Exercise 2: Sum of diameters of Infinite Circles by @mes on 3Speak
In this video, I show that the sum of diameters of an infinite number of circles contacting between two big circles sums to the radius of the bigger circles, thus providing a geometric interpretati...
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2, and it's got quite fascinating. This one, this one's pretty intense, but nonetheless, it's very fascinating. And similar to the one above, we're going to be finding geometric interpretation of, or geometric demonstration of example 6. So the below figure shows two circles C and D, a radius 1, that touch at point P. So here's two circles, a radius is equal to 1 over here. So radius equals to 1, they touch at point P. There, and there's a tangent line across there as well. So now, the T is a common tangent line. C1 is the circle that touches C, D, and T. So there's C1, touches over here, over here, and over here. So it touches C, D, and T, like that. And C2 is a circle that touches C, C, D, and C1. So we have here. touches C1, C2, I mean, D, and C, like that. And then C3 is a circle that touches C, D, and then C2. So these circles all touch C and D, and then the one before it, and then this one is touches the tangent there. So this one touches, yeah, C2, and C and D there. So yes, and it keeps going on and on. So this procedure can be continued indefinitely and produces an infinite sequence of circles, c of n like that. Yes, cn, so that's a sequence there of circles. Now we're asked, find an expression for the diameter of cn, of any of these ones, so the diameter across like that, C or a dn, like that, dn. Yeah, so and thus provide another geometric demonstration of example 6. So, note that some of the diameters of the inner circles are approaching the radius of the larger circles, radius of the larger circles, which is equal to 1. So yes, what I mean by that is, notice we're here finding an expression for the diameter of CN, and that's provide another one. So if we were to get the diameter of this one, diameter of this one, diameter of this one, add those all up. Notice what they're approaching, they're approaching this point P. In other words, they're approaching this right here. They're approaching 1 equals R is equal to 1. This is a sum of the d n like that, from n equals to 1, all the way to infinity. So it's approaching that. So that's what we've got to show. And here I quickly just change color to black. I was applying to blue later. But anyway, so yeah, so the summation of these diameters, yes, it should be approaching 1, which is the radius of the big circle, because that's where they touch right here at P. So, very fascinating stuff. So there's a geometric interpretation, and we're going to prove that it is so. Let's go further. So, let D N be the diameter of C, N, like that. So I have it capital D, so it's the lower one here, like that. So, just to quickly change this to lowercase D, like that, D. So let it be the diameter of C, and we draw lines from the centers of C-I to the center of D or C. So here's the circle D, this circle. see like that so you could choose either one and then using the Pythagrum theorem we write so what we have is so this is the one that I wanted to draw in blue so we have it here in blue like that So notice here what do we have, I just took it from my calculus book, the solution's manual of it. So we have, here's the radius R equals 1 of the big circle D. This one is a circle D. And notice that they all, or this touches the other circle at the point P. So I'll just draw a point P like that. This is our point P. Recall P is right over there. So then, again, the radius here is also R equals 1 there. So this distance all the way across. like this, this distance is, this distance like that, is R equals to 1. Or just for convention, I'll just use a capital R for the big circles, and capital R for the big circles, and there's the, like here, also there's just use a capital R for a big circle like that, R equals 1. Yeah, so in other words, yeah, so at this point getting infinitely small, or these circles getting infinitely small, it's approaching this distance across R equals to 1. But notice that the center of this is always going to go up and up. So the center is just going to go up and up all the way there. This goes like that, goes like that, et cetera. So in other words, what we end up having is, well, this R, this radius, is going across everywhere across here, so we have an R equals 1. R equals 1 from here to here. That's actually from here. So these ones are all approaching, getting closer and closer to R equals 1. This distance is from here, all the way to here as well. So that's D2, and then this one as well from here, all the way across. So yeah, so this one, every single point's going to be from here to there, it's going to be R equals 1 as well. Like that, R equals 1. Let's draw this all the way across like that. We put a 1 there. And likewise, this r equals 1 distance, that's just also equals to 1 there, and there's our tangent