 Hi, I'm Zor. Welcome to a new Zor education. I would like to spend some time talking about Apollonius problems in geometry. Well, as you understand, Apollonius is a Greek mathematician with very long time ago, and he presented a certain number of problems. Well, some of them were solved by different people, including himself. And they have certain commonality as far as the problems are concerned. Solutions are completely different, but the problems do have certain commonality. Well, this lecture is part of the advanced mathematics course for high school students. It's presented at Unizor.com, and I would suggest you to watch this lecture from this website directly, because it has notes, and also there is a bunch of functionality, which allows you to basically self-study the whole course, including exams. Anyway, back to Apollonius. Okay, so the problems which are currently described as Apollonius problems are the problems of constructing a circle, which is tangential, or passing through, in case of a point, which is tangential to three other elements of geometry. Elements are points, lines, and circles. So, if you have three elements of this set, let's say three points, or a point, a circle, and a line, or whatever. So, if you have three elements from this set, then your task is to construct a circle, which is tangential to all these three elements. And in case of a point, it's passing through, obviously. Now, obviously, there are many different cases, like point, point, and point, or point, line, and circle, or circle, line, and circle, etc. So, we will try to address all of them. Now, why I think this is a very interesting topic? Because in certain cases, it actually leads to introduction of a new transformation on the plane. What we had before, as far as the plane transformation, where, excuse me, transformations of symmetry and scaling, which leads to similarity. So, as far as this new type of transformation, it's called inversion or symmetry relative to a circle. That would be a subject of our next, excuse me, of our next lecture. And this lecture is about a simpler cases of Apollonius problems. Now, simple cases are those cases where the given three objects are only lines and points, no circles are given. And you have to construct a circle, which is tangential to all three of them. Well, obviously, there are many different cases, like three points, two points, line, one point, two lines, and three lines. These basically exhaust all the different possibilities of having points and lines as objects which are tangential to a circle. So, let's just do it one by one. So, the first is point, point, and point. And obviously, we have to consider case when a circle goes through these points. Now, obviously, there is no solution if three points are on the same line. You cannot build, you cannot construct a circle which is passing through these three lines, three points lying on the same line. But if they are not, we all know that there is a very simple solution. If you have a triangle and you have to circumscribe a circle around this triangle, what you have to do is to build perpendicular bisectors to every segment. Now, let's address this problem a little bit more accurately. Well, first of all, if you have two points, now locus of the points, so the set of the points which satisfy the requirement of being equidistant from these two points is, as we know, perpendicular bisector of this segment between these two points. Now, we really have to address the proof of this. And proof is really trivial. However, you have to consider two different theorems here. The theorem number one, that every point on the perpendicular bisector is equidistant. And the second theorem is that any other point which is equidistant from these two lies on this perpendicular bisector. To eliminate cases like this, for instance, maybe this is also equidistant. So, to say that this line, the perpendicular bisector is a locus of all points equidistant from these two given points, we have to prove both. That these are equidistant and any other equidistant point actually lies on this perpendicular bisector. Now, both proofs actually are very trivial. Now, in case this is a perpendicular bisector, now obviously, since it's a bisector, this is equal to this, because that's the construction of the bisector. And perpendicular means that these are right angles, which means these are two right triangles which have congruent catchity and another catch is shared among them. So, these two triangles are congruent and that's why hypotenuses are congruent. Now, the other theorem, so let's consider that this point is equidistant. How can we prove that this is on the perpendicular bisector? Alright, let's think about it. If it's equidistant, so these are two points and this one is equidistant between this is equal to this. Now, this is isosceles triangle. In isosceles triangle, the altitude and the median and bisector all are going along the same line as we know, right? So, let's just draw this line. Okay, this is an altitude of this triangle, which means this is right angle, but it's also a median, because it's an isosceles triangle and that's why these two are equal in lengths and that's why this is a perpendicular bisector. So, we have proven that perpendicular bisector is a locus of all points equidistant, which means that if you have three points A, B and C and you construct a perpendicular bisector of AB and then perpendicular bisector of BC, since they are not on the same line, these are not parallel, which means there is an intersection and since this is an intersection, let's call