 Hi, I'm Zor. Welcome to your new Zor education. Let's solve a few problems related to errors. I have seven problems which I'm planning to discuss today. I'm sure you have already spent some time looking at these problems. Well, if you don't, please do it now before you listen to this lecture and try to solve these problems yourself as best as you can. And after the lecture is completed, try to do it again just on your own, solving these problems just to make sure that you remember this. And, you know, next time when the new problems will arrive, you will remember the techniques maybe to use. The whole purpose of solving problems is not really to remember how to solve these problems, but to prepare yourself to solve the problems which you never solved before. That's what creativity is all about, right? All right, so let's solve these problems. Problem number one, if you have a triangle, which is the right triangle, and you draw an altitude towards hypotenuse, then these two segments are related to each other as squares of the corresponding quantity. Actually, that's quite easy. Excuse me. It's all related to similarity of the triangles, because obviously, since this is the right triangle, and these two small triangles are right as well, and they share these angles with the big triangle. So the small triangle is similar to the big one. And another small triangle is similar to the big one, because the angles, the cute angles are the same, right? So from the similarity, we can build a certain relationship. Let's say, let's consider this particular triangle, the small one. So its smaller catatose X relates to its hypotenuse A, as in the big triangle, the catatose which lies against the same its A relates to the entire hypotenuse, which is C. Now, similarly from similarity of this triangle, we can take the catatose Y relates to its hypotenuse B, as let me just mark the same angles that will be better. So Y lies across the single arc angle. In the big triangle, the catatose which lies against the single arc triangle angle is B to its hypotenuse, from which we can derive that X is equal to A squared divided by C, Y is equal to B squared divided by C, and if we will divide one by another, C will be reduced basically, and we will have A squared divided by B squared. So from these two, follows this one. Alright, that's the first problem. That's easy. Next. You have a square, and you have to construct another square so that the area of the first one relates to the area of the second one as m over n, where m and n are given numbers. So m and n are given numbers, and this square, the square number one is given. So we have to construct the square number two, which has the area proportionally to this one using this ratio of proportionality. Well, before actually to address this, I would like to make one very, very simple construction, and we will use it in some other cases as well. Now, if you have number m, how to construct the segment which has the lengths equal to m? Well, that's easy. You have the unit segment, and you attach it to itself m times. Now, the interesting question is how to construct a segment which has the lengths square root of m? Well, here is what I suggest. Remember this theorem that if this is h, this is x, and this is y, and this is the right triangle. So, again, as in the previous problem, all triangles, two small ones and the big ones, they are similar from which we can say that, okay, considering this is similar to this, these two small triangles, we can say that smaller catedus f's to a bigger h. In this triangle is the same proportional as smaller to the bigger in this triangle from which you see that h squared is equal to x times y. This is something which we have already spoken about many different times, many different occasions. We had the square of the altitude from the right angle towards the hypotenuse is equal to the product of two segments which the altitude divides the hypotenuse into. From this, we can very easily derive this. How? Well, let x is equal to m and y is equal to 1. Then what is h? h squared is equal to m times 1, which is m. So, h is equal to square root of m. So, if you would like to construct this segment, all you need to do is you have to construct the right triangle in such a way that the altitude divides the hypotenuse in m over 1 ratio. So, to construct it, you have to do very easy thing. You do first hypotenuse which is equal to m plus 1. So, this is the hypotenuse. This is m and this is 1. Now, how to find the vertex of the right angle? Well, very easily, you draw a circle right here. That's the locus of all the different vertices from which this hypotenuse is viewed at 90-degree angle. So, all these angles are 90-degree because they all are supported by the diameter. But if you will construct the perpendicular, then this particular point would not only be the vertex of the right triangle, but also the one from which the altitude would actually hit this particular point, leaving m units on the left and 1 unit on the right. Then this particular length is square root of m. So, we know how to construct square root of m and I no longer be talking about this but I will use it in different problems. Now, that's exactly what I will do here because if this is a and this is x, the area 1 is equal a square, area 2 is equal to x square and it's supposed to be equal to m over n. Or a over x is equal to square root of n divided by square root of n. Now, we know what square root of m is and square root of m is because we know m and n and I have just explained how to construct square root of m. So, I know the lengths of these two. So, let's just use different letters, let's say p and q. So, we know these are known, I mean not known, but very easily constructable segments. So, a, p and q are segments which we know, we can use them. So, all we have to do is basically to construct the proportional to these two things which is very, very easy. And again, we did it many times before. I'll just repeat it once. So, if you have a over x is equal to p over q where a, p and q are known segments. Well, that's basically very easy. I mean, you can, for instance, you can do it this way. You can have a, p, q parallel lines and that would be your x. So, you put a, p and q parallel lines and you will have, this piece will be x. How to prove it? Well, it's similarity and it's trivial actually. I don't want to spend