 Hi, I'm Zor. Welcome to UniZorification. Today we will talk about construction problems in geometry, primarily related to triangles. I have here just for reminding myself what kind of construction problems I would like to consider today. It will be a set of five or six of them, six actually. And I will basically try to explain how it's supposed to be done. In general, construction problems in geometry are very, very useful. It kind of confirms that you know the theory, but at the same time you know how to solve certain problems. In this case, how to construct certain geometrical object like a triangle. So that's why in the course of these lectures in geometry, I think I will pay a lot of attention to different construction problems. Alright, so a few construction problems which are very, very simple and they are based on whatever the beginning of the theory of triangles actually is. The very fundamental theorems about equal triangles are at the foundation of these construction problems. So, problem number one, construct a triangle by side, angle and side. Angle is in between these two sides. Well, we all know that there is an axiom which says that if two different triangles have the same, have congruent pairs of sides and angle between them, then the triangles themselves are congruent. Well, what it means, it means that we can uniquely construct a triangle by knowing only two sides and an angle between them. Because any other triangle whatever else we can construct using the same three elements, two sides and angle between them will be congruent to any other. Triangle, which means again that construction problem is valid actually. Alright, so how to do it? So let's say you have three elements, two segments and an angle between them. Using these three elements, we have to construct a triangle. Alright, well, what we can do actually is the following and it's very, very simple. If we have this angle, we can take in the compass the distance between these two points on the first segment and basically mark it here. And then take another side and mark another side here. So this segment is equal to this one. It's congruent to different segments are congruent and this is congruent to this one. So now we can just connect these two points together using the ruler, let's say. And here we have a triangle. It has an angle given to us and it has one side which has an equal length since we are using the compass, which means it's congruent to the first segment. And another side, which is congruent to another segment, which means we have actually constructed a triangle by these three elements. Two segments which constitute their sides and the angle in between. Basically that's it, very simple problem. And again, considering called triangles with the same congruent pairs of sides and angle between them, we have basically constructed any triangle which satisfies these three elements given to us. Alright, let's go to the next problem. Next problem is constructed triangle by three sides. Again, in this case it's not an axon, it's a theorem about three sides. If one triangle has three sides congruent to three sides of another triangle, then the triangles are congruent. So, which means that we can actually construct, uniquely construct a triangle by three sides. Well, how to do it? Very simple. If you have three segments, one, two and three, what we can do is take any segment, one of these three, let's watch, about this one, and put it on any line anywhere on the plane. Then, using the radius which is equal to the size of another segment, not equal to this one. We draw a circle using a center as endpoint of this segment which we started with. Then, using another radius which is equal to the length of another segment and the center at a different point of the first original segment, we have another circle. Now, we have two points where these circles are crossing. Well, obviously, since circle is the locus of all the points distanced from the center on the radius, then this segment is obviously congruent to this one and this segment is congruent to this one, because that's how we built it. And this segment initially was congruent to this one, that's how we started. So, the whole triangle is congruent to the three sides of this triangle are congruent to three segments given to us initially. So, is that it? Well, no, there are some nuances here. And the nuances are related to the fact that if you have two circles from two different points of a segment, they might or they might not intercept. Consider this situation. If the segment we started with is too big and two other segments are relatively small, then these particular circles will have no intersections. Or they have something like one intersection, just where they touch each other. So, we don't really have a triangle in these cases. What does it mean? It means that relative size of these three segments which are given to us is important. It was not the case with the previous construction problem where we had two segments and an angle between them, because on the size of the angle we can always put whatever the segments we want. In this case, since we are looking for intersection of these two circles, it might or might not be the case. Which means that the whole problem of construction of a triangle based on three given segments, three given sides, is not always correct. It's not always valid. It doesn't always have a solution. It depends on the relative size of the segments. Now, to be absolutely precise, I can tell you that the biggest segment should be actually smaller than the sum of two other segments. Now, in this case when these two circles are touching each other, then the length of the biggest segment is equal to the sum of the lengths of two other segments. But if this length of the biggest segment is smaller, then there will be two points of crossing. And that's what we actually need. So, the condition for the validity of this construction problem is that the biggest segment should be smaller than the sum of two others. Then we can construct the triangle. Then we have actually two different solutions, but these two solutions are congruent to each other. Because this triangle is obviously congruent to this one. They are symmetrical relative to this axis of symmetry. Alright, so that's the second problem. Let's move on. That's my third problem. Okay, construct an angle equal to a given one, somewhere else on the plane. Well, it's very easy actually to approach this particular construction problem using whatever we have already done before. It's always like that in mathematics, in geometry in particular, you use what you have already accomplished to achieve whatever the new height you want in your logic. So in this particular case, how can I construct an angle equal to a given one at some other place on the plane? Well, very easily. Let's just take two points on both rays which constitute this particular angle. Draw a line. Now we have a triangle. And we know already how to build a triangle equal to another triangle if you have three sides. Right? So basically using whatever the approach I was just explaining on construction of the