it P. I can say that P A is equal to P B because of this and because P lies on this, P B is equal to P C and since this type of equality is transitive, P A is equal to P B, P A is equal to P B equal to P C and therefore, since P A and P C are equal to each other, that's why P should actually lie on the perpendicular bisector to AC as well. So, that's why all three perpendicular bisectors are intersecting in one and only one point, which is a center of circumscribing circle. So, this problem is completely exhausted. It's a very simple, it's probably one of the simplest problems among all these colonial problems. The second one is a little bit more complex. So, we have two points and a line. So, we have two points and a line. And we have to construct this circle, which is tangential to a line and passing through these two points. Okay, the way how it can be done is the following. If we will connect A and B, let's assume there is a point of intersection. If they are parallel, the situation is actually simpler. Alright, anyway, so let's consider that there is an intersection point. Let's call it M. And this point, tangential point is, let's say, T. Now, consider these two triangles. M, A, M, T, and T, B, M. Well, one angle is common, right? So, the small triangle is obtuse and the big triangle. This is obtuse angle. Now, how about the other? Now, this is tangential line. Now, the tangential line, if this is a tangential line, then this angle is equal to this angle. I don't think I have to prove it because I definitely have proven it during the regular course when I was talking about circle and tangential lines, etc. The way how you can prove it, for instance, is this. You have a central angle, which is twice as big as inscribed angle, right? So, half of this angle is equal to this one, right? But now, let's consider, let's say, this is letter P, angle P-O-T and angle A-T-M. These angles have mutually perpendicular sides, right? Because O-T is radius, it's perpendicular to tangential line, M-T, and O-P is perpendicular to the chord, A-T. So, we have these two are perpendicular to these two, and that's why they are equal. So, that's why these two equals. And since we have two equal angles, this is equal to this, and this is a common, these triangles are similar, right? Similar triangles. Which means we can actually arrange a proportion among the sides of these triangles. So, let's say the small triangle, the big one, M-T, divided by A-T. So, these are two sides which form this angle. Now, in this case, big sides which are forming this, the big triangle, the two sides are, the bigger one is B-M, and the smaller one is B-T, right? M-T to, no, that's wrong, to M-T, one second. M-T to A-T, to A-T, okay? Is related as B-M to B-T. No, I'm right, to B-T, right? And also, another proportion is, in the small triangle, A-M, this is the opposite to this, to M-T, equals to, in this case, opposite to this one is M-T. M-T to B-M, right? So, let me just check again. A-M is in the small triangle opposite to this angle. M-T is in the big triangle opposite to the same angle. Now, M-T is the bigger side of the small triangle, and B-M is the bigger side of the small triangle. Now, let's see which one we can use. Actually, I think we can use this one, because A-M and B-M, A-M and B-M are known, and M-T is unknown, right? So, M-T squared equals to A-M times B-M. So, knowing A-M and B-M, we can construct M-T. I'll show you how. And once M-T is constructed, then the whole construction of this circle is, connect these two things, get this point. Then calculate M-T, and you get this point, and then you have all three points, and that's enough to construct a circle. Now, my only open question is, how to construct M-T, how to construct basically one segment, if I know that the square of the segment length is the product of these two lengths? Well, this is, again, a known problem, and I'm sure I addressed it before somewhere. If you take, let's say, a right triangle, this is H, this is A, this is B. This is exactly the situation when H squared equals A times B. Why? Because these two triangles are similar to each other, obviously, right? Because this angle equals to this one, and this angle equals to this one. As mutually perpendicular angles. So this is perpendicular to this, and this is perpendicular to that. And since these are similar triangles, then their ketities are, in this triangle, smaller to bigger. In this triangle is smaller to bigger. And that's why you have the same thing. So basically, you have to construct an H if you know A and B. So this is A, this is B, and this is H, right? Now how to do it? Well, very simply, since this is the right triangle, then all vertices with known hypotenuse are on this circle, right? So we draw the circle on A plus B as a diameter. Now we have A and B, so in this point we just make an altitude until it hits the circle. And that would be the length of my H, that would be the length of my MT. So that's the end of this construction. Now in case A, B is parallel to D, the situation is actually simpler. So in case these lines, this is A, this is B, and this is D, then from, obviously you understand that if these are parallel, now the radius which is perpendicular to the chord divides it in half, right? And the same radius which is perpendicular to this one is perpendicular to this one. It actually goes to a tangent point of tangency. So if you take A, B and you have a perpendicular bisector, it hits exactly at the point of tangency. And that's the third point. So that's simpler, right? Okay, so that's the end of this problem which is