 Hi, I'm Zor. Welcome to a new Zor education. I would like to present a few problems which I can consider like final problems on three-dimensional geometry, and it will probably cover more than one lecture. Today I would like to discuss just one problem, and probably there will be some others. It looks maybe a little bit more complicated than it really is. After all, it's all just Pythagorean theorem and a couple of formulas which we have here. So what's the problem? The problem is as follows. There is a cone, and this is the base, and the sphere is inscribed into this cone in such a way that it touches the side surface of the cone, and it also touches the base of the cone. Now the cone is obviously presumed to be a right circular cone. So the base is a circle, and the apex projects right into the center of this circular base. Now, I don't want to discuss right now the fact that this sphere actually is touching the base exactly at the center. We will discuss it maybe later, and it also touches the side surface of a cone along a circle. It's kind of obvious, and there are certain considerations of symmetry which can be actually used to prove it. I would like to concentrate only on calculating certain things which basically represent this problem, and I will leave these a little bit more obscure maybe considerations to after I will get the result of the problem. So the problem states that the circle has a radius known R, and it's also known that a circle which is the circle of tangency wherever all the points of the side surface are touching the points of a sphere. So these are actually a circle, and again we will be discussing this maybe after all. So what I know about this is the following, that if I will draw a radius from a center to the point of tangency, that would be an angle of 60 degrees. So knowing these two parameters, the radius of a sphere inscribed into a cone, and the fact that tangency is viewed from the center at the angle of 60 degrees relative to a horizon, I would like to find the volume of the cone. Now probably the best way to approach this problem is to convert it into a two-dimensional geometry. So what I'm going to do is I'm going to cut the whole picture by a plane which goes through the axis of the cone and diameter of the sphere, and what will I get as a result? Well, the cone will be basically a triangle, right? That will be my result of the cutting of the cone vertically along its axis, along its altitude. Now the sphere would be actually inscribed into this triangle, and what I know is this is an altitude that this angle is 60 degrees, and this is the radius. Now to find out the volume of the cone, I need two things. I need the cone's altitude and I need the radius of a base, right? Okay, so the altitude is easy from this drawing because if this is, let's say, SABO, and this is the T for point of tangency. So I know that, I also need this one, let's call it C. I know that the altitude of the cone, which is basically an altitude of this triangle, SC, is basically a combination of two segments, SO and OC. Now OC we know, that's R. Now how can we determine SO? Well, obviously from this right triangle, we know the characters OT is equal to R, and we know that this angle is 60 degrees, which means this angle is 30 degrees, and this is also 60 degrees, right? So what is a hypotenuse if this is R? It's obviously the R divided by square root of 3 times 2, is that right? Yes. Now this particular characters ST should be half of the hypotenuse, in which case it's R over square root of 3. So this is SO, and this is ST. And let's check the Pythagorean theorem. This square, which is hypotenuse, which is 4R divided by 3, this square is R squared divided by 3, and this is R squared, or 3R squared divided by 3. If you summarize, you will have this one. So Pythagorean theorem is correct. Everything is fine. So we found the hypotenuse. We found SO. And since we found SO, we can find the SC, which is equal to SO plus another R. So that's the altitude of the cone, right? So it's 2R over square root of 3 plus R. So that's my altitude of the cone. So that seems to be simple, right? Now, how to find the radius of the base, which is basically BC? It's also not very difficult, because obviously triangles SOG and triangle SBC are similar, right? Both are right triangle, and both share the angle of 60 degrees. Now, since they are similar, I can do the following. So I think BC divided by OT equals to, so this is 30 degree, equals to SC divided by SC, and it was against 30, divided by ST. So I know this. Now, what do I know? OT is R, so BC over R equals to SC. SC, I know what that is. It's this one. Let me maybe make it a little easier. What is it? It's multiplied by square root of 3. That would be 2 square root of 3R divided by 3, and this is 3R divided by 3. So it's plus 3, right? So that's my SC. 2 square root of 3 plus 3R divided by 3 divided