 Hi, I'm Zor. Welcome to Unisor Education. I continue presenting certain problems related to similarity. These will be about different sets of points which have certain property. It's called locuses in mathematics, in geometry. So all these problems which I will talk about today will be, okay, what's the locus of points, that and then certain property will be specified. All right. So let's go on. Locus of points, midpoints of all chords originated at a given point like in a given circle. So you have a circle and you have midpoints of all chords originated from X. Well, I think you probably notice that it looks like a circle, right? Well, how can I prove it? Actually very easily. Let's consider this is a diameter, which means this is a center of a circle. Now, if I connect a center of a circle with any midpoint of a chord, so any chord take a midpoint, then this particular radius will be perpendicular to the chord. We know this property from the properties of the circles, which means that from this center, line a segment which will connect it to midpoint of any chord will be perpendicular to the chord. So all these triangles, this one, this one, this one, this one, all these triangles are right triangles with hypotenuse equal to the radius from the given point to the center, which means the circle around OX as a diameter would be basically a locus, because we know that locus of all the vertices of the right angle of the right rectangles with the same hypotenuse, they all lie on a circle around this hypotenuse as a diameter. That's simple. Next problem is basically as simple, but instead of midpoints, we have a point which divides a chord in certain ratio. So we have a ratio given to us, and we have a point on a chord which divides this chord in this ratio. So what's the locus of all points which divide all the chords from this point, including the diameter, by the way. So what's the locus of all these points, which divide each chord from this point in this ratio? Well, it's kind of simple as well, because if you will connect these points and these points, so this is any chord, X, Y is any chord, and X, Z is a diameter. So any chord and diameter, diameter is also a chord, so it's divided by a point in this ratio, and this chord is divided in this ratio. Now obviously these two triangles, X, Y, Z and X, P, U are similar because the sides are proportional and the angle is common. From similarity follows the congruence of these angles, and this one is the right angle because it's supported by half a circle, inscribed angle, half a circle, so it's 90 degree. Therefore, this angle is also 90 degree. So no matter which chord I take, if I connect the point Q on the diameter, which divides the diameter in this ratio, with this point which divides this chord in the same ratio, this angle would be the right angle, which means locus would be a circle around XQ as a diameter. Points was given ratio of the distances from the left, so given Q10. All right, so you have a Q10, and you have point which has this ratio of the distances. So this is point X, this is P and Q, so XP over XQ is equal to M over N. Okay, now first of all let's prove a relatively simple fact that if you connect with this vertex of this angle, then any point on this ray would also have the same ratio of the distances. How can it be proven? Well easily, because if you will take point let's say Y, and this would be let's say A and B. So AYM and QXM are obviously similar triangles because angles are the same. These are right angles and this is the common angle. So they are similar, which means these sides have a ratio the same as MY to MX. But similarly, these triangles AMXB and MYB are similar, which means these two distances are also related as MY to MX. That's why ratio of these distances is exactly the same and it's equal to M over N. Now I have proven that this particular ray, which connects any point with this ratio of distances with the vertex, so any point on this ray will have exactly the same property, which means that it has the distances related as M over N. Question is, have I completed my solution? Did I find the locus of all points? It's very important to find all the points which satisfy this particular characteristic. Well, let's try to prove that no other point inside this angle but not lying from this ray will have the same property. So let's just assume that we have some other point which is not on this ray and try to prove that it's impossible to have the same ratio of distances. Let's say this is point Y. This is a distance. This is a distance. Now, how can you prove that ratio of these distances is not the same as these? All right, let's do it this way. Let's call this A, B, C, and G. So A over B is equal to M over N. And I have to prove that C over D is not equal to M over N. How can you prove that? All right, let's do this. Draw a line through point Y parallel to this. Now, obviously, the distance between these two lines is exactly C. So this will be C as well, right? This particular segment, the distance from this, let's call it D prime. So this is C and D, and this is C and D prime. Now, since this point was on this particular ray, which I know contains old points, which contains the points with this particular ratio of the distances, I know that C over D prime is equal to M over N. Now, what if C over D is equal to M over N? Then, if I assume this, I will have C over D prime is equal to M over N, and C over D is equal to M over N, which means D is equal to D prime. D is equal to D prime, which means this line should be parallel to this particular leg of the angle. But it's impossible, because this line is, by condition, by the construction, is parallel to this one. It cannot be parallel to this if these are not parallel among themselves. So that's the contradiction, which means C over D cannot be equal to M over N. Now, this concludes our proof that only this particular ray which I have drawn in the beginning contains all the points, not just the points, but all the points with this property of having the distances in a certain ratio. OK, the next. Points with given sum of squares of distances from these points to two given points. OK, so you have two given points, A and B. Well, you can say that they have a distance between them, D, because it's given. And now, x. I have to find all the points x, where Ax square plus Bx square is equal to some constant. Well, let's put it C squared. Doesn't really matter. It's a constant that's given to us. Now, you do remember from one of the previous lectures a theorem that if you have a parallelogram, then sum of squares of all four sides is equal to sum of squares of the diagonals. If you do not remember exactly, go to unizord.com, where in notes for this lecture, I'm actually putting the reference to the lecture which contains this theorem and the proof of it. I will be using this particular theorem right now. How? Here's how. What if I will build a parallelogram from this triangle by drawing line parallel to this and parallel to this and call it y? Now, Ax square plus Bx square, if I will multiply it by 2, I will have sum of all four sides, because these are equal to opposite sides in a parallelogram. So I