 Hi, I'm Zor. Welcome to Unisor Education. Today's topic is construction problems in geometry. It's a continuation of some other construction problems, much more elementary, which we did before. We covered something simple like construct a triangle by three sides or something like this. Well, today it will be just a little bit more difficult construction problems. However, it's just a continuation, it's a progress, and I do encourage actually you to first take a look at the nodes and all the construction problems which are presented. Think about them and then watch the video and I will try to do exactly the same thing. Now, I deliberately did not spend any time constructing all these things beforehand to prepare for the lecture. So, I will just think in real time how to do this. Construction problems which I will consider obviously have many different solutions and I will remind you can have your own problems without this. So, without further ado, let me just start. Let me construct a sum of two angles. Alright, let me maybe talk about this in more details if you wish. What does it mean to construct a sum of two angles? Well, what does it mean to construct an angle which is congruent to another angle, to a given angle? Well, in some more details it means the following. You have somewhere on the plane an angle which means it's two rays which have the same vertex which is the vertex of the angle. Now, obviously when the problem says construct let's say an angle congruent to this one, it's not really specifically telling where to construct it. So, I think to make the problem a little bit more precise, I think it should be given as well some other ray. Let's say this is APQ and this is B, let's say M. Now, having this angle on the plane and this ray now more precisely the construction problem to construct the angle congruent to a given one should be like this. If you have an angle and a ray somewhere on the plane construct another ray with the same origin as this one, let's call it BN, such that the angle NBM congruent to PAQ. So, the only thing which you really have to construct in this case is this other ray because this one should be given. Otherwise, the problem is not really specified in all the details. So, when I'm saying that I would like to construct... Now, by the way, before I continue with this particular construction problem about two angles, some of two angles, obviously this particular problem has two solutions because you can construct another ray, this one and prime let's say. And this angle will be congruent to this and congruent to this. I put a little arrow to specify the positive direction of measuring the angle counter the clockwise. So, in any case, even in this particular case when I specify exactly one and two rays, there are two solutions above and on one side and on another side of the given ray which gives basically the resulting angle congruent to the one which I mean. Alright, so basically it's important to specify all the details but in theory you understand that if I just say, okay, construct an angle congruent to a given one, you understand it's somewhere on another place on the plane and that somewhere should probably be given at least by the ray or something. Alright, so let's not go into all these details and just go straight to the construction problem. So you have one angle and you have another one. So what you have to construct an angle which is measured as a sum of these two angles. So angle P, A, Q plus angle M, B, N. Okay, question is how to do it? And again, let's assume if we go to really, you know, very scrupulous details, let's assume that there is some kind of a ray even to you, say CX. And you have to construct another ray, CY, in such a way that this particular angle is measured as a sum of these two angles. Actually, you know what? I made a very small mistake here. I should not really say P, A, Q. I should say Q, A, P since this is a positive direction of measuring the angles. So let's do it properly counterclockwise. So it's angle Q, A, P plus angle M, B, N. So, and this is an angle X, C, Y. So the purpose of this construction problem is to construct CY in such a way that the measure of this angle is equal to sum of these measures. Well, how to do it? Well, it's obvious. You know how to build an angle which is congruent to the given one. That was one of the previous lecture and so you built an angle first from this ray CX. You built one angle which is called, let's say, XCZ. So angle XCZ is congruent to angle Q, A, P. That's number one. Number two, using CZ as the new base ray, if you wish. Construct an angle Z, C, Y which is congruent to the second one, N, B, M. Don't pay attention. This angle doesn't really look much smaller actually than this one. That's my drawing. That's okay. So basically you see that this angle would be congruent to this. Now this angle would be congruent to this. And that's Y. By definition of the sum of two angles, the angle X, C, Y will be congruent to sum of these two angles. Or basically is the sum of these two angles. So that's how we build a sum of two angles. We are talking about measuring, obviously. Its measure of this new angle is equal to sum of these measures of the given angles. All right, so that's, I think I talked too much about this particular problem. It's a very simple one. By the way, if you forgot how to build an angle congruent to a given one, it's quite simple. You convert the angle into a triangle and then build a triangle which is congruent to this one here using three sides. One, two, and three. One, two, and three. All right, fine. Let's go on. We are wasting a lot of time on something which is extremely true. Next is a difference. Okay, if you have again one angle and another angle and you would like to build a difference between these two angles, which means a new angle, the measure of which is equal to a difference between measures of these two angles. Well, obviously, if you have an initial base ray, you build first an angle which is congruent to this one. And then using this ray, you build an angle congruent to this one. But instead of going to the same direction