 Hi, I'm Zor. Welcome to Unisor Education. I will continue talking about similarity problems. This is the lecture number six about problems, which are using the similarity. So, basically, there are certain algebraic usage of theorems related to similarity, which in the previous lecture was just trying to explain how to manipulate with segments to obtain new segments, the lengths of which is somehow related to the previous segments. For instance, how to multiply a segment by two, which basically means how to construct another segment with the lengths twice as long as the previous segment, as a given segment. So, there are many different geometric algebraic problems, which I would like to address here. And obviously, multiplying the segment by two is one of the simplest things. So, let me just go through all these problems and I will try to explain their solutions. Oh, they all means try to do it themselves first and then listen to this lecture. All right. So, the first problem is a very simple one. Oh, one more thing which I wanted to mention. If I'm saying that there is a given segment of certain lengths, actually, it always implies that not only a given segment is given, but also another segment which has the unit lengths. So, I always assume that if you have any segment measured in some units, the unit segment is also given. So, I won't even mention this. And in my manipulation of the segments, when I'm saying multiply by segment by two, it actually means construct another segment with the lengths twice as big. Or adding two segments means again constructing another segment with the lengths equal to the sum of these lengths. So, just bear with this terminology. It's very simple. All right. Perform the following linear algebraic manipulations with given segments. Okay. A, three times A, B, A plus B, and C, A divided by five. Okay. This is the simplest of all the different manipulations which can be done with a segment. This one is obviously if A is given, you just use the compass and each of these segments is A, and this is 3A. That's simple. Now, addition A plus B is also simple if you have another segment than you basically... Again, using your compass on a straight line, you construct the first segment which is congruent to the A and then another congruent to the B. Combined segments will be A plus B. Okay. These are the simplest constructions. Just one little, tiny bit more complex is a division. And here the similarity actually comes handy. So if you have a segment A, how to divide it by five? Well, have any line at a certain angle to this segment and have any segment of any length. Basically, if I'm usually saying that there is always a segment of the length one, you can use this segment, but basically it doesn't really matter. Let's say you use this one. You put five times one, two, three, four, five. So this is of the length five. You connect the ends and construct parallel lines. And as I have proved the theory in that if you have equal segments on this side of the angle and these lines are parallel, then these are equal as well. And since they are equal, each one of them is one-fifths of a original segment. All right. Now let's go to a slightly, slightly more complex problem. I'm giving a linear equation and I'm trying to find a solution, but I'm using segments instead of numbers. If you have a segment A and another segment B and you have a linear equation, X minus A equals B, how to construct a segment X, which is a solution to this equation where A and B are given? Well, simple. If you have an algebra to solve this equation, and obviously the solution is, let me do it strictly by the rules. You add A to both sides. You have X minus IA plus A is equal to B plus A. Now, using the associative property of addition, you add minus A and A, which is zero. So on the left you have X and on the right you have B plus A or A plus B because it's commutative. And this is a solution. And this is exactly what I did in the previous problem, how to construct a segment which is a sum of two given segments, which you have already learned how to do. So that's the solution. Now, let me just emphasize that this is a combination of algebra as far as the solving solutions and geometry as far as construction. Okay, next one. It's as easy as B, 3X minus A equals to zero. Well, again, plus A to both sides and I will skip a couple of obvious steps. It's equal to A. Now you divide by three both sides of equation and again skipping a couple of steps, which are related to associativity and commutative, etc., etc. have X is equal to A divided by three. And this is a problem which we have solved before when I was basically explaining how to divide a segment by five. So this is by three, the same thing. One, two, three. Connect the parallel lines. Each one is one-third. Okay, so again the combination of algebraic solution of the linear equation and geometric property of how to construct a certain segment are both combining together to give you the solution to this problem. I didn't mention it actually in notes to this lecture, but what if you have another equation, where X plus A equals to zero, where A is a given segment. If I will try to solve it exactly the same way, after the couple of manipulations, I will have X equals to minus A over three. But we are talking about segments. Segments cannot have negative lengths. So basically this is not a solution. This is algebraic