 Hi, I'm Zor. Welcome to Unisor Education. I would like to spend some time and talk about circumference of a circle. This lecture is part of the course of advanced mathematics for high school students. It's presented on Unisor.com and that's where I suggest you to watch this lecture from because there are lots of very useful notes to each lecture. Now, circumference of a circle. Now everybody knows, right, that if you have a circle with a radius r, then circumference is 2 pi r, right? Where pi is something like 3.1492, etc., some irrational number. Yeah, everybody knows that. And why am I talking about this? Well, let's not forget that my purpose is not actually to provide you with certain facts in mathematics, but to develop creativity, logic, analytical thinking, etc. So right now my question is, well, how people came up with this formula? What's the history of development of this formula and what can a reasonable person say about circumference of a circle if he doesn't know the formula? So imagine yourself as the person who doesn't know what actually the formula is and you're trying to derive it, something like this. So what should you do? All right, so question is what is a circumference of a circle? I mean we talked about lengths as something which is based on straight lines. So you have some kind of a unit of measurement and then you can measure something like this by counting how many times this unit fits into this particular segment. And we still had problems because sometimes it doesn't really fit the integer number of times. So we have just thought about how rational lengths is supposed to be done. We've divided this unit into tens or hundreds or whatever and we're trying to fit it. So even there we had problems. Now we are talking about a curve. I mean the unit of the measurement doesn't fit to any particular part of it, right? So that's the first problem which we have to face somehow. And what can a reasonable person do or think about if he wants to measure the lengths of this circle? Well, the first time you think about the circle and you want to measure its lengths, it's probably to approximate it. How can it be approximated? Well, let's just think about the following process. First, you divide a circle into six equal parts. Now this is 360 degrees, so each one of them is 60, right? Now the circle by definition is set of all points on the plane which are equidistant from one particular point which is called the center and the distance let's say is R. So what do we have right now? Now if this is a center, let's use the letter O and this is A1, A2, A3, A4, A5, and A6. Now we have here a regular hexagon. Now why is it regular? Well, obviously each radius is equal to R, so each triangle is equilateral triangle and since the angle at the top is 60 degrees, the other triangles are also 60, right? So it's equilateral triangle and therefore these are also R equal to R. All sides of this hexagon are all equal to R as well as the radius. Each one is equilateral triangle. And so it's because these are two radiuses and 60 degrees at the top, so it's an equilateral triangle. So can we say that this hexagon approximates the length, the perimeter of this hexagon, approximates the length of the circle? Well, yes, at some point we can say that yes, approximately it is equal. Can we improve this approximation? Sure, absolutely. Here is the process. Let's just have a perpendicular to each side and connect the points and now we will have, instead of hexagon, we have a 12-sided regular polygon. It's very easy to prove that this is regular, right? Now let's call it B1, B2, etc. Now what I would like to say is that the perimeter of this next one is probably closer to our circle than the perimeter of the original hexagon. The 12-sided polygon looks like it's closer, so it looks like it's a more circular in its form. So I presume that the perimeter of this 12-sided polygon is closer. Well, let's calculate. So the perimeter of the first one, this is my first approximation, and I have 6R, right? Each side was R of the hexagon. Now let's think about the 12-sided polygon and for this I will change my picture and I will have it a little bit more. So these are my A1, A2, this is my center, this is R, this is R, this is R. Now we have dropped a perpendicular here, got the B1. So my purpose right now is to find what is A1, B1, based on whatever I know. Well, this used to be 60 degrees, right? So now this is 30, and this is 30. Okay, fine. How can I calculate A1, B1? Well, let's put this point P and I can say that from the Pythagorean theorem A1, B1 square is A1P square plus B1P square. What A1P I know, that's half of this, which is R divided by 2, but B1P I don't know. However, I can find it out what's B1P from O B1, which is the radius, right? R, and O P. And O P in turn I can find out from the Pythagorean theorem because this is half of the R and this is R, right? So a couple of Pythagorean theorems will lead us to the answer. So let's just do this calculation. First of all, O P. Now O P square is equal to O A1 square, which is R square, minus A1P, which is half of the A1A2. And you remember this was the equilateral triangle, say A1A2 is also R. So this is half. So this is R over 2 square equals 2 3 quarters of R square. So O P is equal to R square root of 3 divided by 2, right? O P square is 3R square divided by 4, O P is square root of this, okay? Knowing O P, we can find B1P1. So B1P is equal to radius O B1 minus O P, which is this, minus R square root of 3 divided by 2. Or if you wish, R divided by 2 times 2 minus square root of 3. That's why B1P1. This is this one. So now it's very easy to determine what's A1B1 from the Pythagorean theorem here. A1B1 square is equal to A1P square, which is half of the R, plus B1P square, this square, right? Which is R square over 4 times square root of this, which is 4 minus 4 square root of 3 plus 3. Is that right? Let me just think about it if I didn't make any mistakes. So equals to 4R square plus 7R square. So it's 8R square, right? 