 Hi, I'm Zor. Welcome to Unizor Education. We continue the course of advanced mathematics for teenagers presented on Unizor.com. I do suggest you to watch this lecture from this website because it contains notes. Well, in this particular case, notes are basically the problems which we are going to solve today and answers and I would definitely encourage you to try to solve these problems just by yourself. Check with the answer if you are correct and that would be probably even a better schooling for you than to just watch this lecture. Anyway, I will present these problems right now. I will try to solve them. I will try not to make any arithmetic mistakes, which I know I can do. Well, basically, let's just start solving problems. It's about cones. I have how many? One, two, three, four, five problems. So let's just go one by one. All right. So the first problem, well, it might have been actually presented as a theoretical material. I did not really mention in previous lectures the concept of a truncated cone. So let's do it now. So let's consider you have a cone and then you truncated it by a plane, which is parallel to the base. Now, this is the radius. This is the overall height. Now, let's consider you have a plane parallel to the base or perpendicular to the x-axis and it cuts this circle. Now, the fact that this is a circle is obvious because this plane is parallel to this plane, which means that if you take any two points here and here, now, these two radiuses are the same, right? So it's very easy to prove that these are also the same by, let's say, cutting the plane through the axis through this point, then it will actually intersect this circle at this point. And from the parallelism of these planes, follows the parallelism of these two lines obviously, right? Because it's two parallel planes intersecting by one line. So it looks like these guys are proportional to the altitudes, right? And the same thing exactly with any other two points obtained by another plane. So that's why we have the proportionality between these and this also like altitudes. And since these are equal to each other, these two will be also equal to each other because they are proportional to the same ratio of the altitudes. Okay, so that's simple, that's a side issue. So every plane parallel to the base cuts a circle. Now, if I will replace this cone with a truncated cone, now the obvious problem is to evaluate its volume in terms of now two radiuses, right? r1 and r2 and the height, which is h. So we need the formula, basically formula of the volume in terms of r1, r2 and h. All right, let's do it. Obviously, we will calculate the volume of this truncated cone as a difference between volume of the big cone and the volume of the small cone on the top, right? So what we don't know about these two things is this piece called h, right? So this is h from the apex to the top of the smaller circle, the smaller base. Well, but we can very easily determine this. Now, consider the proportionality between these triangles and obviously h, lowercase h to r2 equals big altitude to r1. Big altitude is h plus h to r1, right? From here, we can determine lowercase h in terms of r1, r2 and h. And then having all these parameters at hand, we will just calculate the volume of the truncated cone as a difference between two volumes. So from here, h r1 equals h r2 plus lowercase h r2. From here, h is equal to, this goes to this. So h r2 divided by r1 minus r2, right? I just solved this linear equation, transfer h r2 to the left by subtracting h r2 from both sides. And I will have h times r1 minus r2. That's why I divide. Okay, fine. Now, knowing this, we can calculate the big volume. Okay, the volume 1, let's say. It's equal to volume of the whole cone. It's one third pi r1 square times the height, which is h plus lowercase h. So it's h plus h r2 divided by r1 minus r2. So that's my top pyramid, the big pyramid. Sorry, cone, not pyramid, cone. Now, this small cone has a volume 1, so I have to subtract. One third pi r2 square, that's the area of the base. So one third area of the base, and the height is lowercase h, which is this one. h r2 r1 minus r2 equals. So we really should somehow simplify this thing. Well, it's one third pi r1 square times, well, I don't really need this parenthesis. If I will use the common denominator, it will be h r1 minus h r2 and plus h r2. So it will be h r1 divided by r1 minus r2 minus one third pi r2 square h r2 divided by r1 minus r2. Okay. Now, I don't need this anymore. So what could we do here? Well, obviously we should one third pi h divided by r1 minus r2 and open parenthesis. r1 square and r1 would be r1 cube minus r2 cube. r2 cube. Everything else is outside of the parenthesis. One third pi h and r1 minus r2. So this is almost the end of it. You can simplify by dividing r1 cube minus r2 cube as we all know this is equal to r1 square plus r1 r2 plus r2 square. If you don't remember it, you can always try multiply this by this and you will get this r1 cube minus r1 square r2 and then there will be plus r2 square r1 and then there will be r2 square with a plus. So whatever is necessary will cancel out and that will be the remaining. And this is the formula. This is the answer. So in terms of r1, r2 and h, this is the volume of the truncated comb. Number two. Okay, side surface. Okay, so if you have a comb and you cut it, you cut it by one of its generatrix which goes along the side surface, all right, connecting an apex with one particular point on a circle which is the base. You can roll it out into a sector. We're already talking about this. It will be a sector. So this is r and this