by ST. Well, ST, I already actually calculated and wiped it out. That's half of the SO, and SO is this. So it's R over square root of 3. So it's R square root of 3. Now, I hope I didn't make a mistake because I have an answer here, which I'm supposed to come up with. So if I made a mistake, I will not make it right. But anyway, that seems to be as efficient for BC. Now for BC, so there is no R here and R here. So that's my result, right? So BC is the radius of the cone's base, and SC is its altitude. So I have to do basically the volume is one-third of the area of the base. Okay, volume is one-third pi in this radius square, which is what? R square times 3 divided by 9 and 2 square root of 3 plus 3 square. That's what it is, right? Square root of 3 square is 3, square root of 3 is 9, and this is square, this is square. Times altitude. So I have one-third the area of the base and times this one. Times R again, so it's R cubed now. Divided by 3, so it's here. And 2 square root of 3 plus 3, so that would be cubed. That seems to be the result. Well, let's check it out if I have it exactly the same as I have it here. Well, 3 and 3 are reduced. Now this is one-third pi R cubed 9. Now, well, I think I remember the formula of A plus B cubed. A cubed plus 3A squared B plus 3AB squared plus B cubed. So, cube of this is 8 and square root of 3 cubed. That would be 3 square root of 3. So it's 24 square root of 3 plus 3 square of this times this. 4 and 3 is 12 times this, 36. So it's 3 times 36, 108. Plus 3 this by square this. So it's 3 and 2, it's 6 and 9, 54. So it's 54 square root of 3 plus 3 cubed 27. That seems to be the right thing. Equals. Well, obviously one-third every one of those is divisible by 3. So it's pi R cubed divided by 9 here. So 108 and 27 and we have to divide by 3. So it's 36 and 9. So it's 45. 24 and 54, it's 78. Divide by 3, it's 26. Square root of 3. And that's exactly what I have. So that's the answer. So as you see, it's basically nothing as far as the ideas or anything like this. Just a couple of theorems, Pythagorean theorems and the formula for a column. What's probably more interesting and more, I would say, in spirit of three-dimensional geometry is to talk about what is, for instance, a sphere which is tangential to a side surface of the cone. Or what is tangential to a sphere at all with a line or with a plane. So let's just talk about this a little bit. It's interesting. Well, let me just wipe out all this. We don't really need it. Now, sphere which is tangential to a plane or a line. Now, what is a sphere? Let's just recall whatever the definition is. Remember, that's the locus of all the points in a three-dimensional space which are equidistant from its center and the distances are. Everything inside the sphere has a smaller distance to the center, right? Everything outside the sphere has a greater than r distance from a center. So it kind of divides the whole three-dimensional space into two parts. Smaller than r and greater than r. And obviously, the boundary between these two is where it's exactly r, the distance to a center. Now, if I have a line, for instance, in space, now what does it mean? How can I define the line which is tangential to a sphere? Well, I think that the obvious definition is that the common set of points between a line and the sphere should be just one point. So it should touch it at one particular point. And that's it. Because you understand that if the line goes outside of the sphere, then there is no points of intersection, right? If it's inside, there will be two points of intersection, and only if it's tangential to a sphere, there will be one point of intersection. So we can actually define that if the line has only one point of intersection with a sphere, this line is called tangential. Now, you remember from the two-dimensional geometry that if you have a circle and you have a tangential line, then the radius to a point of tangency is perpendicular. Well, it can be proven just because this is supposed to be the smaller distance from this point to any other point on the plane, right? Because all other points are outside of the circle. Same thing here. So all points on this line are outside of the sphere except the point of tangency. So the point of tangency is the smallest distance between the center and the line. And the smallest distance, as we know, is perpendicular. Now, analogously, we can talk about a plane. So if you have a sphere and a plane, we also can define the concept of tangential plane through this intersection. It's supposed to be only one point of intersection between the plane and the sphere. And again, it's very easy using this minimum distance. It's easy to prove that the radius towards point of tangency is supposed to be perpendicular because this is the smallest, the shortest distance between a center and the plane. So all other, except the point of tangency, all