 Hi, I'm Zor. Welcome to a new Zor education. We continue solving different problems which are not really part of the standard course of school mathematics. These problems require certain creativity, both thinking basically, finding some new way which never was presented to you as part of the theoretical course. Now, obviously this is all based on the theory which was presented in another course. So this is called mass plus problems, but the prerequisite course on the same website, Unisor.com, is called mass for teens. This is more theoretical course, although there are many problems presented over there as well, but more to kind of enforce the theory. Now, the problems in this course are more of creative kind. Let's put it this way. And today we will talk about certain geometrical problems. The lecture is called Geometry 02. Now, this course is presented on Unisor.com website, but you might find it somewhere else like on YouTube, for example, or some kind of a search engine or whatever. I do recommend you to go to Unisor.com to watch the lecture. There are few reasons for this. Well, reason number one, don't worry, everything is absolutely free. There are no advertisements, no strings attached. Assigning on is optional if you don't want, don't do it. If you are basically studying by yourself for a supervised study, we do need this kind of a sign-on because we have to establish the relationship between supervisor or parent and the student. But for a self-study, you don't need it. Now, another reason is that every lecture which I'm presenting as a video also has a textual part, which is basically like a piece of a textbook dedicated to this particular lecture. So altogether, all this written material combined into one book might be a very good textbook for you as well. And there are exams on the website which you can take as many times as you want until you get the perfect score. And I do recommend you actually to solve it many times until you will get the perfect score from each exam. So there are reasons to use Unisor.com as your main source of information. And obviously there is a menu which drives every course in a logical sequence, so menus, submenus, etc. Okay, so let's solve these problems. Now, these problems are related to triangles. As far as triangles, I prefer to have certain standard of naming elements of the triangle. The standard which I'm just using, you can use this one or any other, but it just makes sense to have a standard. So the standard which I'm using is I will use uppercase, lesion letters A, B, and C as the vertices of a triangle. Now, opposite sides to the letters to vertices will be called the same letter but lower case like A, B, and C. Then there are certain elements like for instance medians. Now medians which are dividing the opposite side into equal parts. I'm usually using the letter M with subscript which corresponds to the side it goes to. The altitude would be H with the same subscript in this case A. And there is an angle bisector which I will use letter L with the same subscript. So, oh yes, there is a circumscribed circle which has radius R and inscribed circle which has radius lowercase R. So this is a standard which I will try to adhere to. I think it's important for clarity and also it's shorter if you are trying to explain for instance the problems. I will use these letters as the short description of the problem. Now, all these problems today, there are four of them, they are about construction. Construct triangle if you know something about this triangle. Now, in the regular course of geometry you have learned that there are three main theorems about congruency of triangles. You have three sides of one equal to three sides of another, they are congruent. If two sides and angle between them equal correspondingly to two sides and angle between them, then it's congruent. And the third one is the side and two angles which share this side as part of the angles are equal. So it's side-side-side, side-angle-side and angle-side-angle. Now, these three main theorems about congruency lead to three main problems. Construct triangles by knowing these elements. And these problems were considered in the main theoretical course which is a prerequisite for this one on this same website unizord.com which is called mass for teens. Now, in this case I will go a little bit further. So these are again construction problems but not these main typical ones. Some other elements are given but you still have to recreate the triangle. In this case the solution might not be the unique one so maybe you can knowing like certain sides maybe there are more than one triangle which you can create but we will not stop on this. We will just try to come up with some kind of a procedure to create some triangle which has the properties specified in the condition of the problem. So my first problem is you have side, side and median to the third side. So these are three elements of a triangle which are given two sides and a median to the third side. Well, you have to construct triangle knowing these three elements. How can we do it? Well, let's first draw the triangle. Okay, let's go A, B and C. Now you have A and B given sides and we have a median to the third side. So this is side C and this is MC. The question is how can we build a triangle knowing these elements? Now at this point I suggest you to pause the video and think about this yourself. Now as a hint I can just tell you that what's important in construction problems is to reduce the given problem to another one which has already been considered. For example, we know just as an example that we can build triangle knowing it's three sides but we don't know three sides of this triangle. But we do know something which can lead us to that. So here is additional construction which I will do right now which will lead us to reduce this problem to already known ones. Okay, so again try to think about this yourself, pause the video, whatever. Now here is the solution which I am suggesting. So let's draw a parallel line to this one and parallel line to this one. This