 Hi, I'm Zor. Welcome to Unizor Education. Today's lecture will be about problems, solving problems. The problems are in the notes for this lecture as well as in the notes of a few lectures which will be in the future. I think I'm planning for eight different lectures completely dedicated to solving problems. I would definitely encourage you to solve these problems yourself. Try it as much as you can. And only if you feel completely comfortable with either your solution or with fully understanding of what the problem is all about and you've thought about this, then go to the lectures and basically listen to what I suggest as solutions. My solutions are not necessarily the only ones. They might be better or worse than your solutions. But anyway, it's something as a guiding line, whatever you can use. So, without further ado, let's just go through problems one by one. Okay, construct triangle by two sides and the median to the third side. Okay, that can be done quite quickly because I remember this problem. A, B, C, you have a triangle, you have a median. So you need to build a triangle by knowing AB, BC, and BM, which is median. Well, the way how I suggest you to do is continue the median by the same length as it is, get point D. Now, let's consider AB, D. What do you know about this triangle? Well, you know this side, you know BG as well because it's double the median. That's how to build it. Now, my statement is that A, G, and BC are actually congruent and the whole thing ABCG is a parallelogram. Now, how to prove it? Well, let's fight out this. Consider BMC and AMC triangles. Now, they have vertical angles. Now, these two are congruent to each other. That's how we constructed actually the point G. We continued BM by the same length and these two are congruent because M is the midpoint of AC. So BMC and AMG are congruent by side, side, angle, angle, and another side, another side. That's what makes BC congruent to AB. So now, ABD is a triangle which we know because it has a known side, a known side equal to this one, and a known side which is double the median. So we know all three sides. So we can build ABG and knowing ABG, having built it, how can we get the point C? Well, first, let's get M which is midpoint of BG. And then continue from AM to the same distance and that's how we get the point C. So here is our ABC. So we first analyzed the problem using some additional geometrical construction. We found that there is some element which we can start actually constructing because we know all about it. And from this element, which is ABG triangle, we expanded the picture to whatever we want to build. That's it. That's easy. I wish all my problems are as easy. Construct a quadrangle by all four sides in the median segment connecting midpoints of opposite sides. Okay, so we have a quadrangle. We know all four sides of it and we know the median, which connects opposite sides. Okay, all right, how can we solve this? Well, I think we can do it this way. We will take the AB and parallel shift so that B actually slides towards M and then A will slide towards D. Same thing here, C, D, we will slide again parallel so the C slides to the M and D will slide to and let's consider PN and NQ. Would be nice if it's a straight line, right? PNQ, but is it? Well, let's just think about it. Let's consider triangle APN. I bet it's equal to NBQ. Now, why? Well, since N is a midpoint, AM and NDR come great. Since AP is parallel and equal to BM and DQ is parallel and equal to CM and these guys are congruent between themselves because M is a midpoint, so that makes this AP congruent to DQ and parallel because again both are parallel to the same line. So they are parallel between themselves and consider this AG as a transversal that makes these two angles alternate interior, right? Two parallel lines and transversal these are alternate interior angles, which means they're equal. And from this APN and NGQ are congruent triangles these angles because we had an axiom of side, side, angle, angle, and another side, another side. So that makes these angles congruent to each other and since AD, AND is a straight line and these angles are congruent to each other that makes PQ a straight line. Not only that, we also have that the distance from P to N is exactly the same as the distance from N to Q because these are sides in the congruent triangles, APN and NGQ which makes PN a median in PMQ triangle. Since N divides PQ into equal parts PQ is a straight line we have now established because these angles are vertical, so PMQ is a triangle where PM is, sorry, where MM is a median. Now, what else do we know about this triangle? Well, we know that this side and this side are congruent because this is parallelogram, right? So since we know AB, it's one of the four sides that are given without PM. Same thing here. Since CD is equal in parallel to MQ this is a parallelogram, MCGQ by construction. So knowing CD without MQ. So PMQ we have side, side, and median in between which is exactly the previous problem which we have just solved. So now that's the end of the analysis of the problem. We have come up with an element of this picture which we can build right now which is PMQ. Alright, so now going backwards, real construction. So knowing PMQ triangle, we built it. Okay, it's M, well, divided by half, that's point M. Alright, now, look at this, NDQ. What do we know about this triangle? MQ we already know, here it is. ND is half of the side AD which is given, so we know it. NDQ equal to MC which is half of the BC side which is also known. So in this triangle we know all three sides. So having MQ and having these two sides we basically, you know, built the point D. And similarly on this side we can just continue and double the lengths, we get point A. And then parallel to this, and parallel to this we get B, and from B we get parallel to MQ and parallel to DQ, and we have a point C. So A, B, C, D. This is our, it doesn't look exactly like this one because I'm not a good artist. But anyway, you get the point. So knowing the beginning basically from which we can start then we build the rest of the picture. Well, let me just emphasize. First we did analysis of