solution, but there is no geometric solution. So this is an equation which does make sense in pure algebra where X can be any real number or any integer number or any rational number. We can talk about this. Actually, among integer it doesn't exist. If A is integer, but among rational numbers, yes, the solution does exist. But rational numbers is completely different from the segments. Well, actually segments lengths are only non-negative real numbers. Non-negative is important. And this, since A is a segment, so this is a positive real number with a minus, it's a negative real number, and there is no such segment which has a negative length. So basically this is an equation which does not have a solution in geometric terms. Although it does in, let's say, rational numbers or real numbers. Alright. And the last equation, X over 5 minus A equals B. Alright. First, algebra. And A to both sides, X over 5 is equal to A plus B. And again, I skipped a couple of steps here which are related to the associative and commutative laws. And now I multiply by 5, both sides of the equation, and skipping a couple of times, I have 5 A plus B. Now, how to construct this if you have A and B? Well, this is not a one-step procedure. This is a two-step procedure. And obviously you should realize that in two steps you can do it. First step is A plus B. So first you do A and B. And you get, let's say, C is equal to A plus B. Now, second is 5 times C. And again, you know how to do this. Straight line and 5 times the segment C. So this is the two steps' geometric procedure. First, again, algebraic solution of the equation. Secondly, you construct the formula for X using whatever you know how. Okay. Next. Next is something which is related to theorem which is very important. And I'm actually going to prove it again just to make sure that everybody understands what it is. If you have a right triangle, this is 90-degree angle, right angle. And you have a hypotenuse. Let's call it H. And let's call these two segments, P and Q. And these are two-pegged A and B. Now, notice that these two triangles, I can put letters A, B, and C, and H. Triangle C, H, B, and C, A, H. They are similar for obvious reasons because this angle equals to this angle. And this angle equals to this angle. Now, why is that? Well, that's kind of obvious because you have these two angles in sum are equal to 90-degree. And these two angles are also 90-degree. That's why these are equal. Similarly here. These two in sum are equal to 90-degree. That's why these two are equal to... And these two are also equal to 90-degree. That's why these two are equal. Okay, so we have two triangles, C, H, B, and C, H, A, which are similar, which means sides are proportional, if you remember. So let's just write down this proportionality of the sides which are lying across equal angles. Now, in the left triangle, across this angle is H, right? In the right triangle, across the same angle is Q. Now, in the left triangle, across this angle is P. In the left triangle, across the same angle is H. So what we have is H over Q is equal to P over H, or if you transform it, it would be H squared is equal to P times Q. Now, again, when something like this is written, H, P, and Q are lengths. They're not segments. We don't really multiply a segment or square the segment. We multiply its lengths. Or multiply lengths of one segment by lengths of another segment. So that's what it actually means. But in any case, what I would like to point out is that the altitude divides the k-partness into two segments, which satisfy this equation of proportionality, or this one, which is the same thing. It's actually called a geometric mean. Basically, H is square root of P times Q, right? In another way to represent the same formula. So H is a geometric mean of P and Q. All right. Now, just remember this. And now, let me just use this particular theorem, which I could just do, and I did actually prove it before in one of the lectures, to construct a segment the length of which is a geometric mean of these two segments. So how to construct, let's say, H if I know P and Q. So that would be my geometric mean. So if I can, knowing P and Q, if I can construct the right triangle, then its altitude would be exactly the geometric mean. Okay. How to do it is very simple. We all know that if you inscribe the right triangle into a circle, then the k-partness would be a diameter, something like this. Why? Because this is 90 degree. And 90 degree is an inscribed angle, and it should be supported by an arc of 180 degree, which is half circle. So in any case, if you have a k-partness, then you basically have a circle, and every point on this circle can be a vertex of a right triangle. Out of all these right triangles which have this k-partness, I have to choose one which has an altitude dividing my k-partness in segments P and Q. So if P and Q are given, that means P plus Q, which is a k-partness, is also given, I can construct a circle around P plus Q as a k-partness. And now I can say that every dot, every point on this circle can be a vertex of the right angle. Now, how to choose the one which we need? Well, we don't have just the k-partness, we have these two segments as well, P and Q. So we know that k-partness, that the altitude should fall into this point. So let's just draw perpendicular here. That would be our altitude, right? Which means this is our triangle. So