle.. Etc
Common mistake, you can't fit an infinite amount of circles, aside from the possibility of plank length, you will still be adding circles when the universe dies, never reaching even 1%of infinity.
Oh, they were saying "P sub k" not "piece of cake"
He*
@@calmsouls4502 "they" is a gender neutral personal pronoun in the English language, as well as being 3rd person plural. this means it can be used to describe any person (without the need to discriminate between genders) as well as a group of people
Unnecessary correction:
Both use of pronouns are valid@@calmsouls4502
oh, they were saying "let C be a circle" not "let cba circle"
Oh, they were saying ‘c naught’ not ‘sea nut’
These were the kinds of videos I craved back when I was in middle and highschool, there just weren't enough good quality ones. I am certain you'll have a number of passionate young mathematicians learn a lot of important skills from these videos.
This whole time I was trying to think of what the “ideal” case would be like if you started with an infinitely large “circle” that was parallel to your line, so Csub1 was congruent to C, maximizing all CsubN
You can do an inversion transformation at the point of tangent to change it into two parallel lines. en.m.wikipedia.org/wiki/Inversion_transformation
The diameter formula still works as a limit (take D to infinity), yielding D_k = D_0 as expected.
What's cool is that you can choose any initial position for the C0 and the infinte sequence of circles will all still converge to the same point, this is a pretty neat intuition for a how an infinite sequence is determined more by how it converges rather than what it converges on. Each set of circles is following the same general rule, but they all converge to an indistinguishable state in the limit; so, intuitively, it makes sense that you cannot infer the intial circle only by knowing that limit (or by only looking at that indistinguishable state) . Seems pretty obvious in this context but it can be confusing for limits of 1^n for example.
dude. great animation and sound design, great video in general
Did you deliberately start the video with "let C B A"?
this video is ceaselessly, boundlessly amazing
0:00
let CBA
chess battle advanced
Icely mentioned
No the icely is leaking
oh my god
ICELY MENTIONED
chess battle advanced
Very nice question. I had a little play around, i even summed the areas of the circles. lfhcvsqjyo (desmos code) if anyone wants to have a look
for anyone too lazy, the total area is pi/24 * D * psi[3](sqrt(D/D_0)) where psi[3](x) is the 3rd derivative of the digamma function
Let cba circle
0:00
Let chess battle advanced circle...
the choice of defining all the distances in this equation by diameter and not radius was bonkers
maybe because the question is asking for an equation in terms of diameter
When trying to solve the initial problem, I found that the locus of the secondary circles follows the equation of a parabola (x^2/4) This means that you can have a secondary circle of any size. If you go backwards, you *can* continue the sequence past the size of the original circle, however, this requires that the circles alternate sides of the base circle, which isn't as pretty.
I am genuinely surprised you did not use circle inversion to solve this
1:06 Wait 5 minutes.
actually the problem of drawing circles to the right of the C0 can be solved the same formula as the first problem, but with negative indexes
I did something similar to this with stacking them vertically between two adjacent equally sized circles on a line. It ended up being an elegant proof of the sum of the inverse of triangular numbers being 2
The largest circle you could make depends on if you are counting up or counting down. The largest circle circle that you could draw with the two points of tangency would be a congruent circle. Then the question becomes how can you count up so that the final result is a congruent circle where you can draw a line between the two diameters that is parallel with the bottom line. If the second circle has to be smaer it can be reconceptualized as a limit
Bro is back! 🎉🎉🥳
So the outro question boils down to replacing +k with -k
Dk = ( sqrt(D D0) / [ sqrt(D) - k sqrt(D0) ] )^2
Now the impossibility of drawing Dk strictly speaking only happens when `sqrt(D) - k sqrt(D0) = 0`, in other words `D/D0 = k^2` is a perfect square.
To construct these circles start with with D0 = D, you have two identical circles on line L touching each other. Now fit a smaller circle inbetween those two, you'll get D/D0 = 4. Keep going and these circles' diameters will always be a perfect square factor from D. The "impossible" circle would have to touch the two identical circles, thus it has to be a degenerate circle of infinite size, aka. a straight line parallel to L.
If D/D0 is not a perfect square, the denominator `sqrt(D) - k sqrt(D0)` will get the closest to 0 on the positive side at `k = floor(sqrt(D/D0))`.
Let's rewrite Dk = ( sqrt(D) / [sqrt(D/D0) - k] )^2 = D / [sqrt(D/D0) - floor(sqrt(D/D0))]^2
Note that the denominator is just the fractional part of sqrt(D/D0) squared which is less than 1, so Dk > D. After that C{k+1} would have to touch L on the left side of C and all the following circles will form squeezing circles on the left side of C.
