 Hi, I'm Zor. Welcome to Unizor Education. We continue talking about Apollonius problems as part of the advanced mathematics course presented on unizor.com. If you watch it not from this website, I do recommend you actually to go to this website and watch it from it because there are comments and actually the site has the whole educational functionality with exams etc. So, previous lecture was dedicated to relatively simple Apollonius problems when we have to construct a circle tangential to lines and points. Well, points means passing through actually. But again, the most important and interesting part of Apollonius problems are those when the base given elements are among other circles like lines and circles or circles and circles and we have to construct another circle which is tangential to these. And I promise that there is a special approach to these problems based on transformation called inversion. Well, first of all, let me talk about transformation in general. What is the transformation? We have actually learned many of those. We have learned scaling, for instance, which is the base of similarity. We have also talked about symmetry relative to a line when the left part of the plane is reflected to the right one and the right back to the left one. And those which are on the line itself, those points are transformed into themselves. Well, that's a transformation when for each point on the plane you have some other point which is related to it and the transformation is reversible because you can always apply this transformation backwards and you will return back to the original point. So that's transformation of points on the plane. Now, today I'm going to introduce something similar to reflection relative to the straight line. This would be a reflection or symmetry relative to a circle. What it accomplishes? It accomplishes a relationship between the points inside the circle and points outside of the circle. So remember, whenever we're talking about symmetry relative to the line left goes to right and right goes to left. In this case, inside goes to outside and outside goes to inside. That's the base of it. Now, what's very important is that in some way it's kind of magic, math magic if you wish. You probably are surprised when some magician is transformed one object into another right in front of your eyes and you don't know how it's done. Well, this particular transformation can help you to transform circles into lines and lines into circles. Now, when I learned this, when I was basically in high school I was really impressed, I mean emotionally impressed. It's really like a small miracle but in a completely different world. It's in the world of mathematics and it really helps you to solve some more important problems. So that was my kind of magical and emotional attitude towards this particular transformation and I hope you will kind of share it with me. So what is inversion or symmetry relative to a circle? Well, first of all, you have to have a circle. So let's consider there is a circle in the same way as you have a reflection relative to the line you have to have a line to define this transformation, right? So in this case, we have a circle on the plane. It has a center and it has a radius. Now, how the points are transformed one into another? Well, let's consider any point inside P. Now, to construct the point which corresponds to it in this transformation, the point which is a target, the image of this actually, you have to draw a ray from O through P and continue forward and outside of this circle find a point P prime which is characterized only by the lengths relative to this length. So this length O P times this length O P prime should be equal to R square. Now, another point inside Q, for instance. It's a little bit closer to the center than the point P. Again, if we will find on this ray point Q which corresponds to it, again, the requirement is that O Q times O Q prime should be equal to the same R square. Now, but if this is closer, which means it's smaller than this one, then this multiplier should be greater, right? So the closer point is towards the center, the further from the center would be its image in this transformation. Now, how to transform backwards? Well, exactly the same way. If you have any point here, you have this connection between this point and the center and find the point here which corresponds in exactly the same way to our main relationship between main rule of this transformation, the main inversion rule, if you wish. So it's reversible, obviously. It's one-to-one correspondence between points inside and points outside which means that the number of points is, well, infinite here and there, but this seems to be bigger, right? Because this is restricted in just one circle and this is the whole other points on the plane. But still, since there is a one-to-one correspondence, we can say that the cardinality of these two sets of points is the same. All right, so this is a transformation. This is a definition. How to construct an image of a point? Now, it's defined for every point outside. You can find the one inside. And as far as the points inside, for every point inside, except the center. The center is not really participating in this because it's zero lengths and there is, basically, you have to have an infinitely far positioned image. And, well, infinity is abstraction so we're not really talking about it. There is no infinitely remote point. So point O is excluded from this transformation. It's not an image. It's not the source for this transformation. So any other point inside this circle is transformed into a corresponding point outside and backwards. Now, the points on the circle itself are obviously transformed into themselves. Again, similarly to the reflection relative to the line when the points which are on this line are transformed into themselves by reflection, right? And why? It's obvious because this is already R, which means if you connect it and then you have to find the point which satisfies this particular