from the k-partness, which is P plus Q, we draw a circle. Now, the points, the segments P and Q are positioned on this k-partness, and the boundary between them, the dot which separates them is this one, and through this point we draw perpendicular to a diameter wherever it intersects the circle, we get the vertex of the triangle which we need. Well, obviously this is exactly the same triangle and exactly the same altitude. Okay, so if we have two segments P and Q, we know how to build a triangle with an altitude dividing k-partness into these two segments P and Q, which means we know how to build an H which is equal to geometric mean between these two segments. That's what we will use in many problems, which fall. I just wanted to do this theoretical deviation before solving other problems. So construct the segment which is equal to A times square root of three-fifths. It doesn't seem to be obvious, at least from the first kind of one. Sorry, but let's just think about it. If I want to do this, I have to actually reduce my problem to something which is more, I would say, constructible. So let me first transform this into this. I hope it's obvious that this is exactly the same as this. Three-fifths, this is A square, but this is the square root, so it will be A. So now it becomes quite obvious what to do, because this is the geometric mean of two segments, each of which we know how to build, how to construct. If we know A, we know how to construct three-A. So first, we construct B is equal to three-A. Secondly, we construct this, C, which is equal to A divided by five. We know how to do that. And finally, we construct D, which is equal to square root of B times C. And we know how to do this. I was just explaining this is a geometric mean. So that's the solution. All you have to do is to transform the formula which you are given into something which is more geometrically constructible, so to speak. Square root of A times B minus C times D. Again, from the first glance, A, B, and C and D are given segments of this. From the first glance, it's not obvious what to do in this case. Well, but let me again do it step by step, and it will lead us to a solution. I understand we cannot construct the formula like that, just like that. However, we do know, let me just analyze the problem from the end, we do know how to construct this. Am I right? Why? Because this is, or, which is a Pythagorean theorem, right? So if you have P and Q, you can construct X, where P is a hypotenuse, and X and Q are two causative. If you have a hypotenuse and a causative, you basically do this. You construct the straight, you construct the right angle. Here is your Q, and then if this is a P, the shape from here, and this is the radius, the droid circle, this is P. So this is P squared is equal to Q squared plus X squared. So that's how you construct the triangle, and that's how you construct this segment X. So you know how to construct this. It's just basically a construction of the right triangle with P as a hypotenuse and Q as a keton. If you know that, the question is how to convert this into this? Again, step by step. We would like, instead of AB, to have P squared, right? How to do it? Very simply. Let's construct P, which is equal to square root of AB. Then P squared would be A times B. And we know how to do that. This is a geometric mean. Same thing with CG. We construct Q, which is equal to square root of CG, which is a geometric mean. Then Q squared would be equal to CG. Once we have constructed P and Q, we can use this formula to construct our X. So the solution is, again, step by step. First, instead of AB, we construct a segment P, which has this property, geometric mean. Instead of C and G, we construct its geometric mean, which is segment Q. And then we construct a right triangle with P as a hypotenuse and Q as a keton. Square root of AB plus 3. In this case, again, it's not that obvious. I mean, it looks more or less like the previous problem. But in the previous problem, we have the multiplication of two segments, C times G. Here we have just number 3. But let's think about what 3 actually is. 3 is the length of a segment, which has three units actually in it. We are talking about lengths, basically. So I can easily replace it with 3 times 1. And as I was saying, we always have not only A and B, but we also have a unit segment, segment of the length of 1. So let's just build a segment of the length 3, which is this. Call this C and call this G. And we have a formula AB plus CD square root, which is more or less the same as the previous problem. So you construct P and Q equal to geometric mean of AB. P is geometric mean of AB, and Q is geometric mean of 3 and 1. And we know how to do that. And then this is a hypotenuse of a right triangle with a strategy P and Q problem solved. Notice that lots of these problems with square root involved are really reduced to the problems of finding geometric mean and Pythagorean theory. Just two things. Right triangles. Everything about right triangles. Because geometric mean is also right triangle. But construct the right triangle by two segments of the hypotenuse. And Pythagorean theorem is basically right triangle where you know either hypotenuse in the calculus or a couple of calculus. Now the last problem here, which has three different sub-problems, construct segment X, which is a solution of the following quadratic equation. Now