4 and 3 is 7 and 1 from here is minus 4R square root of 3. That's what it is. So my first approximation had side equals to R, six sides equals to 6R. That's my first approximation. My second approximation is side is equal to, let's just think about it, we can reduce it by 4, right? 4, 4 and this would be 2. So it's R square times 2 minus square root of 3. So my second approximation is side equals square root of this, right? R square root 2 minus square root of 3. And now we have 12 sides and 12 sides equal to this times 12 and I already calculated it somewhere, 6.21R. So if you multiply square root of 2 minus square root of 3 times 12, it will be 6.21. So my next approximation with 12 sided polygon is this one. Are we getting closer? Remember 2 pi R, that's what we are thinking that we are approaching. Pi is 3.14, so it's 6.28R. That's approximation. So there are much more decimal numbers here, but this is the first two after the point. Okay, fine, so we've got next approximation. Well, let's go one step more. So instead of 12 sided polygon, we will do exactly the same as before and we will consider 24 sided polygon by dropping perpendicular to each line, to each edge of the 12 sided polygon. So let's assume that this is now b1 and this is obviously not R and this is a1b1 which is this. So this is 6 square root of 2 minus square root of 3. This is a1b1. Now we dropped the perpendicular. So this is an edge of 12 sided polygon and this now will be two edges of c1, of 24 sided polygon. Let's do exactly the same calculation one more time and see what happens. First we calculated, well this is no longer p, let's put another letter, let's say q. So oq, oq square from the Pythagorean theorem, oq square is equal to oa1 square, which is r square minus half of this. Now half of this, so it's r square over 4 and square of the root which is 2 minus square root of 3. That's what it is. That's my oq square. So oq is equal to r times, so it's 1, so it's 4 minus 2, so it's 2 plus square root of 3 divided by 4. Am I right? Sorry, divided by 2. Square root, is that right? So it's again 4 minus 2 plus square root of 3, so it's 2 plus square root of 3 divided by 4. So square root is 2 and square root is, yes, looks okay. Okay, that's my oq. Now c1q is equal to r minus this. So r minus square root of 2 plus square root of 3 divided by 2. Right? This is my c1q, c1q, which is r minus oq, which is this. Think I'm right. Finally, a1c1 equals this square plus this square. a1q is half of this, so it's r square over 4. Let's put ac square, a1c1 square, a1c1 square. Square of this, which is r over 2 and 2 minus square root of 3. Right? I got rid of the square root because I'm squaring it. Plus c1q, which is this one, square. Now let me just, okay, now this is a little bit more complicated. So it's r square minus 2r square square root of 2 plus square root of 3 plus plus square root of this, which is 2 plus square root of 3 over 4. And again, r square. Am I right? Yeah, looks okay. All right. So the common denominator is 4, obviously. Now, r square is everywhere, so we can put r square in front of this. And what's next? Well, so it's 2 minus square root of 3 plus r square, which means since 4 is common denominator, so it looks like it's plus 4. Minus 4 square root of 2 plus square root of 3. Am I right? And plus 2 plus square root of 3. Now what do we have now? We have square root of 3 plus and minus. And now r square, 2 plus 4, 6 plus 2, 8, 8 minus 4 and divided by 4, so it's 2 minus 2, no, minus 1. Square root of 2 plus square root of 3. And finally, a1c1 is the square root of this. This is a1c1 square. So my third approximation, in this anymore, my third approximation is side equals r and square root of this, square root of 2 minus square root of 2 plus square root of 3. Such a big hierarchy of square roots. And 24 size equals 24 times this. And I have already calculated the result. Yes, 6.27. So this thing is 6.27. You see? First approximation, 6. Second approximation, 6.21. Third approximation, 6.27. And remember, we still have to go something like this. So it looks like it does actually approaching our 2 pi thing. So it looks like we are correct. Okay, now let's go back. Why did I decide to do all these calculations? Remember, I wanted to define what is a circle length or circumference, actually, of the circle lengths. And I was saying that, okay, let's start with a hexagon, then divide each side in two. So we will get inscribed into the same circle, 12-sided polygon, then again divide by 2. So it will be 24-sided. And it looks like we are approaching something. Question is, can I say that the limit of this process, whatever this process I have just described, by dividing the lengths into having this perpendicular to this edge and then basically doubling the number of sides, can I say that the result of this process, the limit of this number, whatever this number is, is the length of the circumference of a circle? Well, there are problems with this definition. Problem number one is, I have no idea if the limit