 Hi, I'm Zor. Welcome to Unizor Education. Well, we were considering similarity in three-dimensional space in the previous lecture. So I would like to spend some time solving a couple of problems. Problems are really very simple. We will crack them in no time. And as always, let me remind you that this lecture is part of the course of Advanced Mathematics presented on Unizor.com. And that's the website I suggest you to watch this lecture from. Okay, so the problems. First problem, prove that all cubes are similar to each other. Okay, so if you have a small cube and you have a larger cube, they are similar. All cubes are similar to each other. How can it be proven? Well, let's just think about what is similarity. By definition, two objects are called similar if there is some kind of a scaling which converts one into another which is either another or maybe some other which is congruent to that another. So this is A and this is B. So I have to find such a scaling which is defined by center and the factor that A is converted, is transformed to A' which is congruent to B. This is A and this is B. That's what basically our similarity is defined how. So, to prove the similarity, we have to find the point, the center and the factor of this scaling. Now, the point is any and the factor is if this is A and this is B, then the factor is B over A. Now what happens if I use this factor and any center of scaling, let's say here, X. Well, each point of this cube should be somehow extended if I will be able to draw it properly which obviously this is not proper. My artistic talents are not really that great, something like this. So this is this side, this is this side, something like this. That's this one. So anyway, we are building this cube. It will be something like this. So this would be transformed into this. Now, if this is A, this would be F times A. But since F is equal to B over A, this is B. So now we have this cube which has a side B and every side is B obviously. And this cube and these are obviously congruent to each other. The congruence is usually the ability to transform one into another. So if you have two cubes of the same length of the edge, how can they transform one into another? Well, actually it's very simple. First you pick a particular vertex in one and move it to this one. Then you align the plane where it belongs to, like a base of this cube for instance. If it's a little bit tilted you have to position it. Then the whole base can be aligned by rotating within that plane around this point. And when the base is aligned everything else will be aligned as well. Okay, so that's the simple thing. And the most important is the factor, not the center. Because I could use any center and with another center of scaling I will produce another cube with the size of the edge B. And all of these cubes are congruent to the one which we need. So it's not dependent on the center, it depends only on the factor. The factor should be chosen correctly. And the factor is in this case ratio between the edges, the lengths of the edges. Okay, next, very, very similar. You have to prove that two tetrahedrons are similar to each other. Now tetrahedrons, I'm talking about the regular tetrahedron which is something like this. Where all sides are the same. So every side, and there are four of them, one as a base and three as side faces of this pyramid, every one of them is equilateral triangle. That's what regular tetrahedron is. And now we can take another one which is maybe positioned somehow differently with all edges equal to B. Exactly the same thing. Choose factor equal to ratio of the lengths of the edges. Take any center of scaling and basically scale this tetrahedron into the one which will have A times F, which is equal to B, lengths of each edge. And once you have two regular tetrahedrons with the same size of the edge, they can be transformed one on the top of another. So that means they are congruent. And again the process is very simple. So first we align the vertex, then we adjust the base, the base plane, then we rotate within the base plane to have these triangles coincide and that would be the end of it. Next is again very similar. So I wanted actually to show different shapes which are congruent to each other. So the third one is the sphere. So if you have a small sphere and a big sphere, they are always similar to each other. Now again the way how to do it, how to prove the similarity, if this radius is capital R and this radius is lowercase r, we just have a ratio equals the ratio between the radiuses. You can use center here, although you can use actually any center, but it's easier if I will have it here. And using the scaling with this factor, I will transform every point which is on the radius lowercase r into something which is f times lowercase r, which is R, capital R, and that would be a sphere like this. And these are two spheres of the same radius, even if it doesn't look like that on my drawing, but they are the same radius and obviously two spheres of the same radius are congruent because all you need to do is just to move one center into another and everything else will be in place. So these are three very very simple things where I have proven that certain very regular figures with the same number of edges or the same shape, so we have either cubes where all edges are the same and all angles are the same, we have regular tetrahedrons which have all faces, equilateral triangles and spheres which basically is determined by the radius. So these are all groups of figures which are similar within each group. Okay, now let's use something else, let's talk about volume of similar figures. Here is the problem. Well, let me first give you, it's kind of a homestyle description and then I will abstract it into more mathematical language. So the homestyle is if you have, let's say, an empty glass and you fill it with, let's say, bowls of certain diameter. Now there is certain amount of space within that glass taken by these bowls and there is certain amount of space which is free so you can pour some water and it will be certain amount of water in that glass. Now what if you will replace these bowls within this glass? These are all bowls with the bowls of smaller diameter and then again pour some water. Will it be more water in the glass with smaller bowls actually? Yeah, smaller bowls or more or less. So what's the answer to this question? What kind of scenario leaves more free space, let's say, for the water if you fill it up with big bowls or small bowls. Now let me transfer this particular problem into mathematical language. I would say the following. Now consider you have, let's say, a cube and this cube is filled, well not filled, but we will inscribe into this cube a sphere which will touch every something like this. It's inscribed into each face of this cube. So basically the diameter of the sphere is equal to the edge of the cube. So again, we have certain amount taken by this sphere and certain amount around it which is free. Now instead of this sphere, we will take the one which has, let's say, one-enth of its diameter. And we will take many