equation and if one multipliers R, therefore the second one also should be R. So the points on the circle themselves are transformed into themselves. All right. Now, I will address two things. How the lines are transformed by this transformation, lines which are lying outside of the circle, how they are transformed inside. What happens with these lines? What's the image of the line? And how the circles are transformed? These two things. But let me just tell you in front before everything. Lines outside are transformed into circles which have one of the points which have the circle which passes through the center of the inversion circle, okay? So again, something like this will be transformed into something like this. And circles will be transformed into some circles here. So these are two things which I would like to prove to you. All right. So let's wipe out this and we will go to a concrete theorem number one about lines and circles. All right. So let's consider we have an inversion circle and a line. Now what I will do is the following. I will draw a perpendicular to the line. Let's say this is point P prime. Now this is point P such as that O P times O P prime is equal to R square. Now what I'm going to do is the following. Let's make a circle with O P as a diameter. Now my statement is that every point here would be reflected to point on this circle. That's the theorem basically which I'm going to prove which means that the line is transformed by inversion into this circle. Now how can I prove it? That's actually quite easy. Now this is Q, this is Q prime. So Q prime is any point on the line. Now let's connect Q and P. So what I know about O Q because it's O Q times O Q prime is also R square. So this is equal to this. Let me write it down. O P times O P prime is equal to O Q times O Q prime. From this we see that O P divided by O Q prime O P O Q prime equals to O Q divided by O P prime. Now let's consider these two triangles. O P Q and O P prime Q prime. Well this O P prime Q prime is right triangle, right? Because I said this is the perpendicular in the very beginning. I constructed it this way. Now what can I say about these two triangles? I see that line side O Q O Q divided by O P prime is equal to O P divided by O Q prime. So sides are proportional to each other. Angle is common. So we have a situation when two triangles have one common angle and the lines are proportional. Sides are proportional. Triangles are similar as we know from the rules of similarity. They are similar then obviously their angles are equal. So this angle is equal to this angle and this angle is right angle. So as we see points Q, whenever the Q prime is moving this point Q is moving but always triangle O Q P is the right triangle. So O P is viewed from the point Q at the right angle. And we know that the locus of all the points from which a particular segment is viewed at right angle is a circle where this segment is a diameter. That's very simple. Now by the way, I mean if you don't remember that this is true just try to prove it and the proof is really very very easy. So if you know that this angle is always straight, it's always right then this point belongs to a circle. It's a nice little theory. I mean the reverse is always true. I mean if you have the diameter then every point here views this diameter at the right angle. So you can always use this type of thing just to reverse the theory. So that's why Q always lies on a circle. Regardless of where exactly Q prime is this Q will be on the circle which means that the line is transformed into a circle. When Q goes this way you will get this piece of a circle and when Q goes this way you will get that piece of a circle and this point is transformed into this as our initial construction. So what have we done? What have we proven? We have proven that any line lying outside of a circle is transformed by inversion into a circle inside which is passing through a center of the inversion circle. Now if this line is closer then the P prime would be closer to the radius which means that OP also should be closer to the radius. So this line will go to a little bigger circle and if line is tangential to an inversion circle the circle which is the image of that line would be also tangential to this circle. So that's it. Now obviously there is a reverse transformation from a circle to a line because we know that the inversion is reflexive. So if this is transformed into this this will be transformed back into that. So any point outside would be on this line would be a point inside on a circle and any point on a circle will be transformed into the line. So that's how a problem which states for instance something like if you have these two circles and let's say a point and you have to construct a circle which is tangential to these two and goes through this point. What can you do? Well for instance you can do this. Let's use this point as a center of a big inversion circle. This is the center. It doesn't look like a center. Okay, it doesn't matter. This would look more like a circle. And we will transform everything using inversion. So this circle since it goes through a center will be transformed into this line. This circle would be transformed into this line and point will be transformed into point. Now what we can do? We can draw a circle which is tangential to two straight lines and passing through a point and then transform everything back. Now to transform everything back we know that this line will be into this. This line will be into this circle and this circle which is tangential to this would be transformed into something and that's my next theory which shows that the circle is transformed into a circle. So then you will get this as a result of transformation. Alright, so we need the second theory. We need to talk about how circle is transformed. Okay, so line outside of a circle of inversion circle would be transformed into a circle inside the inversion circle. Now let's talk about what happens with circles. Okay, so let's say we have some kind