exists. Right? It looks like it approaching something, but I have to prove this number one. Number two, what if I, instead of a hexagon in the beginning, chose, let's say, a square and then double its sides, number of its sides, will I get the same number? Right? So we have the problem of existing of the limit. And we have the problem of the uniqueness of the limit, regardless of the process I'm using, hexagon, square in the beginning and then doubling, whatever it is. And now I am in the land of mathematics where I cannot really absolutely rigorously prove you that these questions can be answered quite favorably. So I will just tell you the result, which is kind of outside of the scope of this course, and the result is as follows. No matter what's the process of increasing the number of sides of our inscribed polygon is, as long as the maximum length of every side, maximum length of a side of every polygon is diminishing to zero. So this is the requirement. You know, whenever we were doubling the number of sides, obviously the side itself is going down. This is the lengths of the side. R, this is less than R, this is even less than R, but we are increasing the number of the sides. So as long as the process assumes that the maximum length of the side of the polygon, it doesn't even have to be a regular polygon. So maybe it's not regular, whatever it is, but as long as on each new step, the maximum, the length here, the lengthiest side is going to zero. Then, number one, the limit exists. Number two, it does not really depend on how this process is arranged. As long as the lengths of the longest edge of the polygon inscribed into the circle goes to zero. And as a result of this theorem that the limit exists and it doesn't really depend on the process as long as the length of the polygon is going to zero, then I can say with basically complete positive note that our length of the circumference of the circle is equal to this limit. Since I have proven that it exists, well, I didn't prove myself, but I'm just telling you that there is such a theorem which is outside of the scope of this lecture, that there is a theorem which actually states that no matter what the process is, there is a limit. And that limit we actually call this number 2 pi. Well, there is one more thing here. I have proven this for one particular r. What if I will take a different r? Will I get a different number here? That's a different approach to this, which is we didn't really touch it yet, but let's address it right now. Now, this is all related to similarity. Now, obviously, all the circles which you can imagine with all the different radius are similar to each other. Why? Very simply, circle is defined by the radius, right? So, if I have two different circles, let's say this one and this one. What I can do, I can just have two congenital lines, use this as a center of my scaling. These are perpendicular. Now, if I will use this as a center of the scaling and the ratio between this and this as the factor of the scaling. Now, obviously, from similarity of these triangles, my radius will increase proportionally to the same factor, right? And since radius is increased, and these are the lengths, these are all the numbers, all the points which are on the same lengths from the center, let's say this one, then I can say, if I will take the parallel, I will say that this point will be converted by the scaling into this point. So, whenever I want to find any image of any point, I just connect it with the center, have a parallel line, and obviously, from the similarity of all these triangles, this one and this one, will follow the same proportionality of all the elements. Now, if these two circles are similar to each other, then my process of increasing the number of ages in the inscribed polygon also can be proportionally transformed from this to this. So, basically, each, let's say, a square here, it will correspond to this square here. And obviously, these squares will be similar for the same reason. And since this point goes to this, this point goes to this, etc., which means the perimeters will be similar. So, what I would like to say is that if the circumference of this circle relates to its radius as something like whatever the factor is, then it's exactly the same factor will be here because the circumference, as a limit of this, is defined as the limit of the polygons as they increase the number of sides, right? And each polygon is similar to this one, so the ratio between these perimeters and this perimeter is exactly the same as the ratio between the radiuses. So, if one particular circle has the circumference equal to this, where this number is not this particular number, but some kind of limit, whatever the limit is, then in any other circle, the number will be the same. So, circumference is always proportional to a radius by the same number, or if you wish it's proportional to diameter, which is double the radius by the same number. And historically, it happened that its relationship, its relationship of the circumference between circumference and the diameter of the circle is actually signified by letter pi. So, for a radius it would be 2 pi. Now, before going any further, I would like to kind of illustrate whatever I just said with a different process, and instead of hexagon, I will take the square in the beginning and let's see how it will go. I will do exactly the same thing first. So, I have a circle and I will inscribe a square in it. And then I will double the number of sides by perpendicular to each