counterclockwise, you actually go counterclockwise. If you remember to build, to construct an angle congruent to a given one, you have two solutions. Remember, solution this and this. So basically you can, in this case, construct the second angle using the second solution if you wish. So whatever remains would be the difference. That's simple. And it's all based on the ability to build an angle congruent to a given one, which is done by constructing a triangle congruent to a given one. Given sum and difference between two angles, construct these angles. All right, this is just slightly more difficult and it's a combination of algebra and geometry. Let's talk about measures. Let's say you have one angle which measures x and another angle which measures y. So you are given some of these things and the difference of these things. What you have to find out is you have to build an angles x and y given angles a and b. Well, let's approach this algebraically and consider this as a system of equations which you have to resolve for x and y. Well, how to do it? Well, since both equations can be added together, you would have 2x equals a plus b from which x is equal to a plus b divided by 2. And if you subtract top minus bottom, you would have 2y equals a minus b from which y is equal to a minus b divided by 2. So, given a and b, you have to construct x and y using these formulas. Now, how to do it? Well, very simple. If you have two angles measured a and b, how to construct a plus b divided by 2? This is angle measures a and this is measures b. So first, you build some of these two angles and that was just the previous problem, one of two previous problems. So this will be something like this. And then divided by 2, what does it mean to divided by 2? Well, have an angle bicep. And again, you know how to do it from one of the previous lectures. So half of this is exactly what's necessary. So if this is a and b, this will be x. Now, if you have to build this one, begin a and then b would be somewhere here. So you go backwards, right? This is a and this is b. So whatever remains is a minus b. And again, you construct a bisector of this angle and whatever remains is your y solution. So that's how you build two different angles using their sum and their difference. So as you see, we have combined elements of algebra, which is how to solve a system, a very simple actual system of two equations with two variables, x and y. And the previous problems, which involve how to add two different angles or how to subtract one from another or how to divide an angle by two. Dividing an angle into four, eight and six congruent parts. All right. Now, you know how to divide it in two, right? So how to divide into four? Well, actually, it's the combination of two things. First, you divide it by two and then you divide it by two again. So if you have an angle, you divide it by two by drawing a bisector. Consider the half of this and then divide it by two again, another bisector. And that's how you actually do it even further. So this is one fourth, right, of the original angle. And if you build another bisector, you will get one eighths. And another bisector, you will get one sixteenths of the original angle, et cetera. So any power of two, two, four, eight, 16, 32, et cetera, can be achieved by sequentially dividing an angle by two. You divide the half by two and then the quarter by two and then the one eighths by two, et cetera, et cetera. And dividing by two is just drawing a bisector. So any power of two means just bisecting an angle more and more and more. By the way, some time ago, I was talking about certain construction problems which can be solved using the ruler and the compass. And one of the examples of the problem which cannot be solved and it's proven that it cannot be solved using the ruler and the compass is trisection of the angle. So if you want to divide any angle in three equal parts, this is not possible to do in using the tools of elementary geometry. However, however, what is possible to do is to build an angle which is as close as we want to one-third of a given angle. And this is based on the following thing. In algebra, you're actually supposed to learn something about infinite sequences. There is a sequence. So these are two, four, eight, sixteen, et cetera, powers of two, but the signs are interchanging. The minus plus, minus plus, et cetera. Now, this sequence, if continued to infinity, gives a sum equal to one-third, which means that if we will stop here, let's say, and just build this particular angle, then we will get, well, close to one-third. If we will continue this process longer, we will get closer to one-third. So the more time and efforts we spend, the closer to one-third we can get. Now, how to build this particular sequence of events? Well, that's simple, actually. You start from the angle. You divide it by two, and you get half of the angle, right? Then you subtract one-fourths, which means you have to divide it by two, and this is what remains. Then you have to add one-eighths, which means you have to divide one-quarter by two, and this would be one-eighths. Then you have to subtract one-sixteenth, so that would be this way, et cetera. So the more you continue adding and subtracting, adding and subtracting parts of the original angle, which we know how to do using the ruler and the compass, two, four, eight, et cetera, parts we can divide into. So using plus and minus, plus and minus, which basically, again, we have considered before in the beginning of this exercise, we can get as close as we want to one-third, but we will never, in any finite time, we will never achieve absolute precision of one-third. We will be only close to one-third. Construct a triangle by its two sides and angle opposite to a bigger side. Is it always a unique triangle? So you have two sides, let's say this and this, and this angle opposite to a bigger side. Well, how to construct this triangle? First of all, we start from the angle, since it's even, right? That's not exactly how it is. That can be a little bit more close to reality. Then we have a compass and we cut. This