is h. Now the side, the lengths of this line along the side the generatrix, let's say, would be l. And obviously l2 is equal to r2 plus h2 from the Pythagorean theorem. Now if you will roll it out on a flat surface, then the radius of this sector would be l, right? Because you are using all these points on the circle, they are becoming this arc. And the arc length would be 2 pi r, where r is the radius, right? So if this is the cone, this is the result of its rolling out after the cut along the generatrix. Now what's known about this is the area of this thing is s and this angle is 120 degrees. That's given. r and l and h, whatever else, it's not given. And we have to determine the value. Okay, first of all, let's just think about it. How many parameters define the cone? Well, two parameters, let's say r and h. Everything else can be derived from it. For two parameters, we need two equations, right? So, and these two equations are given basically. There is one information, there is an angle of a 120 degree, and there is another information about the area of the side surface. So basically, we somehow should represent the angle and the surface in terms of r and h, and we will get two equations with these two unknowns, and then we will solve them for r and h, and then we will determine the value. All right? All right, so what can we say about this? Well, first of all, we have a very easy thing here. It's 120 degrees. It's one-third of the full circle of 360, right? Now the full circle of the radius l has the length 2 pi l. At the same time, it's three times greater than this. So it's three times 2 pi r. So what follows is that l is equal to 3r. Okay? That's important. Now the surface area, the side surface area is s. That's given. Now, do you remember how the side surface area is represented in terms of h and r or l or whatever? Well, let me just remind you. I will use exactly the same thing. I will cut it and open it up, right? So I know that the whole circle area is pi l square, right? Now, if this arc has the length 2 pi r and the whole circle has the length 2 pi l, then what I can say is that whatever part of the whole circle, whole circumference this 2 pi r takes, which is 2 pi r divided by 2 pi l. That's the ratio of this area relative to the area of the whole circle. So s divided by pi l square. So what do we have from here? l square l. We have the s equals pi r l, right? Pi r. So this is the formula which we have derived a long time ago. And instead of l, we can always have square root of h square plus r square. But this is the formula. So let's just write it here. s equals to pi r l equals to pi r square root r square plus h square. So we will use it. And this is our second equation because s is given, right? So s is given, which means this is given. And this is another equation. So we have two equations. Now I think it would be easier if I will also have l square here. Do we need l square or not? Actually we probably don't need l square. I'll just use directly this formula. Okay, so s which is given is equal to pi r l. And l is 3r. So it's 3 pi r square. Now from this we can determine r. r is equal to s divided by 3p, 3pi square root, right? s divided by 3pi is equal to r square. So r is equal to this. So we got r. Now from this we can actually define h. Let's put it this way. l square which is equal to h square plus r square is equal to square of this 9r square. Okay, so h square is equal to 8r square. And h is equal to r square root of 8 or 2r square root of 2, right? Because 8 is 4 times 2, square root of 4 is 2, and square root of 2 is remaining. So I have h and I know r here so I can find out what's my h exactly. h is equal to 2 square root of 2 times r. r is this. Square root of s divided by square root of 3pi. So I have r and I have h. All I need is what? The volume, right? Right, so let's just volume is equal to, now it's just, you know, pure technicality. And as I was saying, I always 1 third pi r square which is s divided by 3pi times h which is 2 square root of 2 square root of 3pi and square root of s, right? What else? We can simplify it a little bit, basically, I know. We probably should multiply it by square root of 3pi so we don't have radicals in the, so we will have what? Pi and pi are reducing so we will have s square root of s, 2 square root of 2, square root of 3pi. I multiply by 3pi and in the denominator I will have, therefore, just 3pi times 3 times 3, it's 27pi. Something like this, right? Is this right? 2s, yeah, that's what it is. You can combine square root of 2, it's square root of 3, so it's 2 square root of 6pi s square root of s divided by 27pi. Oh, that's not a very pleasant formula but anyway, we did the problem. We did solve it. That's the answer. All right, 2 down, 3 to go. Okay, this is a very interesting problem actually. Here is what's given. So, given site surface s and what also is given is the following. Consider this as one of the generatrix, so a point connecting the apex with a single point in the base along the straight line. Now, if I will have this plane all AB and I will draw a perpendicular to AB within this plane, that would be distance d. That's also given. d is given. Now, obviously, if you take any other line like this on the surface and connect it with this one and draw a perpendicular, it will be the same d, right? Because these triangles AB prime O and AB O are obviously congruent because they're all right triangles with the common catatris and other catatris, other catatris are radiuses of the circle, right? So, they are right triangles with equal catatris. So, that's why the perpendicular, the altitude to hypotenuse