 Hi, I'm Zor. Welcome to Unisorification. Today's lecture will be also about construction problems where similarity plays very important role. You obviously are encouraged to do all these programs just by yourself. All the notes on Unisor.com contain all these problems. Then listen to this lecture, whether you succeeded or not, doesn't really matter. After you listen to the lecture, I would encourage you to do it again just by yourself, just to make sure that you remember everything. Now, today's problems will be related primarily to algebraic methods to solve geometrical problems. Personally, I think that using algebra to solve geometric problems is a heavy artillery and a last resort when you don't really find a purely geometrical solution, then you resort to algebraic. But look, I mean, if you have to, you have to. So most of these problems will be related to algebraic methods to solve geometrical problems. So number one, right. The first one is actually about certain techniques. Some of them you know, actually most of them you know, some of them might be new. But this is actually algebra of segments. Now, the simplest way is if you have a segment of certain lengths, how to multiply it by NL5 or something like this. So let me just go through all these numerous examples. What can we do with segments? And I'll just explain how to do each one of them. All right. So now, the couple of conditions, right. When I'm talking about some kind of formula like this one, for instance. So what I imply in this case is X is the segment which we have to construct. Segments which have letters A, B, C, D, whatever they are given segments. And M and N are usually integer natural numbers, which for instance in this case means multiply the given segments by some natural number N. Basically, getting a segment whose lengths is N times the lengths of the segment A. So all these algebraic manipulations with segments, obviously are related not exactly to segments, but to their lengths. So in this case, construct a segment, the lengths of which is N times greater than A. All right. So this is basically the first example of this series. It also implies that I also always imply that there is a segment whose length is 1, and it's also given as well as segments like A, for instance, or B or C. So it's always implied that the segments of the lengths A, the lengths 1 is given, units A. Okay. Now I'm ready to go through the problem. So the first problem, multiply the segment by number N, simple thing. So if you have a certain segment A, you just continue the line and using the compass, just put N times this particular segment. Now considering N is a natural number, you can definitely do this. And starting from the beginning to the end, that would be your new segment X, whose length is N times greater. Okay. Now this is just an introduction. There are some slightly more complex problems. A divided by N. All right. A divided by N. Now this is a simple thing to do. So this is your A segment. You can take any other segment, for instance, unit segment, put N times here, connect the ends, and each parallel line to this one, will cut one end of the segment A. The proof of this is in the similarity theorems. Sides of these triangles are proportional, I guess. So that's important. Next. Okay. A times M over N. Well, this is obviously a combination of the previous two. First, you can multiply by M by pushing one segment of the length A after another, and then divided the resulting segment by N with thoughts. Okay. Now considering A, B, and C are given segments how to construct this one. Now, or if you wish, this can be transformed into X over B as A over C, right? So basically, you have to find some proportionality between the first proportional segment, if you are given the three of them. Well, the way to do it is something like this. You draw an angle, all right? You positioned your, let's do it once more, let's do it. Now, you draw parallel and then parallel to this one. So that would be your X. A to C is equal to X to B. So the way how to construct it is, using the compass and any angle, you put the segment C after that B on this side A, you draw the lines, connect end of this, end of this, and the parallel, and that's where you are A, the X. So again, the proof follows from the theorem of similarity. Now, these constructions, these simple others, are supposed to be really like in your repertoire or in your toolbox, if you wish. So because more complicated problems, obviously, would be dependent on these ones. All right, next. Next is maybe a little bit strange. Now, what it actually means in this case, it means A times 1 over B, where 1 is a unit segment. So it's basically the same problem as before, but one of these three given segments is a unit segment. And it's implied, whenever you say something, whenever you see something like this, it's implied that there is a unit segment which is participating in the whole thing. Or if you wish, X over 1 is equal to A over B. And again, the way how you construct it is A, B, and then you have 1 and parallel line, that would be your X. So absence of the segment means you can substitute the segment of the unit length. X equals A square divided by B. Well, this is basically the same thing as before when I have A, B over C, except the two segments are equal to each other. So there is nothing actually to talk about. This is a simple thing. Then X equals A times M square, where A is a segment, M is a number. Well, you can always think about this as A times M times M. This means first you multiply by M, your given segment, A. And the result, you will multiply by M again. So that's a simple thing. Now I'm going through these three little stuff just because other problems might actually use it. And I don't want to refer to something which is not really maybe obvious from the first glance. All right, next. X is equal to square root of A times B plus C times B. Well, this is a slightly more complex