line and having octagon. And then perpendicular to each side and I will get 16 sided, 32 sided, 64 sided polygon, etc. So, how my perimeter is increasing and approaching the circumference of a circle? I would like basically to illustrate that we will do exactly, we will get exactly the same approximation, we will approach the same number differently. Approach will be different because the numbers, every number will be different, but it looks, it will look like it will go into the same limit. Okay, so let's just do one very simple thing. Now, if I have a square, now square, now this is the radius, right? These are four reduces. So, this is r, this is r. So, this is r square root of 2. So, the perimeter, on the first step, perimeter is equal to, so side equals r square root of 2 and perimeter, which is 12, sorry, four sides, it's equal to 4r square root of 2. Now, let's just introduce a process. I will do it slightly different. I will not directly calculate, but let's assume that at some point we have the lengths of this inscribed polygon equal to gm. Now, I will divide it in half and put a double number of sides. Question is, so this will, my a1, a2 and this is my b1, this is p, this is all. So, my question is, how can I describe dn plus first, which is the edge of the n plus first polygon on the n plus first step, in terms of dn? I would like to know this iteration between steps, because if I will have this law of iteration, I can program it in the computer very easily, and the computer will calculate my value of the pi. The longer I will run the program, the more precise I will have the results, right? Okay, how can I do this? Well, again, that's very simple. This is r. Well, actually, since all the circles are proportional to each other, similar to each other, and proportional radiuses and circumferences, I'll choose the radius equal to 1. So, I don't have to carry the r all the time. Now, this is 1. Now, this is dn. So, this is dn over 2. My OP would be square root of 1 minus dn over 2 divided by 2 square, right? That's my OP. Now, b1p is equal to radius minus OP, which is 1 minus square root of 1 minus dn over 2 square. And, since I know b1p and I know a1p, I can calculate d1 dn plus first square is equal to dn square over 4. This is the square of this plus square of bp, which is 1 minus 2 square root of 1 minus dn square over 2 plus 1 minus dn over 2 square, right? Now, this is cancelling this one. And, I have 2 minus 1 plus 1, 2 minus 2 square root of 1 minus dn square over 4, or 4 goes out, cancels with 2, 4 minus dn square, right? So, here is my dependency. dn plus first is equal to square root of this, 2 minus square root of 4 minus dn square. That's the formula, iterative process, which I can start from the d1. And, I know what d1 is. If r is equal to 1, then d1 is equal to square root of 4, right? d1 is equal to square root of 4. Now, how many sides? I have 4 sides. So, the first approximation is 2, I'm sorry, square root of 2. 4 times square root of 2 is equal to, do I have numbers here? I don't, but 1, okay, 141 times 4, 4, 5, 6. Oh, no, I have to have. Oh, yes, 5, 6, 5, 6. Okay, 5, 6, 5, 6, 9. That's what it is. That's my first approximation with a square as perimeter of the square. Now, next approximation, d2. If I will put square root of 2 here, what should I have? I have 4 minus 2, which is 2. It will be square root of 2 minus square root of 2, right? And then, you multiply it, the perimeter will be, circumference will be equal to approximately perimeter, which is 8, right? 8-sided by this square root of 2 minus square root of 2, which is approximately 61229. Now, d3, if I will substitute d2 into this formula, I will get d3. And d3 is approximately, well, I put 16 times d3, which is perimeter, 6.2432. Now, 32-sided polygon, that's d4, approximately 6.2734. You see, we are approaching 6.28, blah, blah, blah. And it's just an extra illustration that's our process, while completely different from the hexagon-based polygon, right? It's a square-based process, but it's also approaching. And the more sides you have, the closer you are to the result. So, basically, what I also have done, and I put it just into notes of this lecture, using this formula and going back to the hexagon, I can check if the formulas for hexagon correspond to this formula. Well, they do. I mean, if you will just put whatever the numbers we have for hexagon-12-sided polygon and 24-sided polygon, you will see that the dependency is exactly the same. If you will take the lengths of the hexagon side, you will get the lengths of the 12-sided polygon. If you put 12-sided polygon edge, you will get the 24-sided polygon, exactly the same as the formulas in the lecture presented before. Well, that's it. And again, the purpose of this lecture was to explain you how people actually came to these numbers. And by the way, I did mention it before, if I will put this formula into the computer and assign my the first as a square root of 2, I will get the approximation up to whatever I want, basically, because the longer computer runs, the more precisely my numbers will correspond to the lengths of the circumference of a circle. Well, that's it. I suggest you to read the notes for this lecture. They are a little bit more maybe elaborate than whatever I put on the board and it's just useful for you to know. This is how the circumference of the circle came to be, basically. That's how people developed this particular area of mathematics and we are all using the results. All right, thanks very much and good luck.