would be always the same, or this one, for instance. So, these are all perpendicular from the center to generate resist. All right, that's given. Well, question is the value. All right. So, I actually first solved this problem like directly without thinking much. And that's what I'm going to do right now, right? So, let's assume that this is R and this is H. So, what I would like to do is, in terms of R and H, I will put the surface area. Now, I know it's pi R L, which is pi R square root of, all right, don't have enough space here. So, here equation is S is equal to pi R square root of R square plus H square, right? Pi R L, where L is this particular diagonal, hypotenuse of this triangle. Now, at the same time, if you consider just this right triangle, H, R, L, where L is this, this is D, how to determine D? Well, obviously, the area of this triangle is R times H, well, one half of this, and it's equal to one half of L times D, right? This is base and this is altitude, or this is base and this is altitude. So, from here, D is equal to R H divided by L. So, that's my second equation. R H divided by square root of R square plus H square. Now, I have to find, from these two equations, I can find R and H, right? And calculate my value. So, I did that. How can we resolve it easier? Well, we can say something like this. R H is equal to D square root of R square plus H square, and S divided by pi R is equal to square root of R square plus H square, right? So, we can substitute this to this. We will have R H equals D S divided by pi R. So, pi R square H, pi R square H is equal to D times S, right? That's interesting. This is basically almost the formula for the volume. It's triple. So, one third of this, pi R square H, which is the volume of the cone, is equal to S times D divided by three. It's too simple, right? Now, and then I was thinking, why this is such a simple formula? It looks like there are complicated calculations, but the formula seems to be very simple. Strange, right? And then, as I was thinking, I came up with a perfect explanation. Let me just draw this better. Okay, here is my cone. Okay, let's take two planes which are connecting two lines, two generators. And consider, this is C, and this is B, this is O. Consider A, B, C, and O in front of it. What is it? It's a triangular pyramid, right? Now, if I will consider A, B, C as a plane, this pyramid is basically almost like a sector of the cone. So, the plane A, B, C, the triangle A, B, C, the base of this pyramid is very much close to the surface of the cone, right? Now, if my points B and C are closer to each other, then this line will almost be like a circle, right? The closer B and C together, the closer segment B, C would be an arc B, C, right? So, that's how we were calculating something like an area of the circle, if you remember. We doubled the number of sides of these inscribed polygons, and the more vertices we have placed on the circle, the closer perimeter would be to a circumference, right? Now, here we have exactly the same situation, but it's in a three-dimensional thing. The triangle A, B, C would be closer and closer to the surface of the cone. So, now, if I will divide my circle into many small pieces, the volume would be divided into small pyramids like these ones, right? And the volume of any individual pyramid would be what? Area of A, B, C, A, B, C, right? If O is apex, so I need the area times the distance from O to this triangle A, B, C, right? Which is like called H, whatever, divided by three. So, one-third of the area of the base, which is a triangle A, B, C, times the altitude, which is from O to some whatever point B. Now, if I will add them all together, all these pyramids, areas of A, B, C together, A, B, C, and A, C, D, etc., etc., they will be more or less like a side surface of the cone. And the H, the distance from center to this plane would be very much, as soon as my B and C are closer, it will be very much like the distance from this center O to the generator, which goes somewhere here. So, the more vertices we will put on this circle, the closer the total volume of all these pyramids will be to the volume of the cone. The sum of all bases, sum of all areas of the bases will be closer to the side surface of the cone. And the altitude of each pyramid will be very much like this distance from O to a generator on the side of the cone. And that's why, as soon as I will go to a limit with the number of divisions going to infinity and sum them all together, the total area would be S and this H would be D. And that's why we have this particular simple answer to this problem. Okay. But I have came up with this only after I saw that the formula is very simple. I didn't really think about this, I just saw the first directive without any. Okay. Side surface is twice as big as base. Okay. So, again, we have a cone and side surface, which is pi RL is twice as big as pi R square, right? Now, this is H, this is R, and this is L. And obviously, L is equal to square root of R square plus H. It's just more convenient to use side surface in this format. All right. So, I've got that. Side surface is twice as big as base. Now, central section through the main axis. So, if I will cut this cone with a plane going through the main axis, through the main altitude, which connects the apex with the center of the base. It will cut a triangle, obviously, right? So, within this triangle, the area of this triangle, let's call it Bc. Area of B ABC is equal to S. Find the volume. Okay. All right. Let's find the volume. Well, in terms of R and H, the area of ABC obviously is what? This is a diameter, right? Since