of inversion circle here. It's center somewhere. Now we have a circle here. So let's draw a line through the center. So this is all, this is M. This is A and B. Now this would be B and A, right? B is further from the center. So the B prime would be closer to the radius because product of O B times O B prime should be R squared and product of O A and O A prime would be as well. Actually, we don't really need this point wherever the center is. Now what I'm going to do is I'm building a circle using A prime, B prime as a diameter. And I'm stating that this circle would be transformed into this one. Let's try to prove it. Okay, let's take a point and draw it this way. Let's put it M and that would be N prime, M prime. Okay, let's connect A to N and N prime to A prime. So what do I know? I know that O N, O N times O N prime is equal to R squared. And the same thing I can say about O A. O A times O A prime, right? Because these points are images of each other. So from this I see that O N divided by O N divided by O A prime equals to O A divided by O N prime, right? From this. Again, O N divided by O A prime is equal to O A divided by O N prime. So what do we see right now? We see that the triangles O N A and O A prime and N prime are similar because proportional sides and common angles. Which means that other angles are also equal to each other, right? So, like for instance, this angle is equal to this angle. And this angle is equal to this angle. Now, what I would like to prove is that if I connect this to this I will get the right angle here. That's what I'm going to prove. And similarly I will connect N to B. Well, I know that A and B is the right angle. I know about that. So if I will prove that angle A and B equals to angle A prime and prime B prime that would actually prove that all these points and prime lie on the circle, right? So that's what I would like to prove. Now, let's consider the second triangles. What I know about the other triangle O N B and O N prime B prime I will have very similar thing. O N on N prime equals to O B times O B prime. Because both are R square, right? From here O N divided by O B prime equals to O B divided by O N prime. So again, O N divided by O B prime is equal to O B divided by O N prime. So obviously triangles O N B and O N prime B prime are also similar to each other. And the angles are therefore the same. Now which angles? This angle O N B, right? And this angle. That's what I know. And basically that is enough to prove that B prime N prime A prime is the right angle. Because, let's think about angle A prime B prime, sorry N prime B prime A prime N prime B prime. One which I would like to prove that this is the right angle. Now, if you edit with this angle plus angle N prime A prime B B prime. These two angles are equal to this one. Because this is exterior angle of the triangle. So any exterior angle has to be equal to sum of two opposite angles, right? So it's equal to O B prime N prime. O B prime N prime. Now, here I can see that angle O N A, angle O N A plus angle A N B, which is the right angle, plus 90 degrees, equals to angle O N B, right? Now, angle O N B, this is my wavy line, equals to this one, O B prime N prime. So these two are equal. Therefore these two must be equal. Now, angle N prime A prime B prime, right? A prime N prime A prime B prime, this one. So double, double arc is equal to O N A equals to this one. So these two are equal. So these are equal. These are equal. So these two must be equal. So A prime N prime B prime is equal to 90 degrees. And basically that's the end of the proof. Which means all points which are images of the points N, like here on the circle, will be points from which this diameter A prime B prime is viewed at 90 degrees. So this is also 90 degrees. And this, as we know, is a circle. This is a locus of points from which given segment viewed at angle 90 degrees. So basically a circle is transformed into a circle. And obviously this circle outside is transformed back into the inside circle, since everything is reversed. And as I was saying before, that allows us to solve the problems of Apollonius, where one of the components, given component, is a circle. That's the way how we are going to approach it. Well, that's it for today. I do recommend you to go to theunisor.com and do two things. Number one, read again the notes. And there are some drawings there, so you can follow basically. There are different letters, I believe. I don't remember what letters I was using in the notes. But I do suggest you to use it to look at it, to read it. And then close the website and try to prove the same thing on the piece of paper, just yourself. These two theorems, that the line outside is transformed into a circle, which goes through a center of inversion. And a circle outside is transformed into some circle inside. Now, obviously we can have a little bit different situation. For instance, if this is a circle of inversion, maybe we can have a circle, not just here, maybe somewhere like here, like concentric maybe with this one. These are all different cases. Now, there are also cases like this, for instance, which is not as trivial maybe as this one. But basically, the situation is the same. It will still be a circle as a result of the inversion. So this part will be inversed into this one, and this one will be inversed into that one. So that's all true. And I think I can leave it without the proof. Because for practical reasons, I will not have cases like this. Because I will have an inversion circle, which either puts everything outside or everything inside of it. But anyway, it's true as well. So any circle, even the circle which intersects the inversion circle will still be transformed into a circle. Or if it passes through the center of inversion, it would be some kind of a line. Well, basically that's it for today. Thank you very much. Try to follow my recommendation about going to this website and trying to do something yourself. That would be a very helpful exercise. Thanks very much and good luck.