 Hi, I'm Zor. Welcome to Unisor education. We continue course, which is called math plus and problems presented on unisor.com Now this is about solving problems and what's important is these are not your usual problems, which you might actually solve in schools where those problems are usually just to illustrate certain theoretical concepts. Now the problems presented in this particular course are not of that kind. These problems need thinking. To solve these problems, it's not just applying certain regular to use algorithm or formula or something else, which somebody gives you and you just apply this formula or algorithm and you come with some kind of an answer. Now these are unusual problems in the respect that you have to really think about how to solve them. Nobody told you how to solve, how to approach even these problems. Well, unless I put some hint maybe in the textual description of these problems. Now all the problems on the website, well actually all the lectures on this website are presented together with textual part, notes, parallel of the video and the notes on the same screen. Well, the notes are basically like a textbook. Now as far as the problem solving, sometimes I put solutions, sometimes I put just a hint or an answer to the problem. During the lecture, I usually present the problem and the solution. So if you are listening to this lecture first, before you're reading the notes, I suggest you to pause the video immediately after I present the problem and try to solve it yourself. It's very important. The whole purpose of this course, Mass Plus and Problems, is for you to think independently first, before you reading or listening to or watching solutions to these problems. Okay, now today we will solve a few geometric problems and there are others. Now I have to mention that there is a prerequisite course for this. It's called Mass for Teens presented on the same website. Now all the courses including this and its prerequisite, the Mass for Teens, including Physics for Teens and Relativity for All. All these courses are totally free. There are no advertisements, no strings at age, signing in is optional. So basically you just consume as much knowledge as you want. We share the knowledge. Okay, so the problems are, I have one, two, three problems. The problem number one is related to a relatively known fact that the geometric average between two positive real numbers is always less than or equal to their arithmetic average. So this is geometric. This is arithmetic averages. And this is a known equation. Now why is this particular equation algebraically can be very easily proven? Because you have to a minus b square is greater or equal to zero because it's a square, right? a and b are let's say it's positive real numbers, which means that a square minus two a b plus b square greater than equal to zero. Which means I will add four. So a square plus two a b plus b square greater or equal to four a b, right? Now this is a plus b square and this is two square root of a b square. Both are positive. So I do the square root from both and a plus b is greater than equal to two square root of a b and two can be brought here and you have exactly this. Okay, so this is an algebraic proof, which is fine. But I would like to prove it geometrically. And that's very interesting actually. If you can just bring together two seemingly unrelated things. Well, personally, I think it's just kind of a beautiful thing in mathematics. If you see that the same thing can be approached from two absolutely different directions, I think it's a very interesting point. So here's the point. Okay, you have a circle and diameter. Click any point and do this. Okay, so this is perpendicular. This is point p. This is point q. This is center of the circle. Now, let's consider this point dividing the whole diameter in two parts a and b. So a plus b divided by two is equal to what? Well, that's the whole diameter divided by two is a radius. So it's basically radius. P o. P o, right? Okay, now I would like to prove that pq is square root of a times b. How can I do that? All right. Well, let's say let's say this is equal to c. pqc. Okay, now consider triangle pqn, which is the right triangle because this is the right angle and triangle pmn, which is also right triangle because this angle is supported by diameter, half of the circle. So this is 90 degrees. But they have a common angle, right? So again, pqn and pmn are two right triangles with the same angle, which means the all angles are the same, which means triangles are similar. From similarity of these triangles, there are some proportionality of the sides. But now consider yet another triangle, pmq and the same pmn. Again, this angle is the same in both, right? Again, pmq and pmn. Both are right triangles and they have the same angle. So again, they are all similar. So what my point is that triangle pmq is similar to the big one and this one is similar to the big one. So all angles are equal between the small one and the big and the medium one and the big, which means small and medium are also similar to each other because they have the same angles. Now, let's just express the similarity in proportionality of the sides. Now, small catatoules A is divided by small catatoules of this medium triangle, which is C, is equal to a bigger catatoules, which is in a small triangle, which is C, divided to a bigger catatoules in the medium triangle B. From which, as you see, C squared is equal to AB and C is equal to square root of A times B. But now, let's just think about it. This is R. P O, this is R. Pq is C, which is square root and radius is arithmetic. Radius is a hypotenuse