we cut the plane through the point O, this is a straight line. So, it's BC is equal to 2R. So, the base is equal to R of this triangle. The height is H. So, 2R times H divided by 2. That's the area of the triangle. And it's equal to S. So, we have two equations. One and two. All we need is, again, it's technicality right now, to find out R and H. Well, L is equal to this. All right. So, let's just do it. Let's do it a little bit simpler. So, from here, I can reduce by pi R and I will have what? That L is equal to 2R. So, square root of R square plus H square is equal to 2R. That's one equation. And the second equation is R times H is equal to S. That's my system of two equations with two unknowns, right? Okay. Let's square this one. We will get R square plus H square is equal to 4R square. H square is equal to R square times 3. H is equal to R square root of 3. We can substitute it into this. So, we will have R square square root of 3, right? Equals S. So, R is equal to square root of S divided by square root of the fourth degree of 3, right? From here. R is equal to square root of S divided by square root of square root of 3, which is fourth degree. Okay. I've got R and H here. And H is equal to R times square root of 3, which is square root of S, square root of 3 divided by square root of fourth root of 3. Now, what is square root of 3 divided by fourth root of 3? This will be square root of square root, which is fourth of 3. Square root of, again, fourth root of 3 times fourth root of 3 is square root of 3, right? So, that's why you divide this by this. You will get this. So, I've got both. But I can do actually, well, I'll simplify it later. All right. So, now all we need to do is substitute it into the formula for the value. 1 third pi R square H equals to 1 third pi. R square is this, which is S divided by square root of 3, right? Fourth root of 3 double, I mean square would be square root times H. And H is square root of S, fourth root of 3. Well, that's the answer. And if you don't want to have this, let me check. Well, you probably should multiply it by square root of 3 to get rid of the radical in the denominator. So, in denominator we will have 3 and 3, which is 9. So, on the top we will have pi S square root of S. Now, what is this? This is square root of 4, fourth root of 3, sorry, times fourth root of 9, right? Square root of 3 is fourth root of 9. Same thing. Divided by 9, 3 times 3, which is pi S square root of S, fourth root of 27, 3 times 9 divided by 9. So, that's the answer. All right. Fourth down, one to go. I hope you're not scared with all these fourth degree roots and stuff like this. Well, that's what it is. All right. Here's an interesting problem. Okay, you have a cone altitude, and you divide the altitude into n equal parts. So, somewhere here you will have a 1, a 2, etc. a k minus 1, a k, and the center of the base would be a nth, right? So, we have n minus 1 from 1 to n, from a 1 to a n minus 1 division points, which divide into n pieces. All right. Now, let's cut with the planes through each point. So, you have n truncated combs. Well, actually n minus 1 truncated, and the top one is not truncated. So, the question is, if you will take the cone number k, what is its volume? In case you know the volume of the whole cone, and you know the number of divisions. All right. Okay. Now, this is actually a very simple problem. It sounds complicated, but it's simple. So, let's have the plane representation of this. Let's cut the whole thing by a plane, which goes through the central axis from a 0 to a n, and it will cut the whole cone in section, which looks like this, right? This is a k. This is a k minus 1. This is a n. This is a 0. Now, obviously, let's call this one r, which is equal to r n. This would be r k minus 1. This is r k. Now, let's talk about these heights. This one would be h n. This one would be h k. This one would be h k minus 1. Now, obviously, you have the similarity of all this, right? So, how can we determine what is the value of r k in terms of r, let's say? Well, very simply, since the distance from a 0 to a k is k times small part, and the small part is actually equal to h divided by capital N, right, where h is the whole height. So, the height, which is h k, is equal to h divided by N times k. Same thing, radius r k, because it's all proportional, right? So, it's r divided by N times k. Now, since you know this, you can find out this truncated cone as a difference between the k's cone and k minus 1's cone, right? So, the volume of the truncated part would be the 1 third pi r k square times h k, right? Minus 1 third pi r k minus 1 square times h k minus 1 equals 1 third pi open parenthesis. r k is k nth of the r, so it's k nth of the r square times h k, which is k nth of the h minus 1 third pi. Now, instead of k, I should put k minus 1 r divided by N square and k minus 1 h N equals. So, what do we have now? Well, we can 1 third pi r square h out of the parenthesis, and actually N also we can put out of the parenthesis, N square and N, N cube. That's what it is. So, let me just write it more accurately. So, it would be pi r square h divided by 3 N cube. Pi out, 3 out, N cube out, r square h, r square h out. What would be inside? Well, k cube minus k minus 1 cube. That's what it is. Now, this is the volume which is given to us. So, the whole thing is equal to v divided by N cube times k cube minus k minus 1 cube. That's the answer. That's it. I do suggest if you didn't do it before, why don't you go through the same problems again and try to solve them by yourself. Check the answers. Answers are on the web page. And good luck. Thank you.