is parallelogram. Now if I will continue this line it will go into this vertex of the parallelogram. So parallelogram as we know has diagonals which are crossed intersect in the middle. So this is divided in the middle and this is divided in the middle which means this is actually part of the median, right? Because median divides in between two equal parts, the opposite side. Now why is it better? Because this piece, let's call this letter P. So PD is actually congruent to CP because again the diagonal is divided by another diagonal into equal parts. So this is also MC. So what do we know about this triangle? We actually know all its three sides because this is also A because it's opposite side of parallelogram and CD is equal to 2MC. So if we know median we can double it. Well you know how to double it. You just have the same using the compass and put another one. So that will be double. Now I didn't mention it but all the construction problems in standard course of geometry are supposed to use the tools which are straight ruler to draw the line and compass to draw a circle. So using the compass you can just take this and put it here and using the ruler you can draw parallel lines. By the way how to draw parallel lines using compass and the ruler? That's actually part of the standard course and I can just basically relate you to the mass for teens course or any other course where it is explained how to construct parallel lines using the straight ruler and the compass. So I consider these to be as techniques which you know basically. So we can build triangle CAD using three sides. Well if you know that now everything else is busy. Then you divide CD into half basically you know this is the length which you can use and draw the line here and basically from here from C you draw the line parallel to this. So what we did is we did analysis and the solution. Analysis was basically drawing this, analyzing the whole thing and realizing that you can draw this particular triangle. Okay now the solution is draw this triangle using three sides A, B and double MC and then draw a parallel line here and line here through the middle point of CD. So that's the solution. Now analysis and solution. So let's move on. Next problem. Next problem is also similar kind of thing but slightly different. So what do we have? We have A, we have MA and we have MB which means analysis again. First analysis. This is A, B, C, this is A and we have median to the A so this is the middle point, this is the median. Okay so let's call it R and we have median to B. So this is B so median is this. So this is MA and this is MB. Now that's what we know. Let's call it Q. So again, this is analysis and what this analysis tells us. Well obviously it's kind of difficult to build this triangle if you know only this side and to median. You don't know this point obviously. So notice, however, we do know certain properties of medians. Again, by the way, addressed in the main course, Mass for Jeans when we were talking about triangles. Now the medians are crossing at a point which divides each median into two to one ratio. Again, if you don't remember it, go back to the previous course. But this is a known fact. It was proven in the regular course which I have Mass for Jeans and there are a few theorems about this but it was rigorously proven. And now I'm going to use it because QR is only one-third of MA. Now QB is two-thirds of MB and BR is one-half of A because this is a median, right? Which means we do know these three segments from which this particular triangle QBR is built. So we can construct it using three sides. By the way, if you know MA, how can I get one-third of MA? Again, that was explained in the main course how to divide a segment into any number of parts. But basically, in a very simple case, if you would like to divide this particular segment into three parts, you have any other segment repeated three times here? Connect the end and draw the parallel lines. So if these are equal, these will be equal. It's a theorem and it was proven, rigorously proven, in the main course, Mass for Jeans and then just using it. So I assume that we know how to take one-third of a segment or two-thirds of a segment. Now knowing this, I can actually basically finish my analysis. I know what to do. I know what I do as a solution. I take one-third of this, two-thirds of this, one-half of this and build this particular triangle. Having built this particular triangle, everything else is easy because you just continue this line and have the same segment here. So now I have my point C. You continue this line and have another two segments and that's how you get point A. And the problem is solved. That's two points of a triangle. Okay, next. Again, my purpose is to present the problems but also to kind of encourage you to think about these problems yourself. So if I'm explaining the solution, it really means that I would suggest you to post the lecture right before I'm starting presenting the solution and try to do it yourself. Thinking is the most important purpose of the whole course. You have to think about certain things yourself. Don't just take information which I'm giving it to you. Try to do it yourself first. And then, whether you succeeded or not, maybe you will come up with a different solution. Fine, great. And by the way, if you can send me another solution to any of these problems, I'll be more than happy to publish it on the website with reference to you. Okay, the third problem. The third problem is really a little more interesting. M A, M B and M C. How to construct triangle knowing only its three medians. See, it's a very non-typical problem. And again, it has a solution. I suggest you to pause again right now and think about this. And here is what I suggest. So let's try again to analyze the situation. Here is my triangle, A, B, C. And let's put three medians. One, two and three. Medians are intercepted in one point. All three medians, again, that was proven. So what we will do is, we will do the same thing as before. The parallel here, parallel here. So we