the problem. We built a couple of extra lines and analyzed what exactly we know in this new object which we have built. And knowing that we can build it from a different perspective we built that part which we know and then expand it to whatever we need. Constructed trapezoid by one of its interior angles two diagonals and a median. Trapezoid, which means this is parallel to this, we need an angle, two diagonals and a median. It will be more convenient for the drawing to have angle on this side. All right, so what should we do? Well, that's actually quite easy. Here it is, A, B, C, D. So what we do is we parallel shift B, C, parallel to itself so C is sliding towards A. This will be B prime. Now let's think about a triangle B prime, B, G. What do we know about it? Well, since B, C is parallel shifted to A, B prime so these are equal and parallel to each other. These are equal and parallel to each other. Which means B, B prime is equal to diagonal A, C, which is known. B, D is another diagonal which is also known. So what's left? B prime, D. Now what is B prime, D? B prime, D is sum of two bases of a trapezoid because this is exactly the same as this. Which means B prime, D is sum of two bases. Now we know the median, but you remember the theorem about the median of a trapezoid, that it's equal to half of the sum of two bases. Now this is sum of two bases. How can we get it? If we know the median, just double the size of the median so you'll get B prime, D. So B prime, B, D is a triangle where we know all three elements. B prime, D is equal to diagonal A, C. B, D is another diagonal which is known and B prime, D is double median which is also known. So we know all three elements of this triangle and that's how we can build it. So let's build B prime, B, D. So we got this triangle, but we have this angle. So we built this angle from here and here is our point C. Angle is also given by condition of this particular problem. So what's left? A. But we have A, C as a diagonal, right? So it's parallel to B, D prime. So we just draw a parallel. C, A would be parallel to B, D prime. That's our point A. So we have A, B, C and D. Here's our trapezoid. And again, extra construction, this line, analysis. We found out what we know about this particular picture of what exactly can we construct. And then we start from it, B, D prime, D and expand it to the whole picture. To come up with additional construction, additional lines, circles, whatever else is the most important part of these problems. And that's what these problems are all about. They're supposed to, basically, they're supposed to develop creativity. That's the most important problem. Okay, next. Construct a quadrangle by three sides and two interior angles adjacent to one of the known sides. Okay, so we have a quadrangle. We know three sides and two interior angles. Let's say this and either the one which known sides, one of them is known. Another angle can be either made between the known sides or one of the unknown sides. Now, let's consider the first case. Well, in this case, that's actually very simple. You don't really need anything extra. You have this segment anywhere. You have on one end of this segment built one angle, on another built another angle. You cut the lengths of these two segments, whatever you have, and connect them. That's how you build it. So this is not an interesting case. What's a little bit more interesting, if it's not this angle is given to you as the second one, but let's say this one. Well, this is actually not much different. What we do is the following. First you build this angle and you cut the proper lengths. Then you build this angle without, you don't know where to stop. But you can have a circle from here using this radius a, b, c, d, a, b, d. From d as a center, using cd as a radius, you can just draw a circle and wherever it cuts into this ray, these are your points. So this is one solution, but maybe another solution, c' is also three lines and two angles. So it may be more than one solution, maybe two solutions, but if this is too short actually, if this segment cd is too short, there might be no solution at all. It doesn't cross this line at all. Or maybe exactly one solution if it touches it in one point. Okay, that's simple. Construct a trapezoid with two bases and two legs. Okay, so we have a trapezoid. We have two bases and two legs, all four sides. And we know which one is which. All right. I think that's easy to write. So have bc parallel shifted so the c is sliding towards d. Now bc and ag are parallel to each other. So basically b' would be here and bc, db' is parallelogram. Because these are parallel, because it's trapezoid, these are parallel because that's how we shift it. So these are four parallel, usually parallel lines, two pairs. Now, but let's consider abdb, abdb' abdb' What do we know about this? We know this side, because it's one of the sides. We know bb' because it's equal to a side cd. That's a leg. Now, what do we know about ab'? Well, ab' obviously is the difference between ag, which is the lower base, and b'g, which is equal to bc, which is upper base. So ab' is the difference between two bases, which means you know it as well. So we can build abdb' knowing two sides and three sides, actually. This side is given, this side is given, and abb' is calculated as a difference between two bases. Well, from here it's easy. You continue this, and since you have both bases, you put bigger base here and smaller base here, and that's your trapezoid. What's the additional construction here? Just draw a parallel line here, shift bc parallel to itself. Okay, construct a quadrangle by its four sides, provided one diagonal bisects one angle. It's just connected to, uh-huh. So we have a quadrangle, we have a quadrangle, abcg. What we know is, we know all four sides, and we know the fact that this diagonal bisects an angle. So these are two halves. So now the question is how to build such a quadrangle just knowing that the diagonal bisects the angle. Okay, here's how. Consider bd as an axis of symmetry, and since this is axis of symmetry, then dc, if it's symmetrically reflected relative to bd, and since these two angles