line at the bottom from starting D1. And another note I want to make is notice the distance from here to the tangent line, from this one, D1, this is going to be right from here, it's just going to be D1 over 2, like that. And likewise, over here, this distance across here. This is D1 over, I guess, let's put a, this is, I mean, this D2 over D2, I'm D2 over 2, like that. And now we're going to use the Pythagorean theorem, so recall the Pythagorean theorem, so recall, I'll sorry, recall Pythagorean theorem, p.t. This is just if you had a right triangle like this, where you had the sides a, B and C for hypotenuse, then what we end up having is the formula A squared plus B squared equals to C squared. Now, in our case, what we end up having is, well, we have a right angle triangle right there. So this distance across here, all we're done here, I'll draw this in blue, actually, so we have it like that. This distance is going to be the difference. So this is going to be one minus. D1 over 2, like that. And we also know that this is R is 1 all the way up to here, and then this is just the radius over from here to here. That's just going to be our D1 over 2, like that. Yes, so D1 over 2, and then we have the first side is 1 there. So in other words, we can write this as, so thus, or put an arrow like this, so thus. or actually just put dust instead without it. So we have the line from here to here is just 1. That's just the radius. So, A is 1. Well, yeah, 1 squared, which is just 1. Plus, now B squared is over this, this is B. That's just 1 minus 1 minus D1 over 2 squared. And now what this equals 2 is, well, the hypotenuse is over here, R equals 1, plus. d1 over 2 then just square that all the total so this is the hypotenuse 1 plus d1 over 2 squared because again yeah so we have this height here on this hypotenuse and this one here is this uh is side a side b and there's side c there's what we have so now what i'm going to do is i'm going to actually move this across i'm going to move this over to this side, then what we end up having is, well, 1 squared is just 1. So 1 equals 2, 1 plus the diameter divided by 2, or the radius of a small circle, the D1 over 2, all squared, minus, now we have this, 1 minus D1 over 2 squared, like that. And now, if we recall from earlier video on difference of squares, or link to it, but also, yeah, basically what it states is, if you have a function like this, or if you have equation like this, not a function, just an equation, a plus b, times it by a plus b, or a minus b. And if you were to foil this out or expand it, so you'll have a times a, which equals to a squared, then a times b, which is going to be negative a, b, and then likewise from here, b times a, it's going to be plus. ab and then b times negative b that's going to be minus b squared so this ends up canceling so when we have this right here we end up having is a difference of squares of a squared minus b squared yes in other words we have a difference of squares here but we can write it in a way that removes the squares so there's no squares here this is a plus b times a minus b so in our case so thus we have A equals 2, well, A is just equal to this inside here, so A plus D1 over 2, like that, and then B, all right, and B, I'll just put a comma, and then B is equal to 1 minus D1 over 2. So thus what we end up having is, 1 equals 2, well, this is just different to squares, A squared minus B squared, which is equals to a minus b or a plus b a minus b and we could just fill those inside and add them up so this equals to we have a is one plus d one half like that and then b plus b is one minus d one half like that and then b plus b is one minus d one half like that and then times it by a which is 1 plus d 1 half minus over here. So we're going to minus, we have to put a minus 1, and then minus becomes positive there, plus D1 half, like that. And now we can do is cancel some like terms. So here, 1 plus 1 adds up to be 2. So we end up having as 2 like that, and then we have here, D1 over 2 minus D1 over 2. These just cancel. So we're left with just a 2 there. That's all we have there. Then on this side, we have is, well, 1 and 1 cancel. So this cancels, 1 minus 1. This one just adds up D1 over 2 plus D1, that just becomes, well, D1. So they just add up. So this just equals 2, yeah, 2D1. And where, yeah, so we have 1 equals 2D1, like that. So in other words, this just, yeah, just move the 2 over to the other side. we end up having is, let's move this over like that, so we have D1 is equal to 1 over 2. So there's the diameter of the first circle. D1 equals to 1 over 2. And, yeah, I just want to make a note here, so number of this diameter is D1 equals 1 half, but also from the earlier exercise, notice that the difference between the function x equals 0, which equals to 1, is, I'll just put the equals to 1, like that. Let's put equals to 1 by definition. So we just have a straight line like that. The difference between that and the line right here, Y equals X, this is just half, this area is just