have basically two congruent triangles. And I will continue this. And I will have two medians here and here. So it's two basically congruent triangles divided by A, B. They are basically, you can turn one into another. And you will have this picture. Lines parallel. I mean everything in this particular case, in this triangle is similar to whatever is here. Okay. Let's call this point P and this point Q. And what I suggest is to consider triangle Q, P, B. What do we know about this? Okay. This QP is one and two and each one is a third of a median MC. So QP is equal to two thirds of MC. This is MC. This is two thirds. This is one third. And this is another one third. So it's two thirds. Similarly, P, P, B is two thirds. P, B is two thirds of this median, which is the same as this one, which is MA of MA. And finally QB is two thirds. QB is two thirds of this median, which is MB. So I'm using the fact that medians are intersecting in one point, that they are cutting the median in a ratio of one to two. So it's one third here and two thirds there. And then everything else is just obviously parallel. You can definitely prove this to a very, very tiny detail. I'm not doing it because I'm assuming that certain things are obvious. So if you draw these parallel lines and you have this median and this median, then the lines will be parallel and stuff like this. These are all details which are kind of obvious, easy to prove, but I'm just skipping them because what I am actually interested in to analyze the whole solution first by doing this additional construction. Now when I know this, I can build triangle QPB using three sides. Knowing this, well, we just continue. So how can I continue this? Well, easy. Let's continue this PB for instance by half of it to another point so that's how we can get this point. Let's continue this line for another half of it so it will be this point. Let's continue this to the same two thirds and we will get this point and then to another point of this then we draw these lines and we will have point A. So that's basically a solution. What's interesting is really to follow this particular solution to prove any point. So whatever I have just glanced through as an obvious thing, like for example, this will be a triangle fully congruent to this one. I mean, obviously it's two parallel lines, parallelogram, etc. But it all goes to certain, maybe small, maybe simple, but proofs which basically you can follow through. And if you can, that will be very, very educational. Okay. Now the next problem is it's related to altitudes. So what we have is we have a side and two altitudes to other sides. So this is triangle ABC. You have this side and you have two altitudes to other sides. This is HB and this is HC. This is B, this is C, right? So what's given is this and two altitudes. Altitudes. Okay. How can I construct the triangle using these three elements? Okay. Here it is. Look, this is perpendicular, right? This is altitude, so this is the right angle and this is the right angle. So if I will use C as a center in this case HC as a radius and draw a circle obviously this line would be tangential at this point because it's perpendicular to a radius and the line which is perpendicular to a radius at the end would be tangential. Again, something which has been considered in the regular course called mass proteins and proven so you can just use it. Now, what does it mean? It means that if I will have C as a center so I do have the segment A as a given, right? So we can just take a line and put this segment on the line and that's how we have points BC. Our unknown is point A. Okay, so I draw a circle using the HC as a radius. Now I do have the point B and all I have to do actually is to draw a tangential line to a circle. How to do this? Well, again, that's considered as a basic construction problem in the regular course but I can actually tell you since this is the right angle this would be three points on a semicircle. This is a diameter so I have to basically draw another circle using half of this BC, half of A and in the middle have a center and draw a circle and then where these two circles are crossing each other that would be this point. Now, absolutely similarly, this line would be tangential to this circle with the center B and radius HB and from this point I can draw a tangential line to this one. How? Well, by having this again as a circle well, it's actually the same circle and it's just another point of intersection. It's a circle which has a center in the middle of BC and the radius is equal to half of A. So it's the same circle as in the first case and it will cross this circle in some other point and that would be my this point. What to do next? Well, from C you go through these two points and get one line, from B you get through this point another line and their crossing would be A. So that's a solution to this third problem when you have side and two altitudes. These are four not very difficult problems but what's very important is to do certain things yourself or at least to spend some time thinking about this. So what I suggest you to do is all these problems are explained in the textual part of the course so you go to Unisor.com you choose math plus and problems course you go to geometry this is geometry 01 and in the textual description I actually give only the problem I do not put solution as I do in many other cases. So since there is no solution try to whatever you remember from the lecture or come up with your own solution but try to solve all these four problems again. It will be very very beneficial for you. If you have some time just thinking about this if you remember what I was talking about great if you don't come up with yourself but again you have to analyze the problem first and reduce this problem to constructing something which like in this particular case constructing one circle, another circle, a third circle and these are the steps you have to take to come up with the final solution. Do it yourself, very very important. Ok, that's it for today, thank you very much and good luck.