of this triangle P, PqO and Pq is the catatouse. Hypotenuse is always greater than the catatouse. So perpendicular is always the shortest distance between the point and the any other point on the line. This is perpendicular, so this is the shortest distance, much shorter than this one, which basically means that C is less than radius. C is square root of AB. In case P would be in this position, radius and and this particular Pq would be, if P is here, then O and Q are coinciding, in which case A and P are equal. So in case A and P are equal, you will have equality. In all other cases, you will have the C is greater, is less than or equal to, is less than or equal to R and that's why C is square root of AB. It's less than R, which is A plus B over 2. Genetic average is less than or equal to Arithmetic average and the quality only if A and B are equal to each other, which means Q and O coincide and P is also, P-O is also a radius. All right, so that's the first problem. Next is, I would like to prove something which is very obvious. If you have a triangle and the median Cm is median and angle C bisector, which means these two angles are equal to each other. We have to prove that this is a socialist triangle, which means CA is equal to CB. Well, let's think about it very simply. Now, since Cm is a median, M is middle point between these two. So these two segments are equal to each other. Now, this is a common side between these two triangles, CmA and CmB. Now these angles are the same because it's angle bisector because it's also not only the median, but also an angle bisector. So what do we have right now? We have two sides and an angle. Isn't that sufficient to say that the triangles are equal and therefore CA is equal to CB? Absolutely not. Now, there is a problem with, in the same course, I think it's Geometry 01. So it's in the same course, which is basically about this particular case. I was asking in that particular problem, is it possible to have two sides and all angles of one triangle to be equal to two sides and all angles of another triangle? But Triangles are not equal to each other, not congruent to each other. And the answer is yes. Why? Because this angle is not in between these two lines, two sides which are equal. The theorem, which is proven in the in the regular course of geometry about triangles, about congruency of triangles, quality of triangles, is if two sides and angle between them are correspondingly equal to two sides and angle between them of another triangle, then triangles are congruent. Now this angle is not between these two and that's sufficient basically in some cases to have triangles totally different. I mean they are similar to each other because angles are the same, but not equal to each other. One would be bigger than another, so to speak. And again, the problem geometry 01 in the same course exemplifies this particular case. So this is not a good way to prove what I'm going to prove. We need to find somehow a different proof. So now it's a good time for you to basically pause the video and think about it yourself. And I will present a solution here. By the way, it's a solution. It doesn't mean it's the only solution. There are probably other cases and other ways to prove this, but I'll prove it this way. So let's do it this way. Let's expand the CM, continue it beyond point M to point Z on the same distance. So CM is equal to MZ. This is equal to this. Okay, and connect it. So what do we have right now? Right now these two angles, this one and this one, are vertical and therefore they are equal to each other. Now these two sides are equal because CM is a median. So what do we have? We have triangle CMB and AMD. They have sides and angle between them. So in this case, we can use this theory, which means triangle CMB. I will use equal instead of congruent. In my view, they're supposed to be basically equivalent. I know there is some difference, but I don't think it's essential. It's equal to D triangle D, M, A, D, M, A and CMB. These two triangles, these two, are equal congruence to each other. By sides and angle between them, angles are vertical. Okay, great. Now what do we have? Well, we have from equal, in equal triangles, opposite to equal sides, there are equal angles, right? So this angle should be equal to this one. But this one is equal to this because it's an angle bisector. So we have triangle C, A, D, triangle C, A, D with two equal angles. And in the triangle, from two different and opposite to two equal angles, there are equal sides. So triangle is Asosos. C, A, D is Asosos because these angles are the same. By the way, if you forgot how to prove that, think about this. It's a very simple proof. Which means this is equal to this. Okay, but this is equal to this in turn. Right? Because again, opposite to equal angles, in these two triangles, there are equal sides. Which means since this is equal to this and this is equal to this, therefore these two sides are equal to each other and the triangle is Asosos. So what's creative about the whole thing? Creative is that we have extended this line, C, M, to the same distance. So C, M is equal to M, G constructed this triangle, proven that this triangle and this triangle are congruent. And from this we have angles are the same and therefore sides are the same. C, A, D is Asosos. So if these two sides are the same and these two sides are the same, then these two are the same. That's number two. Number C is a little bit more unusual, I would say. Let's say you have inside this convex polygon. Now, first of all, what is convex polygon? There are different equivalent definitions. One of the definition, for