 Hi, I'm Zor. Welcome to a user education. We continue solving problems. These problems are combined into a course which I call Math Plus and problems presented on Unisor.com. So on Unisor.com you can find this course and prerequisite one, which is Math 13s. Now I consider solving problems to be much more important for development of students' mind than studying theory. Well, the problem is you cannot solve problems if you don't know the theory. So first you have to study the theory and there are some standard problems within the theoretical material which I presented in the prerequisite course Math 14s. Now these problems which I am presenting right now in this course are not standard. They are not exactly to check how well you know the theory. These problems are to basically force you to think. And that's the most important part of the education in school, to teach you to think. I believe that studying theoretical knowledge which is a prerequisite for this one is important, but practically you probably will not use almost like any of these things which you have studied in your real practical life. When something is going in college, etc., that's the real skills which you will learn to use in practice. Whatever you have in high school, for example, unlikely. However, without this you cannot really go into solving problems. And without solving problems which are purely artificial, like for instance today, we will go through geometrical problems. But without solving these problems, most likely you will not be well equipped to solve the real practical problems. So this is a training grounds right now. The problems, purely artificial problems which are presented in this particular course Math Plus and problems, they are training grounds for your mind to teach you to think. And that's what's very important. Now today we will have three geometrical problems. They are dedicated to construction, calculation and proof. All these qualities are extremely important in solving real practical problems in your future life. Okay, so let's go to problems. Problem number one. Okay, you have to construct quadrilateral A, B, C, D, knowing all four sides and angle between A, B and C, B. So this angle. If you will continue these, you will have an angle. So four sides and angle. Now how to construct this quadrilateral? Well, again theoretical material presented in Math 14's course, as far as construction problems, there were some simple construction problems, like for example, constructed triangle knowing three sides, something like this. Now the quadrilateral is something new. It's not really directed towards practical theorem which you have learned in a regular math course. There are no theorems. This particular problem is based upon. However, what's important is to reduce this problem into something which is related to some theoretical material which you have. So as a hint, that's basically a hint. You have to do some kind of additional construction here, which will allow you to build, to construct a different element, something like a triangle, which you know how to, how to construct if you know three elements, like two sides and angle between them or three sides or a side and two angles. So this is something which you know. So you have to reduce this problem into something which is known from the theoretical material. So again, this is a hint. This is where you probably could just pause the video if you're looking, watching the video and think about how to do it. Now try to think about this for extended period of time because it's really kind of a search. This is exactly what's necessary to train your mind. You are searching for a solution. So what kind of other geometrical figure you can really construct using these three sides, four sides and angle between A, B and C, D? Now I'm suggesting my solution, but I'm sure there are some others. But look, if you know the size A, B and C, D, well, you know others as well, but right now you know these two sides and angle between them. So that kind of reminds you, okay, two sides and angle, that reminds you a triangle which you can build using these two sides and angle between them. So let's just reduce this quadrilateral problem, construction of quadrilateral to a construction of triangle using these things. But where is this triangle? Well, very simply. You can just construct from B, build a parallelogram. I did not intend this to be parallel to this one. Let's consider it this way. Okay, so you construct a parallelogram. So this would be, let's say, point P and the D point would be here. Okay, so A, B, C, D. You find point P so that B, P, D, C is parallelogram. I mean, how to do it? Well, very simply, you just build parallel line here and you know the line C, D. So you put it here and that's your point P. Now, let's consider triangle A, B, P. B, P is the same as this one. A, B, you know and angle between them. That's the angle between them. Since these two are parallel, angle between A, B and B, P is the same as A, B between A, B and C, D. So triangle A, B, P, you can construct using two sides. This is given. This is equal to this one, which is also given and angle between them. Okay, so let's start from this. You have A, B and you have point P. You build this triangle knowing this side, knowing this side and knowing angle between them. So what's next? Well, if you know this, then how to build