time on this. All right. We did it before when we were talking about similar triangles. Next one. All right. So, you have two triangles which share one angle. Let's put it this way. This is one triangle. This is another triangle and they have the same angle. Everything else is different. All right. So, what I have to prove is xavcd. Okay. Well, I have to prove that. So, triangle xavc has area. Let's call it a1. And area of triangle xbd will be a2. Now, the theorem says I have to prove that a1 over a2, so the ratio of the areas is equal to the ratio of products of the sides which form this particular angle. So, a1, so it's xa times xc. In this triangle, in the small triangle, xa and xc are forming this common angle. And in the big triangle is xb times xd. That's what I have to prove. Well, the obvious way to solve this problem is let's just draw a couple of altitudes. And what we have is a1 is equal to what? This is xc times the altitude. Let's call it m and n. m times am, right? And a2 is equal to the base xg times altitude. Well, actually, in both cases, I am wrong. I have to divide it by 2. So, in order to save some space under the board, instead of dividing by 2, I doubled the area. Because it's a triangle, so it's half of the product of this times altitude, or this base times this altitude. So, double area is equal to the product of the base and altitude. All right. Now, we do have these altitudes, right? Because in this, we don't have altitude. We have the sides. But let's think about it. am and bm. If you consider these two triangles, xam and xvm, they're both right triangles, right? Because these are altitudes. So, these are right angles, which share the angles, which means that they are similar. And the similarity actually says that the ratio of these two cateches is the same as the ratio of these two cateches or these two hypotenuses, right? So, in this particular case, I need the hypotenuses. So, let's divide this by this. What do we have? We have 2a1 divided by 2a2. Now, actually, 2 can be reduced. We don't really need this. It's equal to xc times am divided by xg times bn equals xc divided by xg times am divided by bn. Now, this is proportionality between these two cateches of these two right triangles. And as I was saying, they are proportional to the corresponding hypotenuse. So, I can rewrite it as xcxg times, instead of am to bn, I can put xa over xb. And that's exactly what we have to prove. a1 to a2 as xaxc and xbxg. That's exactly what's necessary to prove. That's it. Okay, what's next? Similar triangles have areas proportional to their sides. Alright, so you have similar triangles. So, all angles are the same. All sides are proportional. Now, we have to prove that the area of one divided by the area of another is like, this is a, this is a prime. a square divided by a prime square. You know what? Let's do it better. Let's call it a1 and a2 a1 divided by a2 square. Okay? How can we prove it? Well, quite elementary actually. Obviously, in the similar triangles, everything is proportional. I mean, everything can be in any linear element. For instance, obviously, altitudes. So, if you will do this, for instance, or any other altitude, h1, h2, b1, b2, c1, c2. Now, obviously, we know that a1 over a2 is the same as b1 over b2 and c1 over c2. Now, what I'm actually telling is it's the same as h1 to h2. Now, how can that be proven? Well, very easily, because not only the big triangles are proportional, are similar to each other and elements are proportional, but also this part, for instance, this right triangle is obviously similar to this right triangle. Why? Well, because the angle is the same, right? And since these are right triangles, the triangles are similar. If they are similar, then their hypotenuses, for instance, are in the same ratio as their gadgety. And this is the gadgetus right now, right? So, h1 to h2 is b1 to b2. That's how it can prove. Now, having this, I obviously can derive this. How? Well, very easily. Let's just wipe out this piece and just use this. Now, what is area 1? Area 1 is c1 times h1 divided by 2. A2 is c2 times h2 divided by 2. Their ratio, their ratio, a1 over a2 is equal to c1 h1 to c1 c2 h2, right? 2 will be reduced. But again, c1 to c2 is the same as a1 to a2 and h1 to h2 is also the same as a1 to a2. Here you have a1 squared and a2 squared. That's exactly what's necessary to prove. So, all is following from proportionality of any linear element. Medians will have exactly the same ratio. Bissectors will have exactly the same ratio. Rages of inscribed circle here and here will have exactly the same ratio. Rages of circumscribed circle will have the same ratio. So, if two triangles are similar, everything, every linear element will be proportional and the proportionality will be exactly the same. If you remember, I have defined similarity from scaling. So, this is scaling and all elements are scaled. Now, all area-related elements are scaled, so to speak, twice, which means we are supposed to use the squares. Like a2 or area of this, for instance, will also be proportional to the square of any linear element. I can say that, for instance, the area of this relates to the area of this as this square of this altitude to the square of that altitude. So, areas are proportional to the squares of linear elements. So, again, all these similarity-related problems, they all derive from the principle of similarity, which is based in scalability. Next, well, next is exactly the same thing, but not for triangles. If you have two similar polygons, something like this, then, again, their areas are proportional to a square of ratio of linear elements. So, if this is a1 and this is a2 and a1 squared to a2 squared is what the ratio of their areas. Now, how to prove that? Well, there are probably more than one, definitely there are more than one ways to prove it, but if you consider, for instance, you can always divide any polygon in triangles and since every element, every triangle as part of this polygon, which is similar to this polygon, would also be similar to the corresponding triangles and every triangle has the same ratio of, every pair of triangles has the same ratio of their areas as the square of the sides, of the corresponding sides and the ratio of corresponding sides is exactly the same everywhere. So, if this is b1, b2, then b1 to b2 is exactly the same ratio as a1 to a2 because, again, these