are congruent to each other, it will coincide with ad. The point c would be somewhere here, c'. Now, similarly, a would also be on this line somewhere here, a little higher, here, a' I'm really a bad archist, something like this. Sorry about this, we will make this a slightly better. So our quadrangle is abcg. Now, if we reflect the whole picture relative to bd, since these two congruent angles, my image of c will lie on ad, and image of a will lie on cd. But why is it true? I mean, it's kind of obvious, because these are perpendicular, right? So if these angles are equal, and these are equal to each other, and this is common, so these triangles are equal, so that's what makes everything equal. All right, so that's fine. So we don't really need to prove this. This is obvious. But now, what do we know about pieces of this picture which we can't really build? How about abc'? Ab is a known side of the quadrangle. We know all four of them. bc' is equal to bc, which is also known. How about ac'? Well, ac' is obviously equal to ad, which is known, minus c'g, which is equal to cg, which is also known. So we know ac' has a difference between two other sides. So the difference between these two sides, which make an angle from which the diagonal actually sticks out, this difference is the third side of this triangle. So we know abc' by all three sides, and we can start building from it. Since we know ag, it's one of the sides, we put the d-point here, right? Now, since we know that bg is bisecting an angle, so we built an angle equal to this one. And knowing cg, we get the c-point. That's it. And to start. So, again, to summarize what was important here, since this is bisector, it's important to reflect the whole picture relative to this diagonal as an axis of symmetry, and then find that the triangle abc' is basically known by all three sides to us in different building. All right. A billiard table has a rectangular shape. There are two bowls in some position in the direction of the movement of one bowl in such a way that reflecting from all four sides, it hits another one. All right, so let's say you have a rectangular billiard table. Everybody knows how to play billiard table, right? So, we have a bowl somewhere. We have to hit it in one particular direction, so it reflects from this particular side, this particular side, this particular side, this particular side, and this particular side, and hits this bowl. So our purpose is to find how to hit it so after reflecting from all different sides, it will hit the bowl which we need. All right, since I definitely know this problem, I knew it long, long time ago, I'll go straight to the solution without actually much discussing. There's nothing to discuss here, actually. But let's think about it this way. Let's reflect the whole picture relative to this. This is A and this is B. So B would be here. Now, you know the laws of reflecting of the bowl. These two angles are the same. Now, since B prime and B are reflection of each other relative to this axis, that means this angle is also the same. Which means what? That these two angles are vertical. And since this is a straight line, this is a straight line. Okay, that's interesting, right? Now, we have, from here, we have to actually make only one, two, three reflections and hit the bowl B prime. If my purpose is to hit the bowl B prime, I have to only come up with three reflections, not four. That's easier, right? So my first step is, okay, let's reflect from B to B prime relative to this particular side. And now, we have to solve the problem of the same reflection, but only with three reflections. What's next? Well, obviously, it would be... You know what? I have to really make it smaller, otherwise I will not be able to make all the reflections. I hope it will be visible. So, from here, I hit this way, this way, this way, this way, this way. Okay. This is my A, this is my B. Now, I reflect it first, the whole billiard table relative to this side and got point B prime. Let's use B1 instead of B prime because we will have B2 and B3, etc. B1. Now, you see, this is a straight line now. Let's reflect the whole picture relative to this. Now, what happens? This line, this is B2. B2, not B2, B2 is symmetrical to B1 relative to this line, which makes this a straight line since these two angles are equal to each other. These are equal to each other because it's symmetrical. That's what makes this straight line. Now, from A, I have to do only one, two reflections to get to B2, right? Okay, next, reflect relative to this guy. B2 will be somewhere here, which is B3 and let me assure you that from A to B3 it's important to just have one reflection. Finally, the whole picture is reflected one more time. Now, B3 will become B4 and this is the solution. This is analysis, basically. The solution is reflect once, twice, twice, and the fourth time. And wherever you've got the fourth, after the four reflections, wherever you've got the B4, go from A to B4 straight and that's what makes four lines reflected, four segments, actually, which constitute the way how the ball is moving. It will hit exactly the point B which you have here. Well, that's for those who like to play billiard according to certain mathematical principles, which nobody does, of course. Anyway, this is the end of the first session about the problems. There will be other sessions. Don't forget Unizord.com is the website where everything is contained. Parents and supervisors, please pay attention to the ability to control the educational process of your students and children by enrolling them into certain subjects and basically checking the score and their exams and making a decision about pass or fail because if somebody doesn't really have good results on the exam, he can always go back to the lectures, do it again and repeat the exam again. You can repeat it as many times as you want until you will get it right. And again, these problems are extremely important. So try to go through all of them yourself in all the lectures wherever it says problem number one, problem two, etc. Each one contains a set of problems which are very, very useful. Thank you very much and good luck.