example, is that all inner angles are less than 180 degree, less than pi. Another definition is that from any point you can connect this point to all the vertices, basically to all the points of the whole polygon and the connecting segment would be inside the polygon. So between two points of the convex figures. So let's say this point and this point. If you put the line, it will be completely inside the polygon. And that's basically a more general definition, not only about polygons, but about any geometric figure, closed geometric figure on the plane. It's convex if two points can be inside or on the border connected with a segment and the segment would be completely inside. So any closed line will divide the plane into two parts, inside and outside. So whatever is inside, and I'm talking about non-self intersection lines, closed but without self-intersecting. Inside, outside. So if any two points can be connected with a line completely inside, this is a convex. So let's assume we have a convex polygon. So my question is, let's just put all the different diagonals. All the possible diagonals. And let's just assume that no diagonals are no three diagonals are intersecting in one point. Okay? Okay. My question is, how many points of intersections are? So that's my first question. How many intersection points here? My second question is, you see there are different angles here. This is an angle. This is an angle. This is an angle. This is an angle. So my question is, what's the sum of all these angles? Well, to tell you the truth, when I first saw this problem, I was kind of puzzled. And it's not easy to basically kind of approach this problem. So in this case, I suggest you to pause the video and think about yourself. So let's start first with the number of intersections. How can we calculate this number of intersections? Well, apparently it's not really very difficult because any intersection, let's say this is formed by two diagonals from four different points. So if you take any four points, let's say this point, I'm talking about vertices. This, this, this and this. So this is a quadrilateral. Quadrilateral, this is a diagonal and this is a diagonal and this is a point of intersection. So any quadrilateral, which can be formed from these, in this case, six vertices of the polygon would generate one intersection point. So if you take any four vertices and draw two diagonals, any four vertices and draw two diagonals, you will have an intersection. Which means that the number of intersections is basically a combinatorial problem. How many different groups of four points I can choose from and given? So if you have n points given, and if you remember the course of combinatorics, which I do present at the mass proteins course as a prerequisite. So I'll just use the formula. If you don't know the formula, go back to any textbook or my lectures in mass proteins on combinatorics. Number of times you can pick four out of n is n factorial divided by four factorial times n minus four factorial. Remember factorial is right. One times two times three times z times n. Now let's call it lowercase n. So this is the number of intersections. Okay, great. So now the next problem is how much what's the sum of all the angles which are formed by all these. Now angles are either inside between two diagonals and we can summarize that sum of these is 360 degrees which is 2 pi in radians, right? Plus n goes at every vertex. Now, but we do know some of these and some of these. Some of these is 2 pi times n. 2 pi times lowercase n. Because every point of intersection gives me sum of angles 360 degrees, 2 pi radians. Now, sum of all the inside angles in the polygon is, well, I hope you remember, it's n minus 2 times 180 degrees. Now, do you remember why? Well, let me just remind you. If you have any polygon, take any point. I'm talking about convex polygon. Every triangle is 180 degrees or pi radians, right? So if you will combine all these angles, you will have n times pi. But you shouldn't really count all this, which is 360 degrees, which is 2 pi. So minus 2 pi. So sum of all the angles, inner angles of a polygon is always n minus 2 times pi. So if you will add them together, you will have an answer. That's sum of all the angles. Now, in one of the future lectures, I will use this result to calculate how many different areas I have. If I put all the diagonals, if I have some, how many different areas I have. Now, that's a little bit more practical problem, if you wish, because sum of all angles, I don't see it as a practical problem. But by how many different domains I will divide a polygon. If I will put all the diagonals, that seems to be really kind of a nice geometric problem. And I can solve it using this in one of the future lectures. Okay, that's basically it. So what I suggest you to do is read the notes for this lecture. Don't read the solution even if I put it there. There are some hints, maybe. You can use it or you don't use it up to you basically and try to solve the problems yourself. The whole thing actually makes sense only if you solve problems yourself. Now, whatever solutions you already have, they should really contribute to your repertoire, of your approaches, your techniques how to solve problems. And that's what's very important. Then, if you will master all these unorthodox problems, in real life, you will have the problems as well. Some problems which nobody else solved before you. And you will be prepared to think about these problems creatively and come up with your own solutions. That's the whole purpose of this course called Mass Plus and Problems. That's it. Thank you very much and good luck.