know that 2C squared double this is equal to D squared. This is the given distance between A and D. And let's call it Z plus Z squared. Sum of all four sides, which is double this, which is 2C squared, is equal to sum of the squares of diagonals. Now, this is a known thing. This is a known thing. This is given. This is also given. It's a distance between. So I can very easily determine Z squared. Now, again, one of the previous lectures was actually showing how to build a segment with the length Z if I have other two things. Basically, Z squared is equal to, that would be the right thing. So if you have a right triangle, this would be the k-partinous, and this would be another k-partinous, then this is another k-partinous. So that's how I built Z using this particular construction. Well, half of this is the distance from X to P. This is the midpoint of this. So what I'm saying is that if you have sum of these 2 squared, given, it actually gives you the distance from X to the midpoint between A and B. No matter where my X is located, this is still the same distance. That's why the locus is actually a circle with this center in the middle of AB with this radius calculated using this type of thing. So that would be the locus of all the points. But this is the characteristic property of the property of sum of these 2 squares is equal to a given number. That's it. Next is the difference. OK. So now we have AB and X. We have AX squared minus VX squared is equal to something. Let's see, C squared. How to do this? Well, let's draw an altitude of this triangle. Now, obviously, AX squared is equal to, let's call it H, A, and V is equal to A squared plus H squared. This is the right triangle. Similarly, BX squared is equal to B squared plus H squared. So if you have a difference between them, as given, so I subtract C squared, that's the difference, is equal to A squared minus B squared, because H nullifies each other. So now, what also I know. I know that sum of these, so A plus B, is equal to the lengths of the AB, let's call it G. So G is also known very well. Well, from these two, I can very easily find A and B, because this is like system of equations. So let's use our algebra skills. A minus B is equal to A minus B times A plus B, and that's C squared. And A plus B is given, so it's A minus B times G. So A minus B is equal to C squared over G. So what do we have right now? We have two equations with two variables. Add them up together, and you will have two A is equal to G plus C squared over D. So that's where the A you can find. So knowing D and C, you obviously can find A using this equation. Now, you remember how to build C squared X is equal to C squared over D, because this is the mean proportional. And D is just the given thing. So then we just add them together, and you get double A divided by half, and you will have it here. So basically, what I would like to say is that if difference between squares of these distances is given, then this point is fixed always, because A and B can be calculated. I have just gone through the calculations, but that's not really necessary to do. That's for constructing this particular point. But if you don't really know where this point is, but you can still say that there is some point here, and all the points X are aligned on the perpendicular through this point. How to calculate this point, that's basically how. So locus is the straight line perpendicular to AB, which passes through one particular point, which has the same properties. Square difference between squares of these two things is equal to a given number. OK. So we can have one problem. Points that divide in the given ratio, all segments, from outside, from a given point, OK. Here is, so let's say you have a circle. You have a point somewhere. And then, if you connect any point on the circle with this one and divide it in some given ratio, this segment. So this is point which is given. This is the center. And this is the point A. And this is X. So AX over XP is equal to a given ratio. So we need all the locus of all the points X, which divide all these segments in the proper ratio. Well, including this one, which is intersecting at the center, which means BC is a diameter. So let's say it's somewhere here, point Y. Now, there are actually two different segments which you can derive from this thing. AP and A prime P. That's why I probably should do this as a B prime. So PA is one segment, PA prime is another segment. So that's basically another point which we are interested. And this point, X prime, is dividing A prime P. A prime X prime is divided in the same ratio. And same thing here would be Y prime, which divides in this ratio, E prime P. So now what should we do? Let's connect this to this and this to this. Obviously, since ratio is the same, then these triangles are similar to each other. So let's also connect this to this and this to this. And that's also a pair of similar triangles. Now, finally, if you connect this to this and this to this, you also have similar triangles. ABP and XY prime P. And you know that in similar triangles or all angles are actually equal to each other. Now, from all these similarities, obviously, you have similarity of AB prime and XY prime, because these are parallel to each other, which means angles are equal between parallel lines. So what does it mean? This is the right angle, because it's supported by half a circle. What it means also is that this is the right angle. Y XY prime is the right angle. Now, Y and Y prime are two fixed points. One divides BP in the proper ratio. Another divides B prime P in the proper ratio. So these points are kind of given. And no matter where other point is located, you always have, so either it's this one, doesn't really matter. You always have this particular angle as the right angle. This is another drawing. A prime B is the right angle. That's why Y X prime, Y prime is also right angle. So no matter which segment you will divide in this proper ratio by point X or X prime or anywhere else, you will have this Y Y prime a hypotenuse of a triangle, the right triangle, Y X Y prime, which means that the lobus is a circle around Y Y prime as a diameter. That's it for today. I would definitely encourage you to go through these problems again by yourself. Notes contain, use or dot com contain all the problems. Try to solve them yourself just to make sure that you kind of inculcate in your mind these different problems. And again, the more problems you will solve, the better you will be equipped to solve new problems. Because actually after a while, as I was saying before, the tools are relatively finite. So after a while, you will meet the problem which is similar to something which you have already solved. I would also like to encourage you to register as a student or if you're a parent who would like to control the educational process of your children, register as a parent. And as a student, you can take exams as a parent. You can basically see what exactly the scores of all the exams of your children are. And just to help you to have some understanding of where exactly your child is as far as studying math is concerned. OK, that's it for today. Thank you very much.