particular point. So this is A, B, C, and this is A prime. Now this is B prime. So we just have an angle equal to congruent to a given one. Then on one side of the angle, we basically measure the segment one side of this particular triangle. And then using this as a radius with this center, we just draw a circle. Well, now here what's important, a circle can actually cross in two points. Well, that's not exactly correct drawing. I'm sorry for this. But in theory, these are equal, and this is equal to the C. So we have C1, C prime, and C double prime here. So both triangles A prime, B prime, C prime, and A prime, B prime, C double prime are triangles which only one of them actually is congruent to this one. This one, because this angle will be equal. But this is a completely different triangle. However, you can build a triangle not always congruent to the given one using two sides and an angle. All right, now it actually depends on the lengths of this particular side. You see, if this is the biggest side, then what's important is that it will actually cross this line in the second point which is beyond the beginning of this angle. And this is not exactly the triangle which we want because the angle is not given. It's a supplementary angle to the given one. So if this is the biggest side in the triangle, then there's only one solution. This is not a solution because the triangle will be different. This triangle will be not exactly the one which we need because the angle will not be the given one. So if the angle opposite to the biggest side of the triangle is given, then there is one solution and only one. Now, what if it's not? What if it's another angle? What if this is this angle and these two sides? Well, here, again, we will have different problems, different solutions actually. First, you start from the angle and you, using the compass measure, your side which is given to you. And then, using this radius, you start circling. And now, as you see, we can have either two solutions or one solution or no solution depending on how big actually this particular side is. So not all the combinations of segment, segment and angle will give a solution or solution. Sometimes you can have no solutions at all if this is too short for this particular angle or it could be too long and then you will have two different solutions or it will be just equal in which case you will have only one solution. So that's why there is no such thing as a theorem about congruent triangles where two sides and an angle not between them participate. If it's one of the opposite to one of these sides' angle, then it all depends. Sometimes triangles can be congruent. Sometimes not. Sometimes actually, as I was saying, there are no solutions in this case. So that's why there is no such theorem as angle side-side in theory. There is no such thing. There is only like side-angle side or angle-side-angle, but there is no angle-side-side. And this is why, basically. If it's the bigger side, then there is a solution. If it's not a bigger side, then there is none. Or zero. I mean zero or one or the two solutions. Which means it cannot be congruent. Okay. Construct an isosceless triangle by base and one of two other sides. Okay, so we have a isosceless triangle. You have a base and one of the sides. Now, how to construct it? Well, very simply. Let's divide it by two. Can we have half as a leg and then this as a hypotenuse, and this is the right triangles which we can build. And that's why we can build half of our triangle using right angle, half of the base, and hypotenuse. And now I'll do exactly the same thing. That's how we do it. That's simple. Construct an isosceless triangle by base and one of two congruent angles. Okay. Same thing. Very simple. Reducing to right triangles. So if you have a base and one of two equal angles, how to do it? Again, using this right triangle approach. First, we do half a base here. Then using this ray, we built an angle congruent to the given one. And then basically another half will be on this side. So we built an angle congruent to this one. That's how it's reduced. Construct a isosceless by one of two congruent sides and an angle between congruent sides, the vertex. Okay. How to construct a triangle using one of the sides, congruent sides and this angle? Again, very simply by reducing the problem to right triangles. This is half the angle. So we basically have it, right? So what we do is we have the line first, then we built an angle which is congruent to half of this and then we cut this length on this side and then draw a perpendicular and continue with the same length on the other side. Construct a isosceless by one of two congruent sides and base angle to form the base. Okay. Again, very similar. So you have a isosceless triangle and you have side and angle. Well, divided in two parts. How to build this right triangle by hypogenous and an angle? Well, obviously, if we know that this is 90 degree minus this one, that would be easy. But if you don't know that the sum of two angles, two acute angles is equal to 90 degree, you have this angle. Oh, very simple. Sure. You start with this angle. Okay. You measure your hypotenuse and draw the perpendicular down and then do it simultaneously. That's it. Very easy. So as you see, I just hesitated for a second before I found the solution, which does not really depend on the fact that sum of two acute angles in the right triangle is equal to 90 degree. Because we did not really cover this in the lectures which precede this particular exercise, these particular problems. Okay. Fine. So that's it. Don't forget to look at the website unisor.com. It has many other interesting problems and for parents and supervisors, that's very convenient tool to supervise and watch the progress of your students. You can enroll your students into this or that particular part of this whole educational material. They can take exams. You can examine their scores and basically mark while pass or fail. And if it's fail, then they have to really do it again and again until they basically pass all the exams. All right, so good luck and thank you very much.