thing. Now, if you remember the theorem, the Pythagorean theorem is where P and Q are a catheter and X is an apotonous, you know that, which actually means X is equal to square root of B square plus Q square. But you don't have this. You have this, right? So why don't you have first find P square which is equal to AB and similarly Q square which is equal to CG. If you can find segments P and Q which satisfy these equations, then you convert this into this, right? Because then you can use P and Q as two cartridges of the right triangle. And then X would be an apotonous. All right, so now basically we have reduced this problem to these problems, both similar to each other, actually. So let me just refer this problem to the next one, which is this, how to do this. OK, here I can actually refer you to a theorem which you probably remember. If you have a right triangle and an altitude towards apotonous, then E square is equal to A times B. Now the proof of it is really very simple, because all these triangles, and the big one actually as well, are similar to each other. So if you will put the proportionality of the corresponding sides, for instance, this angle is equal to this one. This one angle is equal to this one. So in this triangle, P over B, P over B, P lies against this double arc, and B against single, is equal to here A over double arc to P. And that's where you get P square is equal to AB. So this problem, which is actually next on my list, is the way how you solve this one. You first find P using this equation by building the right triangle, and then the Q in exactly the same fashion. How can you build that particular right triangle, which I could just erase it, unfortunately? How to build right triangle where an altitude divides apotonous into A and B parts? Well, easy. Use A and B, put them together, and use it as a diameter of a circle. Since this is the right angle, it's supported by a diameter and half a circle. And then in the point which divides parts A and B, you draw a perpendicular, and wherever it hits, that's basically your top point, top vertex of this particular right triangle. And this is your altitude. So that's how you build this one. And using this, you reduce this problem to this one. And this you know how to do, because once you know P and Q, you just have a right triangle. This will be P, this will be Q, and the apotonous will be X. That's done. Next, X is equal to square root of, or m is a number, basically. Well, again, let me use the unit segment, because what it actually means, it means this. Meaning this is a segment of the length, m, which is m times greater than unit segment. And this is the unit segment. So it's basically the same thing as saying this one. So these are two given segments, or given segments, yes. This one has the length of m times unit segment. This is the unit segment. And so you have to find X square, which is exactly like the previous problem, where I had, not X. I had P square equals to A times P. In this case, A is m and B is 1. So you have exactly the same problem, and you use exactly the same method to solve this problem. So for instance, somebody is just asking you, draw a segment which has the length of square root of 5, for instance. Well, that's how, basically. You use 5 and 1 as two different segments. You draw a circle perpendicular, and that's where you get your square root of 5. Next, that's it for these exercises. So this was some kind of preliminary first part of this lecture, where I just enlist all the different kind of tricks which you can do with segments. The lecture algebra segments, which precedes these construction problems, explains some of these in a little bit more details. OK, so let me now go to other problems where, as I promised, the algebraic calculations will be involved. OK, calculate the lengths of a second to a given circle from a given point. Given circle, given point, you have to calculate the lengths of a second. If tangent, so you have a tangent from this point has lengths a, and ratio between internal and external parts of a second is m over n. So this is your second. So this is point p. This is mn. This is tangent. So pn over mn is equal to m over n. So this is given, and the lengths of the tangent is given. OK, so why the lengths of the tangent is given? Let's just think about it. I mean, obviously, it's related to a known theorem that the square of the tangent is equal to a second times its external part. So I know from the theorem which I have already proven before that a square is equal to, well, let's call this part x and this part y, if you don't mind. It's equal to x plus y. It's the whole second times its external part. On the other hand, I have this ratio, which means what? pn over mn, pn over mn, which is y over x. Yes, y over x is equal to m over n. Well, this is algebra. This is the two equations with two unknown variables, x and y. All we have to do is we have to solve it. And that's not really very difficult, so let's just do that. And after we solve it, we will have some kind of a formula where x and y depend on a, m, and m. So what do we do? Well, we probably, easier probably, if I will use x from here. It's x times m is equal to y times n from this rate. x times m is equal to y times n from where we get x is equal to y times n over m. Now that we can substitute to this guy. And what do we do? We have a square is equal to open parenthesis y times n over m plus y times y. OK. Now if you wipe out this, we don't really need this anymore. Now this is really a simple thing. This means a square is equal to. We can get y outside, so it will be y square times, I think I had n over m, right? That's not a mistake, n over m. n over m plus 1. Or a square is equal to y square m plus n over m. You transfer this into this part. y square is equal to a square times m over m plus n. And finally, y is equal to, let's have square root of both, a times square root of n over square root of m plus n. All right. Now, if