the the rest of this quadrilateral. Okay, here is how we can do it. You do know A, D and you do know P, D, which is equal to BC. So from this point, you have one circle radius AG and from this point, you have another circle with a radius PG is equal to BC and that's your point D. So you have A, you have D, now you have two points C. Again, you know this side, so it's this circle and you know this side BC, you have this circle and that's your point C. And that's your quadrilateral. Okay, so what did we do? We have reduced more complex problem to a smaller problem, usually some kind of a triangle, which you know how to build using three elements. And then from this, you grow into an entire quadrilateral. So that's the first problem. We have three problems. This is construction problem. Next problem is also geometrical but calculation. Here is the given. Now, there is a circle and there is a regular polygon inscribed into this circle. I will use square as a regular polygon. It can be five-sided, six-sided, ten-sided, twenty-five-sided, polygon. As long as this is regular, which means all sides are equal, all angles are equal, everything is regular. So this is given. Now, question is, but let's say the radius is R. Okay, so question is, let's take any point B here and let's call A1, A2, A3, A4. What I need is P A1 square plus P A2 square plus P A3 square plus P A4 square. Some of squares, the distances between P and each vertex of this regular polygon. Some of squares. Now, I have two different solutions. One in case the polygon has even number of vertices, another a general solution. So I will present both of them. And again, this is the point where you can actually pause the video if you watch it and think about this yourself. Now, for even number of vertices, the situation is kind of simple and you can do it geometrically. Look, if it's even number of vertices, then you can always have the line which connects two opposite vertices of this polygon and it goes through a center. So it will be a diameter. Obviously, it's only in case this is even number of points. Now, this is four. Now, what if it's six? Well, if it's six, it would be like here, here, here, here and here. And again, the opposite would be always going through a diameter. Now, in this case, let's take this square plus this square. Two opposite sides, square of this plus square of this. But this is a right triangle because this angle is supported by diameter. Which means these are cateches of the right triangle. So pA square plus A1 square plus pA3 square would be equal to square of a diameter, which is four R square, right? Now, this is these two opposite. Now, I can do these two opposite sides, A2 and A4. And do again, right triangle. Some of squares of these will be, again, square of a diameter. So pA2 square plus pA4 square is also equal to R square. If I will sum them together, I will have whatever I need, which is 8R square. Now, what if I have N vertices? And N is even. Well, that means that this number is N divided by 2 plus 1. This is A1. This is A2 and 3, etc., AN over 2. And this is AN over 2 plus 1, etc. and this would be AN. So I can always combine these opposite sides and say that the sum of squares would be equal to 4R square, right? And how many of these diameter will be? Well, it will be N over 2, right? Because each diameter connects two points, two vertices. There are N vertices. So if we divide by 2, it will be N over 2. So the sum will be 2NR square. Now, this is very easy solution. And it looks like the answer is 2NR square. That's the sum of squares of distances from any point, any point, by the way, on the circle to all vertices. By the way, even if point P coincides with one of the vertices, it will still be exactly the same. You can check it out. Okay, now, what if it's not even? Then it doesn't look exactly as this is, right? So instead, we offer a general solution. So again, let's start with the circle. And let's say we have 5 vertices. A1, A2, A3, A4, and A5. And point P. Now, I will use vector algebra. I will call this vector A1. This vector to my point P will be lowercase P. This will be A2. This will be A3. This will be A4 and A5. Vector. Now, what is square of a distance between P and P and AI? Well, I can always consider vector PAI, this vector, and have a square of this vector. Now, square of this vector is basically scalar product by itself, right? That's what square of the length is. Now, from vector algebra, you know that if you will multiply vector by itself, well, actually, A times B scalar product is actually by... It's a scalar. It's lengths of the A, lengths of the B, and cosine of angle between them, right? But if it's the same angle, I mean, if it's the same vector, angle is equal to 0, cosine of 0 is 1. So basically, it's a multiplication of both lengths, which is the same length because it's the same vector, so it's A2. So if B is equal to A, then this is A2. So this is known fact. Again, if there are any problems with vector algebra, I can't help you here because this is part of the course which I consider to be a prerequisite, so you have to know vector algebra. All right, so you know what is vector, you know what is sum or difference between two vectors, and you know what is scalar product. So P, let me take another. So P, A, I, vector. What is this vector from a standpoint of central vectors? Well, from P to