half the square. So it's going to be just one half. So just wanted to make that interesting note. So the first 1D1, the diameter of this circle is going to be 1 half, similar to the first consecutive area between the, or first area between the consecutive curves of the previous exercise. Yeah, so going further, so similarly for the second circle, what we end up having is we can also have over here, put this in blue, like that. This is now another triangle, right angle, so we have, this is going to be one. This height from here is going to be, this height, like, from here to there, is going to be, well, we have to subtract one, subtract this by, You're going to subtract this by all the way. Well, we know this one here, this height all the way across, is going to be just D1. This is D1. This is going to be this height from here to here. That's just going to be our D2 over 2, like that. So we've got to subtract D1 minus D2 over 2. So that's this side of the triangle all the way across from here to there. And then likewise, this one is going to be R equals 1, plus this radius D2 over 2. And I'll just put another arrow like this over here as well. Like that. So we have another right angle triangle, so thus similarly for the second square we have. A squared is one, so the adjacent plus the opposite, so 1 squared, which equals 2, again equals to 1, let's put, let's leave it there, 1 squared, plus, yeah, when 1 squared equals 2, and then let's just look at it again, so this 1 plus this, I'll just write, yeah, let's write this for completeness here, so we have that. 1 minus D1 over, oh yeah, 1 minus D1 minus D2 over 2 squared, and then this equals to, yes, remove this is plus equals 2, and then the hypotenuse, which is just going to be 1 plus the diameter, I mean, yeah, so we have the radius of the big circle plus the radius of the small circle, so D2 over 2. like that squared, move this around, again get the difference of squares, so we get a 1 equals 2, a 1 plus D2 over 2 squared minus 1 minus D1 minus D2 over 2 squared, squared, and again this is the difference of squares here. So it's difference of squares, so we end up having is 1 equals 2, and then write this out, so we have A, which is 1 plus D2 over 2, yeah, a plus B, which is over here, so we're going to have a plus 1 minus D1, minus D2 over 2, all squared, and then this is going, I mean, not all square, we're going to multiply this by, again, A is 1 plus. D2 over 2, and then we're going to have to subtract this inside term, which is the B, subtracted by minus 1, then these ones become positive. So plus D1, plus D2 over D2, like that. And then just cancel the terms that we can, so we can cancel this one out with this one. So we end up having is, is 1 equals 2, and notice what we have is 1 plus 1 is 2. Then we have this extra minus D1 there. And likewise on this side, what we end up having is the ones cancel, like that, and then we end up having is a D1, like that, plus D2 over half, oh yeah, D2 over 2, plus D2 over 2, that just becomes D2, like that. And now what I'll do is I want to solve for D2, so move this over to this side. So what we end up having is A, and then also keep D2 by itself, so D2, and then move this all over as well, so D2 equals to 1 over 2 minus D1, because then we move this over, and then we also subtract by D1, also move this as well afterwards, minus D1, like that, just move it. There, so that we have this. What we can do is multiply common denominator to simplify this further. So times it by 2 minus D1 over 2 minus D1, and it's the same common denominator, and this all equals to 1 minus, right here, 2D1 minus, it's going to be plus D1 squared. all over 2 minus d1. Now this one here, we know that one, that's just, this one, we could just factor this out, or write this as a combined square. This one is just, well, 1 minus D1 squared. That's because, well, if you have 1 minus D1 square, we have 1 times 1 is 1, and then 1 times negative D1. is then do that twice and add them up, it's going to be negative 2D1, then D1 times D1 is going to be squared, but negative D1 times negative D1 is yeah, positive squared like that. And this just divide this by 2D1 like that. So this is our D2, the diameter of the second one, dependent on the diameter of the first one. And again, note, we have 1 minus D1 squared. equals to 1 minus d1 for completeness, 1 minus d1, and then we can multiply this out, so 1 times 1 times negative d1, negative d1, this times that is negative d1, and then negative d1 times this is going to be plus d1 squared, so these just become 1 minus 2, d1. plus d1 square, which is this right here. So yes, put the arrows there. And here I just circled it around and made it a bit neater. So we just arrow there to there. And yeah, likewise, for the third circle, we will get 1 equals 2. And then here I just have it written as difference of squares, so just get straight to over here. Let's scroll up. Third circle, if you add another one, it's going to be, now we're going to have to subtract by the fold diameters of the