spheres of this diameter and we will fill it up the same way. Now for instance, n is equal to 2. Then that means that I will have two bowls across each edge and four bowls, it's something like this. And then four bowls on the next level, something like this. So it will be eight bowls, right? Two bowls across each side. Well, they're supposed to touch each other of course. So now the question is we have taken certain amount of space by these bowls and certain amount of space is left free and what's greater amount of free space here or there or any other end, what happens in this case? Well, let's just think about this. If this is your diameter of the bowl, you have n bowls which can fit along this edge, right? If before you had this diameter, which means edge is equal to diameter, now the diameter is one-enth of that. So you will have n bowls here. So on each layer you will have n square bowls, right? How many layers do you have? Well, again, n because it's n here, n here, n here, right? So you have n times n times n. So you have n cube different bowls. So you have n cube bowls. Okay, that's fine. Now, what about the volume of each bowl? Well, if the size is one-enth, remember that the volume is proportional when we are talking about similar objects. The volume is proportional to the third power of the scaling factor. So what's the scaling factor from here to here? Well, obviously, n, right? I said this is one-enth of this radius. I have to multiply the radius or diameter of this by n to get this one. So my scaling factor spheres are always similar to each other. So there is a scaling between this and this. Scaling factor is f equals to 1n. Now, the volume factor, the volume would be equal to the result of the scaling. That's why v2 is equal to factor to the third degree times the original volume. So original volume with this diameter is obviously 1 to the n cube of this volume, right? Since the factor is n. But the number of our bowls is n cube. So each one is 1n to the third power of the volume of this one. But there are n to the third power of these bowls, of these spheres. So as a result, if you multiply these two, we will have exactly the same volume of the original sphere. So no matter how many smaller and smaller bowls you fit into, you accurately fit. I mean, you don't have to leave the gaps or whatever else. I'm presuming that we are doing it geometrically correctly, layer by layer by layer, n square by n square by n square. So if you do this, then the volume which is occupied by these n to the power of 3n cube of smaller bowls is exactly the same as one big bowl. Which means that the free space will also be the same. So no matter big bowls you put into this glass or small bowls, whatever the amount is left for the water will be the same. Provided this done accurately, etc. Alright, so that was the problem with filling up certain reservoir with smaller or bigger objects. Now let me tell you this. If you will make this problem even more realistic, let's say you have some kind of a box, let's say. You can fill it up with big stones and you can fill it up with, let's say, sand. Now there is some kind of a notion that sand actually fills it up better than stones. But that's not true. I mean, yes, with certain amount of abstraction we can prove that stones are regular obviously. In practical life it cannot be ideal. But if you do it with bowls, with spheres and accurately put it in, then the amount of free space will be exactly the same. But that was something which you might ask a child what exactly is the difference in the amount of free water. Alright, problem number five. Okay, it also can have a practical kind of implementation. Let's say you're in a bar and you're drinking a martini. And this is a martini glass. Now you are with a partner and she is drinking from the same martini glass, but she is asking the barman to fill it up only half height. So that's what she has. And you have full glass of martini. Well, question is how much less she will drink? Well, let's think about it. This cone from this half down is obviously similar to this one. And to prove the similarity is very easy. You just have this as a center and two as the factor. And this cone will grow into this one, right? Because every line on this circle will take the corresponding place on this circle. So the factor is two. Now the volume as we know has the factor of F to the third degree, right? So if this volume is V, and the factor is two from here to here, or reverse factor is one half, if you wish, then the volume of this is one half to the power of three, which is one eighth. So, I mean, it sounds like, well, fill up only half, that it's like half. But no, it's not half, it's really one eighth of martini, which she will take. So that's what's important. You are significantly reducing the amount if this is a cone. Now, if it's not a cone, if it's a glass like this, arithmetic is not working. Because this is a cylinder and half of this, although a cylinder as well, but these two cylinders are not similar to each other. The cones, half of a cone is. Because there is no, if you take a center, for instance, I know where. Here, for instance, and you will, well, stretch it by what factor? Like two, for instance. Well, then this cylinder will be converted into the cylinder of twice as high, which is something like this. And twice as big a circle, not like this. So it will be fat cylinder, right? If you will take this bottom half and use the factor of two in scaling, you will have from here this, not this cylinder. So in this case, half a cone is similar to the full cone, but half of a cylinder is not. So that's a big difference. All right, and the last one, which is kind of related to this counter example with the cylinder which I just made, question is if you have two right rectangular prisms, under what circumstances these are similar to each other? Well, let's just think about it. The right rectangular prism is basically defined by three parameters. Two parameters define the rectangle at the base, and the third parameter defines the height. And they're all different, obviously. Now, in this case, you have different lengths. Now, if we want to scale this into this, and the length of each segment is multiplied by the same factor of scaling f, then c prime is equal to c times f, b prime is equal to b times f, and a prime should be equal to a times f. Only in this case, we can have this particular similarity. Or if you wish, forget about f, but this can be expressed differently. a over a prime is equal to b over b prime is equal to c over c prime. If this is the case, then we can choose factor f equals to this, and then every center of scaling will do, as long as we will use this particular factor of scaling to convert this into this. So that's the answer, basically. This is the criteria for two right rectangular prisms to be similar to each other. And that completes this very small set of problems. I do suggest you to try actually to repeat the whole logic just for yourself in any case. Obviously, the problems are on unizord.com in the comments for this lecture. So thanks very much and good luck.