two points and a line. Now two lines and a point. Okay, let's consider the lines are intersecting and the point is somewhere in the middle. And we have to construct a circle which is tangential to both lines and passing through the point, all right? So let's call it P, let's call this one A. Now, what can I say? In this particular case, what's important is that if I will use a scaling with a center at P, my this line and this line during this transformation of scaling will be transformed into themselves, right? Because any point will be stretched or squeezed but it will be still on the same line, right? Now circle will, let's say circle will grow and point A will go to A prime. So that's what happens if I, let's say, stretch it by a factor of 2 in this case or something like this, all right? Now obviously the circle will be transformed into a circle. Why? Because the length of any segment is proportionally increasing during the scaling, right? So the length of this, which is the radius, it's the same for all these points, right? So if this is all prime, image of my center, then these segments will have the lengths of these segments multiplied by the scaling factor, all of them. And since these are the same, all of them are multiplied by the same factor, we will have again an object with all points equidistant from some central point, which is a circle. Now tangency again will be transformed into a tangency because it's only one intersection point. One intersection point is basically transformed into one intersection point. So the circle will be still, the bigger circle will be still tangential to the same line. So what I can suggest right now is the following. Let's just build any circle which is touching these two sides, these two lines. So that's simple, just do an angle bisector, right? You know that. And any point on this angle bisector can be used as a center. Okay, so we've got this circle. Now we draw this line Pa and it will be a prime intersection with this circle. So now we know actually the factor because Pa we know and Pa prime we also know since the circle has already been built. And it's exactly the same factor as between Pm and Pm prime or Po and Po prime, etc. So using the same factor we can just squeeze from this center and get this point and get this point. This center of the circle which we actually need because if we will squeeze everything in such a way that A prime goes back to A, then O prime will go to M and M prime to N. So what's sufficient is to construct this circle, draw the line through this given point A until it intersects this one and then squeeze to whatever the factor is needed. Okay, variation. Now obviously there are different variations because you might actually instead of this point A take this point A prime to double prime. In this case we have to stretch it a little bit and it will be a circle like this. So it will still touch both lines and it will still go through this point A. So there are different solutions, two solutions in this particular case because you can take either this or this point in that new circle which we have created. Now if lines are parallel, well the situation is exactly the same. Instead of angle by sector we will have the midline which is parallel to these two and do basically something similar because if this is your point A you know actually where the center is supposed to be located and we know the radius because the radius is the distance, it's the same. So using this radius you can just find these two points and these are two different circles, centers, this one and this one which constitute the solution. So this is again simple. Okay, so that's my second problem when we have two lines and one point. And finally for this lecture three lines, well three lines again is very simple thing. They are not parallel obviously because the parallel lines is not possible to create a circle which is tangential to three parallel lines. But if they are not parallel lines it's a triangle, right? And for the triangle if you need to build a circle which is tangential to all three lines, well that's angle by sector obviously, right? But now let's think about it a little bit more. We have more angles. You see lines are this, they are continuing. So not only these by sectors we have. We also have these by sectors, these by sectors, these by sectors. So we have new points actually, right? So we can have a circle which is tangential to all three lines from the outside, right? So we have one, two, three, four, four different solutions. We have four different circles depending on which angle we use as a bisector, right? Okay, now to prove that bisector is a locus is very similar to the one which we were using for inscribed, I mean rather circumscribed circles in the very beginning. So I don't want to spend any time on this, it's simple. So basically these are three simplest problems among Apollonius problems. These are only where points and lines are used as given and the circle is supposed to be constructed. Then the next lecture I will probably spend talking about the transformation of inversion which is symmetry relative to a circle which would allow us to reduce problems when the circles are given, one, two or three circles are given, back into the lines and points problem. All right, that's it for today. I suggest you to read the notes for this lecture on Unisor.com, very useful. And the notes are not like complete proof or anything like this. I left a couple of statements like prove it yourself. I do suggest you to try to do it. It's very, very good exercise. Thanks very much and good luck.