first two, and then half of the next one. So we're going to be exactly the same thing. So we're going to be a difference of squared, of the hypotenuse, which is going to be one plus one half of D3. Let's put a D3 like this, over two squared minus, then the sideways height. So the sideways, if you add another zero on another circle there, we're going to have diameter D1, D2, and then half of D3. So just like this one, D1, and half of D2. So this is going to be the difference. Once you just deal with it like that. So we're going to have 1 minus D1 minus D2 minus D3 over 2, like that, exactly the same as we have over here. When we just have the Pythagran theorem, then get difference of squares like that. And now what I'll do here is, I'm just going to call this D star, or D1 star. So if we call that D1 star, so what we end up having is we get, yeah, so we have, we will get this, and going all the way across here, this is going to be now a D3. So instead of dealing with, yeah, so basically we're going to have D1, it's exactly the same thing here, except D3, and this one's going to be D1 star. So. this is going to be exactly over here, D3 equals 2, 1 minus D1 star, and then all squared over 2 minus and then D1 star. Like that. So if you were just followed, we'll get eventually to here, but again, D1 star is not actually by itself, you just replace it D1 minus D2. So what I end up going to have is, d1 minus d2 like that or better yet take the negative out and put a plus like that make it more general so d ys one minus d1 plus d2 all squared over two minus and then here as well d1 plus d2 And now you can do this for infinite number of circles. If you add another one, we're just going to have a D1 star with a 3 of them, or D2 star or whatever. It's going to be exactly the same thing. All we're doing is adding another term here. So this is just two terms in one, so D1 minus D2, whereas this one here is just D1. So they're just separated by, well, yeah, this is difference. You can just split this in half or whatnot. So what we end up having is, So thus in general, we have, let's write this down, so the next diameter of the circle, the next circle is going to equal to, write this in the giant bracket here, 1 minus, so here is going to be 1 minus the summation here of, this is dn plus 1, so this one goes up to just to n, so this is 3, that goes up to 2, so another is this summation from i equals to 1, up to n, and this is going to be the diameter, and then all squared, like this, and then all divide this by 2 minus this same summation there. N, I equals to 1 of D, D ,I, like that. So, let's just box this out. So now, if we actually calculate D2 and D3 from the formulas above, or just this formula, any of them we choose, we find that they are, because we could use the general or the one we just wrote above, so D2 is like that, 1 minus D1, but anyways, let's just use it, this general one here, so we have, we would actually calculate it, we'll get D2, remember D1 is equal to 1 half, remember that, so D1, remember D2 is equal to 1 minus the sum from I's 1 to n, which is, well, 1, so because 1 plus 1 is 2 there. So we just have d1 squared over 2 minus d1, which equals 2, and d1 is equal to 1 1⁄2. So we have 1 minus 1 1⁄2 divided by 2 minus 1 1 1⁄2. So 1.5, that just equals to 1 1⁄2, 1⁄2, 1⁄2 squared like that. So 1 1⁄2, and then the bottom is 2 minus 1. This is common denominator, 2 over 2. So we have 4 minus 1 is going to be 3 over 2, like that. So this is going to be 1 over 4. So 1 over 2 squared is 1 over 4. So we have 1 over 4. We're all like this. This is going to be times by 3. Yeah, I'm going to put this, just remove the fraction, move them around, times it by 3 is at the bottom. and the 2 goes on the top. So what we end up having is now is this cancels, this cancels, we have a 2. So 1 over 2 times 1 over 3, which equals 2, 1 over 6. Like that. So we can just circle this as 1 over 6. And likewise, for 3, D3, this equals 2, this equals to 1 minus, And yeah, this is going to be the summation of the 2 there. Yes, so it's going to be 1 minus the sum of D1 minus D2. So you can just take out the negative like that. Just make it like this, multiply it in size of D1 over D2. Our D1 plus D2 is to put a subtraction all squared. This can be now 2 minus D1 minus D2. So we have a subtraction there of the sum. Now this equals 2, 1 minus 1 half minus 1 over 6, which is the previous one there, all squared over 2 minus 1 .5 minus right here is 1 over 6. So now in dealing with this, let's again multiply to get the common denominator. So what we end up having is... I'm going to multiply this 6 over 6 to get the common denominator. So 1 is just going to be 6 over 6, which equals to 6 over 6 like this. 6 over 6, minus this 1 be times it by 3, so 3 over 2 times 6. That's 6, like that. And then this one, just leave it like that, 1 over 6, like that, all squared. over here, this one's going to be six times, yeah, 6 over 6, so it's going to be 12 over 6, this one's going to be 3 over 6, multiply by 3, 3 of 6, and minus 1…