you know m and m and you know a, how to construct this particular y. Because as long as you know how to construct y, your problem is actually solved, if you remember. So you have a point. This is x and y. So if you have y, and you have the point, basically, draw a circle, and wherever it intersects our circle, that's your, basically, point which you are looking for in the straight second would be what you need. Well, again, you can do it step by step. First, you can multiply by square root of m. Now, what is square root of m? As I was telling, square root of m is something like p square is equal to m times 1, right? Then p is square root of m. So you built this particular segment, p, and you know how. Now, another segment is q square, which is equal to m plus n, also times 1. That will be this one. q is equal to square root of m plus n. And finally, you do y is equal to a times p over q, which again, we know how to do it. This is the fourth proportional if three segments are given. So construction is based on this formula. First, we do the formula algebraically. And then, having the formula, we know how to construct it. Now, people might argue that this is not, well, geometrical enough. And I agree. It is not. Now, if you know how to do it purely geometrically, fine. So much better for you. Now, this particular approach is in cases when you don't really know. As I was saying, this is some kind of heavy artillery, which you don't know how else it can be solved. So you go for algebraic solution, which might or might not be simple enough. But in this case, it's durable, basically. That's what it is. Next, construct a segment x with lengths having a ratio to a length of a given segment s. So x should have a ratio to s, which is given, like squares of given segments p and q. So s, p, and q are given. So you have to construct x, which has this particular property. Well, let's just think about it. You can actually change this into this. x times q squared divided by s is equal to p squared, right? Now, what I will do first, I will do this. So it will be x times y is equal to p squared, where y is equal to p squared over s. Now, I know how to do this, right? We did it before in my first preliminary set of constructions. So first, I do this. Now, when I do have this y, I can do this. Since I know y, I can now build this one. So I'm referring back to whatever the beginning of this lecture I was talking about, how to construct these different things. And basically, that's the approach. Do some manipulations, and you reduce this problem to a couple of those which you have already known how to do it. OK, next. Find a point outside of a given circle such that tangent from it to a circle is measured as half a second from it through a center of a circle to an opposite end of the diameter. So you have a circle. You have to find a point x in such a way that the diameter, which is drawn from x through the center, is twice as big as tangent. Well, if you know the circle, you know the radius, right? So basically, you can consider this to be radius. These are three radiuses. Now, to find this point, I actually have to find this distance, all right? So let's call it which letter do you prefer, w. All right, so I have to find w, which has this particular property that xm is twice as big as xa. Well, now we have many different choices. For instance, I can use the Pythagorean theorem in this triangle, oax. So r plus w is a hypotenuse. r is a cathetus, which means the Pythagorean theorem gives me the lengths of ax, which I can then, after I express it as a square root of r plus w, sorry, square minus r square, it should be twice as big because we double it. And it should be equal to this one. And this one is r plus r plus w, so it's 2r plus w. So this is an equation which we can use to find our w. Alternatively, which is probably the same thing, we know that square of a tangent is equal to product of a second by its external part. So different equation would be, so this one would be 1 half of 2r plus w, right? I know that this is half of this, this is 2r plus w. So the square of this thing should be equal to the product of this length, which is 2r plus w by external part. So that's another equation. And most likely these are exactly the same equations. And whatever you solve will give you exactly the same result. Now, which one do you prefer? Well, let's just take the person who doesn't really matter and try to solve it. Now, how can I solve this equation? Well, actually elementary, if I will square both sides of this equation, I will have a square w in the second degree, which means it will be quadratic equation relative to w. I wish I can do it very accurately. I might actually make a mistake, but let me try to do it. So the square of this would be 4r square plus 4r w plus w square is equal to square of this is 4 times this, r plus w square, which is r square plus 2r w plus w square minus 4r square, right? All right. So 4r square plus 4r w plus w square is equal to 4r square and minus 4r square. So r square will go out. So I'll have 8r w plus 4w square, all right? So what do I have? Let's move everything to the right. So we'll have 3w square 4r w plus 4r w minus 4r square equals 2 0, right? So this is a quadratic equation for w. Well, what's next? Next we have to solve it. I hope I didn't make any mistakes on the way. But even if I did, basically the purpose of this lecture is for you to understand the concept. And I'm sure you will do it accurately yourself. So w is equal to 6 minus 4r plus minus square root of 16r square minus 4 times this times this. This is a negative, so it's 4 to 8r square, right? I think I'm right. Now minus obviously is out because we are talking about segments. So w is equal to minus 4r plus, this is 64. So it will be 4r 8r, which is equal to 4r over 6 or 2r over 3. So that's our answer. So this particular problem has this solution. If this particular segment, which I call w, is 2 thirds of