triangle by its three sides, we construct the triangle with these three sides. Somewhere else, wherever you want. Basically that's it. So let me just emphasize again. It's good, it's valid, it's the right approach in mathematics to say that this particular problem can be reduced to another problem which has been already solved by a very simple manipulation. Like in this case, we just draw a segment between two points on two rays of the angle and say, okay, the problem is solved because we know how to solve the problem of building a triangle by three sides. And the story, what's next? Angle, side, angle. Alright, so if you have a triangle and you have an angle, a side, and another angle to make it this way. Well, what it actually means, we have to transform all these three elements into a triangle so that this particular segment would be its base. This would be, let's say, left angle and this would be the right angle. Well, okay, how to do it? It's quite easy actually. What we can do is we can have this original angle, one of these triangles, let's say this one, and have this segment constructed on one side of this triangle. So this is the segment which is congruent to this one. Now what we have to do? Now we have to build an angle which is equal to this one. Okay, how to build an angle equal to this one? Well, we know how. We just solved this problem and basically we just use it. If you want, we can do it again, but basically what I did, I took two points here, converted this angle into a triangle and constructed triangle equal to this one by basically making a triangle equal to this triangle by three sides. So this is one side. This is another side and another side. So I have all these circles that are crossing somewhere and so I get this point and this angle and then just continue this angle until it crosses another way of the first angle and then I get a triangle. I mean, if you want, we can actually use exactly the same segment here to put this point in which case this point will coincide with this one and basically the same thing. Well, it doesn't really matter. It doesn't make the problem much easier. It just reduces the number of points. So you have from here, you have to draw a couple of circles here to get the point here. But in any case, we get exactly the same triangle which will contain the base equal to congruent to our segment which we started from and two angles on both sides. Okay, so that's the third problem. Third or the fourth? That's the fourth problem, I'm sure. Alright, next is how to bisect a segment. So you have a segment. You have to draw a perpendicular bisect. Now, to do it, it's really quite easy because again we can rely on certain theorems and one of the theorems was that in the equilateral triangle, the median and the bisector and the altitude from the vertex down to the base coincide with each other. So what I will do is I will use certain radius. Let's say we can take the radius which is equal to this segment. It doesn't really matter. What does matter is that it should be big enough. And using this radius, I'll draw a circle, one and another. And they cross in two points because I took the whole segment actually as a radius, so the others cross in two points. And what do we have right now? Let's consider this triangle and this triangle. They are both equilateral. Why? Because I took exactly the same radius from here and from here. So this line is equal to this and this and this. So these four segments are of the same lengths congruent to each other. Okay. If that is true, then if we will draw the line between this point and this point, which we already have, and just connect these two points together, we will have this line dotted line, which is what I would like actually to prove that this line is a perpendicular bisector. But again, it's sufficient to prove that in both triangles this line is the bisector and the median and the altitude. Well, let's think about this. What's very easy to see is that this angle equal to this angle. Why? Because this triangle is equilateral with these two sides equal to each other. And in the equilateral triangle, angles at the base are the same. Okay. Same thing. I can state with these two angles. They're also the same. Now, the triangles themselves, this triangle and this triangle are congruent to each other because three sides of one are congruent to three sides of another. So all these four angles are equal to each other. And therefore, this line is an angle bisector of this angle. And since it's an angle bisector, it's median, which means this point is the middle point of this thing, and altitude, which means it's perpendicular. And that's exactly what we need. We need a perpendicular bisector of a segment. So it crosses in the middle perpendicular to this segment. And the story, we have constructed our perpendicular bisector. So again, we took this radius and made two circles from here and from here, and just connected the line with the crossing points. Okay, that's done too. And bisect an angle. All right. What if you have an angle and you would like to bisect it? Well, there are actually many different ways. And what I can suggest is the following. Take any radius and draw a circle when crossing these two rays. So you have basically an equilateral triangle here. These two lines equal these two angles equal congruent to each other. And now what you can do is you can continue using exactly the same technique as before and construct the perpendicular bisector to this segment. Now, well, very simply, we take this radius and have another circle here and here. This point we already have. So now we will have just yet another point, and we have exactly the same picture. Now connect these and it will be perpendicular bisector to this particular segment, which is at the same time medium and angle bisector, which means these two angles will be congruent to each other. And that's how we bisect this particular angle. Well, that's basically it for these elementary construction problems. Now, what is interesting is that in the future, when we will discuss certain construction problems, well, I would say more complex ones, I will definitely use something like, all right, let's divide this segment in two halves, or let's bisect this particular angle, or let's draw a perpendicular line to this particular segment at this particular point or whatever. And I will not really stop explaining how to do this. These are elementary bricks, if you wish, from which the whole building of geometry is built, construction geometry is built. So if you design a car, for instance, you don't explain how to make a tire. Basically, you say, well, we'll take the tire, such and such tire, and put it here, right? Same thing with construction problems in geometry. You just use whatever the techniques have already been established. And these are just elementary techniques, which I'm just offering, and there are many others. So whatever the new little technique will be, we'll definitely explain it. Other than that, I'll just use it. So whenever I need it, I will just basically use whatever is necessary. Well, thanks very much. That's it for today.