That was really a "piece of cake".
I was just watching the grid video after the full house one.
considering D != D_0 (more generally, there exist no k such that D_k = D/4), max amount of circles we can construct:
n = ceil(sqrt(D/D_0))
We can quick check the trivial case D_0 > D, and see that only 1 circle can be constructed, before we can no longer make new circles, without overlapping old ones.
i think it will work until one circle is as big as the original circle. as long as the last circle is smaller than the main circle you can add another. it may require it to be very large though
ok now assuming the big circle radius is 1 for simplicity, what do the sum of the area of the circles converge to as the amount of circles approach infinity, i know it will be less than 1 but whats the value
Hi! the sum of the areas of all of the circles would be pi/24 * D * phi[3](sqrt(D/D_0)) where phi[3](x) is the 3rd derivative of the digamma function, D is the diameter of the Big circle, and D_0 is the diameter of the first circle in the chain. lfhcvsqjyo is the desmos code if you want to have a look around
@gamerboy7224 my goat
Star Size Comparison 😬
Great quality
I have another challenge- find the value of D0 such that the circle C0+1 “squeezes” C with the same proportions as C/C0, allowing a fractal “anti-squeezing” series
Giving it a go, and I like your animation style. The circumference is irrational. If there is a rational L then you must be measuring the two inner axes of the circle, and the circles will fall inside so measuring the area is the sum of all the circles'. The line must be irrational to follow the described pattern, which means there is no pattern to the circles' diameters yet and they are under pressure. A pattern must be related to some external factor, and not yet disclosed.
Animation style is manim by 3b1b
Waiting for the, best full house in holdem🎉
holdem balls
Interesting solution!
Thanks for sharing.
It would also be interesting to suggest a method of making these circles.
Another question: Equation of locus of centers of those small circles
x^2/2D, a parabola with the center of the big circle, D/2 as it's focus
As I see it from the thumbnail; I'd have to conclude it'd be in kinda a curved shape with a flat bottom so long as we are only looking at a flat mathematical plane.
You should trust me. I'm a very good armchair scientist
Maybe it would depend on the makeup and density of the circle. But it definitely would depend on the shape you're trying to squeeze it into, would it not?
What does circle fit?
This can be solved using Descartes' theorem (generalized to generalized circles) which states that the curvatures of 4 touching generalized circles in the plane satisfy
(k_1+k_2+k_3+k_4)^2 = 2(k_1^2+k_2^2+k_3^2+k_4^2)
In this case, we have (using radii instead of diameters):
k_1 = 1/R
k_2 = 1/R_0
k_3 = 0 (this is the line)
So we get
(1/R + 1/R_0 + 0 + k_4)^2 = 2(1/R^2 + 1/R_0^2 + 0^2 + k_4^2)
And after some computation
k_4^2 - 2(1/R + 1/R_0)k_4 + (1/R - 1/R_0)^2 = 0
So
k_4 = 1/R + 1/R_0 ± √( (1/R + 1/R_0)^2 - (1/R - 1/R_0)^2 ) = 1/R + 1/R_0 ± 2√( 1/(R R_0) ) = (1/√R ± 1/√R_0)^2 = ((√R ± √R_0)/√(R R_0))^2
We're getting two solutions here, but we want the smaller circle so we want k_4 to be bigger than k_1 and k_2, so lets take the positive solution
k_4 = ((√R + √R_0)/√(R R_0))^2
So R_1 = (√(R R_0)/(√R + √R_0))^2
And if we want the previous circle, it would simply be R_{-1} = (√(R R_0)/(√R - √R_0))^2 (note that if R=R_0, then this is not well defined, instead if we look at curvatures we can see that we get that the previous "circle" is a line parallel to the original one)
I learned this while working on a visualization of an Apollonian gasket ( basically do the above computation over an over and over and over and over and over, but not really since once you have 4 circles you can use some linear transformation to get the fifth, and in fact you can calculate the centers in essentially the same way! )
Could we derive the rule
>0
=0
By increasing circle size
I thought you were going to ask what is the total area of all infinitely many circles.
Great video! Also, where is the question from?
This one was pulled from the book "Parabolic Problems: 60 Years of Mathematical Puzzles in Parabola" by David Angell and Thomas Britz. There are so many great problems from Parabola!
Soooo, I solved it, but on radiuses(so had to half on start and double the result) and forgot that R = sqrt(R)^2 so I ended up with the ()^2 in the denominator and R above it, also took me a longer way because of bad ideas I got and mistakes
It was fun
I have no freaking idea.
Pie? I know circles are involved with pie, but what kind of pie? Do they prefer meat or apples?