other points on the plane are outside of the sphere, which means they are further than r, and only the point of tangency is equal to, on the distance equal to r. Okay, so these are the concepts of tangency between the sphere and the line and the plane. But I was talking about a cone. How can a sphere be tangential to a cone? We still have to define it somehow. I'm sure we understand intuitively how it works, right? So you have a cone and the sphere is somewhere here and it's just here, basically, right? It touches all the side surface. Well, here's how I can define it. The sphere is tangential to a side surface of a cone or tangential to a conical surface, if you wish, if every generator, which is line connecting the apex with all points on the base, the base is directors, right? And all these lines are generators. And a side surface of a cone is basically a set of all these lines, if you wish, right? So if my sphere is tangential to every generator of this particular cone, then I can say that it is tangential to a cone, to a cone's side surface. Now, at the bottom, it's a plane and we already know what tangential actually means. Now, so basically, we kind of defined what exactly a sphere which is inscribed into a cone is. Now, again, intuitively obvious that this point where the sphere is tangential to a plane must be a center of this circle, which is base of the cone. And also, all the points where our sphere is tangential to a cone, the side surface of the cone must actually be along some kind of a circle. So it should be flat and it should be equidistant from some point here. Well, let me just, you know, talk about this as seeming to be a little bit more difficult concept. How can I prove, basically I would like to prove it, that all the tangential points between the sphere and the conical surface are lying within one plane and they're all making a circle. How can I prove it? Well, actually, I don't think it's very difficult because let's think about it this way. Again, let's talk about the section of this. Now, this is how this is looking. Now, all these points of tangency belong to a sphere, right? So they're all equidistant and these two are perpendicular, as we were talking before, since these are lines, generatrix, and these are radiuses of the sphere, right? So these are equal. Now, this is right angles, right? Because it's perpendicular to a tangential line, the tangent. So we have, basically triangles are congruent because they have radiuses are the same, right? And they all share hypotenuse. So every triangle which is formed by epics of the cone, center of the sphere and the point of tangency, every triangle S, O, A, where A can be any point of tangency, all these triangles are congruent to each other. Okay, so that's fine, which means these two are equal, right? Now, if that is true, and if I will draw a perpendicular from this point to this and from any other triangle, it doesn't matter that this is opposite triangle, it can be any triangle, it can be a triangle with a point here. So if I will draw from any point of tangency a perpendicular to the axis, I'm supposed to get into the same point. Why? Because, look, I deliberately put a different point. Again, these triangles are supposed to be congruent because these are also right triangles because I said this is perpendicular, right? Now, these angles are the same because these triangles are congruent, right? And these two calculatives, these two calculatives are the same because, again, from bigger triangles, right? Because they're congruence. So that means that this distance supposed to be the same for all the points of tangency. So they're all projecting into the same point. So all these points of tangency are on a perpendicular plane to an axis. That's what's important. And obviously, these two are equal in lengths. So we have the distance from this point to this point is constant for all the points of tangency. That means it's a circle, right? So basically that's what I wanted to talk about as a kind of epilogue, if you wish, to this particular problem. The problem is not difficult as I was saying, but these considerations need to be clarified because I was using them before on an intuitive basis. And I would like to have some more or less solid foundation behind some intuitively obvious things like the line of tangency is supposed to be a circle and some other kind of intuitively assumed things which we did during this problem. All right, so basically that's the first problem among a series of, I would call, final problems for 3D geometry. There will be more. And as usually I recommend you to review this problem just by yourself, especially there are a couple of calculations. It would be nice if you can do it just by yourself. And think again about the definition of these concepts of tangency between the planes or lines and the sphere. All right, thanks very much and good luck.