 Hi, I'm Zor. Welcome to Unizor Education. Continuing the course of advanced mathematics for high school students, it's presented on Unizor.com. That's where I suggest you to watch this lecture from, because there are notes. And in this particular case, and well, as many other cases, especially when I'm presenting the problems, I do suggest you first to go to the website and try to solve the problem yourself. Regardless of whether you will succeed or not, only then, which this lecture. So I'm going to present one of the so-called final problems on solid geometry. And the problem actually is very simple to state. And quite frankly, I was surprised that this particular property actually exists. And that's what I'm going to prove that this property exists. So again, it's something like unexpected, I would say, a result if you didn't know about this before. I can compare it to many other unexpected relationships in the world of mathematics. Like, for instance, whenever I have learned about Pythagorean theorem, that the square of hypotenuse is equal to sum of squares of catechia, I mean, it's a surprising fact. If you don't know about this beforehand and somebody tells you that this exists, well, you said, well, that's great, but it's unusual, I would say. So this is also kind of unusual. Here is what it is. We have a cone, the right circular cone we don't really consider any others. And there is a sphere which is inscribed in it. So it touches the side surface and also the bottom of the cone. And we have already discussed in one of the previous problems which I have presented that this particular sphere has to have a center on an altitude from the apex of the cone. And the line of tangency of the sphere and the cone surface, side surface, is actually a circle. So this is a circle on, this is the line of tangency. And this is the diameter of the sphere. So the problem is, given the ratio of side surfaces, both cone and the sphere, in case of a cone, that includes the base. So given that this ratio of their surface area, find the ratio of the volumes. Well, that gained kind of a strange thing because I didn't really say anything about the cone. Cone can be this way or cone can be that way. So basically the problem states to find this ratio regardless, which means it's supposed to be the same. So let's say you have this ratio of surface area gamma between, let's say, the sphere and the cone. So this is surface area of a sphere divided by surface area of a cone. So what would be the ratio of volume of a sphere divided by volume of the cone? Well, since I didn't really say anything about what's given, it's supposed to be, well, the same for any kind of a cone. So let's just assume that we have certain cone which has the radius r of the base. It has h as its altitude, its height. Now, obviously, that defines the cone. Now, if the cone is defined, then the sphere inside is also defined. We have to really find the radius. So this is the radius, let's call it lowercase r. And first of all, we have to find the radius to be able to compare volumes and surface areas, etc. Now, this particular problem I have already solved in one of my two previous lectures, but let's just do it again. So what I do is I cut the whole thing by a plane which goes through the altitude of the cone. Now, the cone will be cut by a triangle, that would be a triangle, right? And the surface of a sphere would be cut by this plane and the result would be, since it goes through a center, the result would be a circle and for obvious reason, this circle is inscribed into a triangle. So what do I know about the radius of the inscribed circle if you have dimensions r and h? So this would be lowercase r, this would be lowercase r. Well, we can do it differently. We can find lowercase r differently. In the previous case, I was talking about comparing the area of a triangle with area of three smaller triangles if you connect the center with vertices. Now let me just offer another solution. Obviously triangles SPB and SOQ are similar because both of them are right triangles. Now this is an altitude of a socialist triangle and this is a radius towards the point of tangency between the line and the circle. So these angles are right and they have a common angle which they share. So they are similar which means we can use some proportionality. So what's the proportionality? Let's say you compare the proportion between the hypotenuses. One is SO, another is SB. And this is supposed to be compared with the same ratio between smaller categories, let's say lowercase r and capital R. Now what is SO? SO is H minus lowercase r, right? So H minus lowercase r. And what is this hypotenuse, SB? SB is, this is H and this is R, so it's square root of HR squared plus H squared. So I have this equation from which I can determine lowercase r and that would complete all the elements which I need to calculate the areas and the volumes, right? So let's just multiply. So lowercase r square root of r squared plus H squared equals to HR minus lowercase r r. So r goes here is equal to HR divided by, that would be the plus. So r plus square root of r squared plus H squared. Which is exactly the same formula I had in the previous lecture but derived differently using the area of the whole triangle divided into three areas of these triangles. This, this and this. Where I have perimeter multiplied by the radius and divided by two, the area. Alright, fine. So we've got that. Now I don't need actually anything, please. And now I just have to go through calculations which is kind of boring but nevertheless that really would supposed to be done. So let's just compare the area surface of the sphere with the area surface of the cone and then the volume because I have all the parameters h and r define everything. Lowercase r I have expressed in the terms of h and r and h and r are completely unrelated to each other. It can be anything. So basically that's it. Alright, so let's just do the straight algebraic manipulations. Now the area of the sphere is, what was the formula? Four pi and radius squared, right? Now this is the radius so it's four pi and square of this which is h squared r squared divided by r plus square root of r squared plus h squared squared. Okay, what is the surface area of the cone? Well that's the base, right, plus the side surface. Now the base is pi r squared pi r squared and the side surface is, it's a perimeter times this generatrix divided by two, right? I mean if I have a small triangle here it will be this times height and divided by two and if I will have a lot of small triangles the height will be the same as generatrix and some of these would be the lengths of this circle. So it's two pi and divided by two, right? So it's two pi r divided by two would be pi r and times generatrix, that's again square root of r squared plus h squared. So that's my formula, that's gamma. Well let's simplify it. I can obviously reduce by pi, I can obviously reduce by one r here so this is r, this is r and this is r and what remains for h squared r divided by r plus square root of r squared plus h squared. This is the second degree and that would be another, so it's cube. So that's my formula for gamma. That's the ratio of surface areas. Well now let's compare the volumes and we'll see if we will get something which can be expressed in the terms of gamma. Alright, now the volume of a sphere is 4 third pi and the radius cube which is this. So let me do it this way. Pi h cube r cube and divided by r plus square root of r squared plus h squared cube and divided by the volume of the cone which is one third area of the base times altitude. So it's one third pi r squared h. Well three and one third can be reduced, pi and pi can be reduced, r squared and r cubed can be reduced and h would be h squared. And what do I have? Well amazingly enough it's exactly the same thing. So as you see the ratio of the volumes is exactly the same as the ratio of area of the surfaces between the inscribed into a cone sphere and the cone itself. Well again I can only say that I'm surprised to see whatever it is it is. I would prefer you basically do exactly the same calculations just by yourself and that would be it for today. Thank you very much and good luck.