the radius, then the tangent would be half of the whole second length. By the way, it's kind of a complicated problem and the way how I solve it, but the solution is quite easy. It seems to be the right solution. Now, can it be done? Here is another interesting thing. Can it be done geometrically? Can I prove that 2 thirds of the radius is exactly the distance if this is 2 to 3rd, 2 3rd of the radius? This is the radius, this is the radius. Can I prove that if this is 2 3rd of the radius, then the tangent is exactly half? Well, we can actually check because the whole length would be what? r plus r plus 2 3rd of r. So the whole length is r times 1 plus 1 plus 2 3rd, which is equal to 8 3rd of r. External part is 2 3rd. So the multiplication of the whole second by its external part would be 8 3rd r times 2 3rd r, which is 16 3rd of r squared, right? No, sorry. See something's wrong. Now, I know that this length is, square of this length is this particular thing. So what's the length, the square root of 16 9s of r square, which is 4 3rd of r? And as you see, this is twice as big as this. This is the total length, and this is my tangent. So this is kind of a basically checking procedure that I did the right thing. Or you can use the theorem of physical law to check the same thing. OK, next. Construct a second from a given point to a given circle that is divided by the circle in a given ratio. OK, so very similar. You have, but now you have a point. So what you have to do is you have to draw a second in such a way that a b over b c is equal to m over m. OK, let's call this piece x, this piece y. And since my point is actually given, I can also say that the tangent is given. Now, what does it mean that the point is given? Well, in particular, there is this distance from the center of the square. And there is a radius of the square, because the circle is also given. So I can basically calculate using this distance from the center and the radius. I can calculate this from the Pythagorean theorem. Or construct this particular segment. Let's call it m. So a m is very easily obtained from the fact that point is given and the circle is given. But now I know that the whole length of the second, which is x plus y, times its external part, x, is equal to a square, right? That's the known theorem. And at the same time, I know this ratio, which is x over y, is equal to m over m. So from these two equations, again, basically it's very similar to one of the previous theorems, previous problems, actually. I can very easily obtain the values of x and y from this system of two equations, which is too unknown. So let me do a very quick substitution. So and then I will just drop it. So from this, y is equal to x times n over m, right? Ym is equal to x and ym, right? So this is the consequence of this. Now I will substitute it to this and I will get x plus x n over m, x is equal to a square, right? x outside x square 1 plus n over m is equal to a square. We have exactly the same as the previous problem. So x is equal to a square root of n divided by square root of n plus n. That's it. That's simple. Algebra helps. It's not easy to construct it purely geometrically. Well, at least I didn't even think about how. Algebra seems to be a simpler solution, right? Next. Given two circles of different radiuses, a and b. So this is a, this is b. And the distance between their centers, d. From this to this, we have d, all right? Expressed as an algebraic formula, the lengths of the seven from a point of intersection between center line, so I have to continue the center line, and common exterior tangent. So I have to basically find, algebraically, the position of this x. The best way is to call this distance to the nearest circle, x, all right? So now we have to, now this is perpendicular, and this is perpendicular obviously. So I have to find x, which is the distance from a point x to the nearest circle. If I know the radiuses, the distance between centers, right? Well, probably the similarity of these two triangles is a good way to do this type of thing. So similarity is x plus b, right? The distance from point x to the center is x plus radius, which is b divided by this hip-hop. Which is x plus b plus b is equal to ratio between the radiuses, right? Because these triangles are right triangles, these obviously are right angles. So hip-hoptonuses are proportional to these radiuses, which is b to a. Well, that's it basically. And this is the equation. So it's ax plus ab is equal to bx plus b square plus bd, right? From here, we get a minus bx, a minus bx is equal to b square plus bd minus ab, from which we conclude the expression for x. Now, how to construct this? Well, actually, simply. First, you do a minus b. Let's say a minus b is a minus b is your c segment, right? And then you represent this as b square over c plus bd over c minus ab over c. Each one of these, you know how to construct. And then you summarize these two and subtract that segment, and that's your solution. So that's the algebraic approach to geometric problems. I think this is my last problem, right? Yes. OK, this is my last problem. Don't overuse this, let me put it this way. Most of the geometrical problems actually can be solved using purely geometrical methodology. Algebraic methods are not, how should I say, not beautiful enough, not cool enough, or whatever word you can use. This is the last resort. When you don't really know what to do, then you can try to do it algebraically. Well, that's it. I do encourage you to do exactly the same problems again. By yourself, go to unizor.com. And in the similarity, geometry similarity, this is a lecture called Construction 3. And there are hints, actually, in the notes. So they might actually help you to make it easier for you. Now, if you register, you will be able to take exams. So I do encourage you to do this. It's always good to know how good you are. And well, basically, that's it for today. Thank you very much.