it's quadratic equation. Just slightly more difficult. And okay, so the A is X square minus 2X minus 3 equals to 0. Alright, so how to construct a segment, the length of which is solution to this equation. Well, let's solve it. You know what, it's easy for me to just use the formula for solutions of the quadratic equation. But since I don't like it, I'll do it differently. I remember the full square formula. Remember X minus A square is equal to X square minus 2AX plus A square, right? So I'm trying to use this where A is actually equal to 1 in this case. So it will be X square minus 2X plus 1 minus 4. And this is X minus 1 square minus 4 equals to 0. Which means X minus 1 square equals to 4. Which means X minus 1 is equal to either 2 or X minus 1 is equal to minus 2. Now since X is a segment, well actually the length of the segment, it's not negative so this is completely out. So we have only one solution, X is equal to 3. And I don't really have to basically do some crazy manipulations. All I need is a unit segment. And from this unit segment I basically multiply it by 3 and get 3. And that's the solution of my equation. So I have solved it algebraically, got rid of the negative solution which is not applicable to my geometric means, and basically built the solution from a very simple manipulation. So that's an easy one. Now slightly more complex. X square minus 4A square plus B square equals to 0. So I have to find this. Well, again I can probably use some formula, but in this case I can do it slightly more, slightly simpler. I'll just transfer X square is equal to 4A square minus B square. I skipped all these intermediary steps obviously, which means X is equal to square root of 4A square minus B square. Now if A and B are given, it's almost like the Pythagorean theorem, except I have to really make a tiny improvement. Instead of 4A square I will put 2A square minus B square. Now it's a pure Pythagorean theorem where 2A is a hypotenuse and B is a casual. So if I have A and B, all I have to do is from A I built 2A and then use it as a hypotenuse of the right triangle and B as a casual. And that gives me the second casualness, which is X. Trying to avoid formulas as much as I can. However, in the last problem I will not be able to do it, X square plus AX plus B square is equal to zero. Here, unfortunately, it's not as obvious how to do it in some special way. So let's just use the formula as it's supposed to be used, which is so X first and second to different is equal to 2 minus A plus minus square root of A square minus 4B square. So that's the formula basically. I'm just using the formula, which I happen to remember. I mean, if I don't remember I can always do something like extraction of the full square from this. It would be X plus A divided by 2 square. So it would be X square plus AX plus A square over 2 square. So now I have to subtract it and then plus B square, which is basically the same thing. That's how I find X plus A over 2. And this is more or less the same Pythagorean theorem. Anyway, so I can do that. Or I can just go straight to the formula. It doesn't really matter. Now, obviously I should reject the minus here because it produces the negative result. Actually, you know what? I think I'm wrong here. I think I should put minus here. Because with plus and minus here, my result would be negative given with a plus sign. Yes, this is the right way to do it. I mean, this is the right equation. The equation with a plus sign doesn't have a solution in a geometrical sense. But this one does. Okay, so all we have to do now is to, again, as I said, reject this negative thing. Oh, but wait a moment. Now the negative thing you don't really have to reject because both positive and negative will give you positive result. All right, fine. So let's build both of them. So x1 is equal to a plus square root of this and x2 is equal to a minus square root of this. Now, how to construct these solutions? Well, again, step by step. First, you do this. So how to construct c? It is equal to a square minus 4b square. Well, elementary. In this case, a calculus of a right triangle with a as a hypotenuse and 2b, this is 2b square, and 2b as another calculus. So that's how you construct c. Then, in this case, you construct a plus c and then divide it by 2. And here it's a minus c. And then divide it by 2. That's simple. So as you see, what we can do with these methods, you can solve geometrical problems using algebraic methodology. So if somebody has a problem when there are certain segments which satisfy certain equation produced by whatever ways it means, doesn't really matter. And if that equation is basically solvable, like we are solving this quadratic equation, then you can genetically construct the segment which is basically a solution to this equation. Well, that was the meaning of this lecture, that this is an application of algebraic methodology to geometric problems. Now, the notes for this lecture are obviously on the web. This is the website. The website is open for everybody. And I do encourage you to not only look at the lectures, not only try to solve the problems, but also register and go to the exams as well, especially I'm encouraging parents to basically work with their students, their children to guide them through the exams where you can actually check the results and supervise the educational process. That's it for today. Thank you very much.