are similar geometrical figures, which means everything is related to a scaling, the same scaling factor. So, that's why, basically, the ratio of the areas would be a square of the ratio of any linear element. That's easy. Now, if you have some particular case of the polygons, then, for instance, you have the regular polygon and you can inscribe a circle. Then, again, if you will scale it, if you will have a similar regular polygon, then, as I was saying, all elements will be scaled by the same scaling factor. Now, including, obviously, the radius of the inscribed circle. So, if you have this, for instance, this is the center of the regular polygon. So, you can very easily prove that this particular one, if you have a similar, well, forgive me for not being precise. So, if this is the similar polygon to this and then you have the corresponding triangle, which is similar to this one, that's very easy to prove then that the radius of the inscribed circle is also proportional using exactly the same scaling factor. It can be proven by many different things, but including, for instance, the similarity of the triangles and the similarity of altitudes, for instance, which we were just proving before. So, that's why the ratio of the sides is exactly the same as the ratio of the inscribed circles. And, therefore, what we can say is that if you have two regular polygons similar to each other and for similarity, by the way, it's sufficient just to have the same number of sides, the same number of vertices. We did prove this theorem before. So, if you have two similar regular polygons, then their areas is proportional to a square of inscribed, of the radius of the inscribed circle, which is the same thing as the radius and the ratio of square of radiuses of circumscribed circles, which is the same as the ratio of squares of corresponding sides, et cetera. I mean, all linear elements in the similarity case are proportional with the same ratio of proportionality, which is the scaling factor. Okay, I have the last problem to explain. Now, here is the problem. If you have a triangle, what I'm asking you is to divide it using the lines parallel to the base, in certain number of parts, equal in area. So, the area of this trapezoid is the same as area of this, same area of this, same area of the top triangle. So, let's say you have a number m into which you have to divide the area. Well, you can actually think about this as a practical problem. I mean, if you have some triangular piece of land, which you would like to divide among five people in such a way that the area would be the same, so they can grow something on this area. Okay? So, how to divide it evenly? I mean, evenly in terms of area. Well, let's think about it. If you have n, then this piece is one nth, this piece is one nth, this piece is one nth, and this piece is one nth, right? So, if this is a1, this is a2, this is a3. So, a1 cuts the piece which is equal to one nth of the area, right? a2, if you count it, if you measure it from the top, a2 measures two times one nth, right? a3 measures a triangle which is three times one nth of the original. So, basically, what I'm saying is that the position of the point, ak, is such that it measures k nth of area, of original triangle, right? The point a1 measures a triangle which is one nth. Point a2 cuts a triangle which has an area of two nths, etc. So, point ak cuts the piece. Now, we know the original triangle, which means we know all elements, including the sides, etc., right? Now, we know that the area of the triangle which is cut by the point ak is such and such. Therefore, from one of the previous problems, number four, if you have single triangles, their areas are proportional to squares of their sides, right? So, the area of, let's call it case triangle. Case triangle is triangle which is cut by the point ak, ak divided by the area of original triangle, as, let's call this point x, and this is point a, as x ak square divided by x a, all right? I hope that's obvious, right? This is the side of the triangle which is cut by the line, and this is the side of the original triangle. And we were saying that since areas are proportional to the squares of the sides, which is the problem number four, I can write this equation. Now, this is what I'm given. This is k over n. So, the areas are supposed to be related as k over n. So, in this particular equation, x x a k square divided by x a is equal to k over n. I didn't know, sorry, square. I'm sorry. I basically have x a, I have k, I have n, and all I have to find is x a. Now, obviously from this, I can say that x a k is over x a is equal to square root of k over square root of n. We know how to calculate, how to construct these two. So, this is just the force proportional if you have three other pieces, and that's the solution. So, in any case, all you need to do is for each k from one to n minus one, you have to construct square root of k as a construction problem, which we know how to do it, I explained in the beginning. Same thing with square root of n. And then find the force proportional which satisfies this particular proportionality, which we also know how to do it from the similarity of triangles. Well, that's it. I hope everything was well understood, and I do recommend you to go again through these problems. Do not forget these little techniques, like for instance if you have, for instance, square root of five. How to construct a segment which has the length of square root of five, for instance, or square root of any number, using again the unit length, et cetera. How to construct the proportionality, x over a is equal to b over c, for instance, where x is unknown and a, b, c are unknown. So, these are techniques which you probably have to remember, and they should be in your repertoire of the techniques which you know. So, please go again through these problems by yourself. By all means, we are invited to listen to the lecture again. And again, don't forget that Unizor is a relatively comprehensive course for advanced mathematics, and I do encourage you to basically go through the whole course. As a registered student, if you have somebody to supervise your process, somebody to enroll you, or actually you can enroll yourself if you can register as both student and the supervisor, go through the exams. Exams are very important. Well, that's it for today. Thank you very much.