We'll never know.
meat
kind of looks like a golden ratio relationship
I want a piece of cake now.
How do we know that we can put infinite circles to the left? Eventually C will touch the line..
Because for any given circle C_k along the chain, you can easily prove that there exists a circle C_k+1. By induction this means theres infinite circles
@gamerboy7224 fair
pretty neat!
Plug in a negative value?
This puzzle reminds me of a one in VIsual Complex Analysis by Tristan Needham. I wonder if inverting this problem with respect to the bigger circle makes it easier
Edit: I think the problem becomes harder if you invert it in a circle :)
Was this a BMO1 question?
My friend gave this problem in class and we both solved it🗣️🗣️🗣️🗣️🗣️🗣️
banger animation
5:36 "How many circles will we be able to add before the next circle can no longer touch both the line L and the circle C?"
So I haven't done the algebra but I'm pretty sure that never happens. To see this, invert about any circle whose center is the point of tangency between L and C. Then the image of L is L, the image of C is a line parallel to L, and the image of the C_n's is a sequence of mutually tangent circles sandwiched between the two parallels. Seen from this perspective, it's obvious that no matter which direction you squeeze the circles into, they'll both result in infinite chains.
It's obvious that it is impossible to fit circles indefinitely going the other way, just look at it for 5 seconds, it's impossible.
bruh, he literally show u an example in the video
Neither of you read the construction I described, so here's the id of a desmos graph demonstrating it: djipnet4am (youtube won't let me comment the link for some reason). What you're both missing is that after the circles get big enough, they wrap around to the other side of the line.
@@freshrockpapa-e7799 it's possible. As circle at 5:42 is possible too. We have not limitations for diameters of circles, have we?
@@boderaner No matter how big you make the diameter of the next it won't simultaneously touch the floor, the previous circle, and the C circle. Even if this isn't immediately obvious to you, consider the geometry of having the previous circle as big as the C circle. How can it simultaneously touch the floor, the previous circle and the C circle? It's just impossible. If you think it is possible, draw it in paper. At most you will be able to make one circle as big as C (probably not even that), but I guarantee more than that won't fit.
An infinite NUMBER! Sheesh.
Infinite u didn’t specify it could touch either more than once 5:53
Nice thanks
Sphere is a twist away of itself, but different size.
Small sphere is a twist away from big sphere.
What
@themathhatter5290 I taught of sphere and shadow. Projection. Also changed comment to be clear. I hope.
What does 'twist' mean in this context?
@jurd-e3d + in a word. Not rotation x.
+ is axySyMMetry, death,potENtial, AChirality, r-nose-ance, scalar, twist, helice,propeller,dielectric, lon-digit-unal sound(closed system),aDDition (MIXing MM I CS, STr8 line really, because Archimedes screw is upDown, unlike str8 line Aries mushroom of rising empty Shu, eddy or vortex or spinor, 2 opposite swirls whirLings giving 2 atoms or 1 electron. Pi.
Plus, plasma, syMMetry diMMing(saDDLe miDDle ) contains itself and its opposite.
X is bobbin, chiral life, rotation, heat, transverse sound,magnetism,light illumination, phi.
More on twist in tilings. Also quanta magazine on how two spheres are connected. Sphere is a sphere in any dimension.
Mixing ...ST-andi-ng STationary STiLL time vibration..anda or ence is ellipse or cenote.
EN or e9 is hole pore..evaPoration or voLaTile, eversion. Tion is ohi light 89.
1+1=4 means 1 straight back and forth.
Atom + atom = 4 atoms of voID or vcUUm.
Light + light = darkness ShaDow.
HeliCity= knots + links
Euler eq connecting exponents and angle..
Etc
@@junak777I’m still not sure what you’re trying to get at
GREAT
mandelbrot
Why even bother with D, why not just use R?
D means diameter, R is just radius and D = 2R. Probably why
Chess battle advanced
chess battle advanced
Woot
Descartes' theorem (:
me who has not once thought about how circles squeeze, seeing this video: yeah how _do_ circles squeeze anyway?
2520
HOW ARE THERE SO MANY ICELY FANS HERE
yt algorithm deliberately and conveniently recommending this video to Icely fans maybe because it heard the same phrase from the start of the video
Common mistake, you can't fit an infinite amount of circles, aside from the possibility of plank length, you will still be adding circles when the universe dies, never reaching even 1%of infinity.
get lost dude, the planck length doesnt exist in mathematics dummy. Also nothing actually forbids things from being smaller than the planck length
Would this be fun in higher dimensions - maybe longer video
Graphics and equations here are nice 😎🤌