A, I, what is it? Well, that's A, I, vector minus P vector, right? Because P plus P, A, Y, P, A, I would be A1. So this is for any I. I is vertex number. Okay, so square of this would be square of this, which means scalar product between A, I, minus P was itself. These are all vectors. Equals to A, I square minus 2P times A, I plus P square, right? Well, I don't need the parenthesis. So this is one of them. Now what we need is we need some of these for all I, from 1 to n where n is number of vertices. So this is sigma, this is sigma, and this is sigma. I is equal to 1 to n. But look at this. What is A, I square? A, I square is the length of the vector A, I, which is radius of the circle. So this is radius, square. Same thing with P. P square is radius. So what we have is r square plus r square. And that's n times, so it's 2n r square minus sigma P, A, I. I from 1 to n. So this is one component. This is another component. Now what is this particular component? Let's go back here. Sigma P, A, I is equal to these are two vectors. Well, obviously I can take P out of the sigma and put P scalar product with sigma A, I. Okay. Now what is sigma A, I? Sum of these vectors as vectors. Now it's vectors now. Now it's important. Well, consider this is a point and these are forces. And they are actually going into n different directions, symmetrically. I mean, you feel the result must be 0, right? This is supposed to be 0. So it's P times 0 is equal to 0. Now, how can you prove it? If you have a regular polygon and you would like actually to have all these vectors into all the different vertices, why is it 0? Sum of these vectors is 0. Very simply, if it's not, let's shift the whole picture by this angle. Every vector, A, I, will go into next A, I, which means it will take exactly the same position. Which means it will be exactly the same composition of vectors. And the result should be the same. But if this is not equal, if sum is not equal to 0, whenever we will turn it, this vector will also turn by some angle. So this vector will be different. So the sum cannot be non-zero because if you will shift, it's a symmetry. If you will shift by the angle equals to this one, this sum will also shift. And it should be exactly the same because all vectors are turning into themselves. That's it. So it's 0. So the whole thing, which is remaining, is just 2n r square, exactly the same as we had with even number of vertices purely geometrically. And that's the same result. Okay. Now, the last problem is also kind of the same with a shift. Okay, if you have an equilateral triangle, we will build another one. A, B, C, D. So it's two triangles, equilateral. And this is a common base for both of them. E. So I will connect this and take the middle of it. I will connect this and take the middle of it. So let's say it's M and N. Now I consider this triangle, C, C, M, N. You have to prove this triangle is also equilateral. Now, again, pause the video. Think about this as long as you want. And I will give you the solution. Solution is actually very easy. Consider triangle A, E, C. And consider triangle B, C, D. If you will turn A, C, E by 60 degree, this is equilateral triangle, right? Now, C, E would take place of C, G, right? A, C will take the place of B, C. Because it's also 60 degree. So triangle A, E, C will be turned into B, C, D. Middle point of A, E would go to middle point of B, C. Obviously, because it's exactly the same length. So this point will go to this, which means these two sides C, M is equal to C, M. And the angle between them would be also 60 degree because we have turned by 60 degree. So we have two equal sides and 60 degree in between. So it's equilateral triangle. Very simple. So we have three different problems here. One was construction. And you had to really think about how to have some kind of other element, like a triangle it was, which you could build. And from there you grow into big quadrilateral. The second one was calculation. Maybe we were using either geometrical calculation for n-sided polygon where n is even number, or general solution based on vector algebra. And the third one was just the proof. And we have proven it by moving, basically, by shifting, by turning, rotating, whatever picture. And obviously this is one of the methods to prove something. So whenever you're saying that two geometrical figures are congruent to each other, I will use the word congruent rather than equal. Congrant means you can move one into another without changing the lengths of the sides or angles between the sides of one figure into another. So these are three different kind of ways you can structure your problem. And again, the most important is for you to think about these problems before you're watching the solutions. Now, even if you did watch the solution, now it's a good time to basically repeat everything. Look into the textual part of this lecture and textual part you can go into Unisor.com course called Mass Blast Proteins, choose geometry, and this is geometry 05. Now, I put only the condition of the problem and sometimes a hint, but not the solution which I'm presenting here. So it's a good opportunity for you to think about solution again. You remember probably whatever I was talking about, but it's a good time to really repeat it again. And that I suggest you to do. Okay, that's it for today. Thank you very much and good luck.