 Hi, I'm Zor. Welcome to your new Zor education. I continue with construction problems for circles. Don't forget that there are many other lectures before this one. This is number six of construction problems for circles. There are five others before it, and I'm going to use the results of the previous lectures obviously. So if I can reduce my problem to a previous one, I'll just refer to this and you will have to look for the same kind of a simpler problem in the previous lectures. Now, in any case, it's absolutely mandatory for you before you're listening to this lecture to think about all these problems yourself, spend some time if you solve it great. If you can solve it, at least you will spend some time to get into the problem, to realize what different constructive solutions at least come to your mind. So it's very important as a process of thinking, process of trying to find the solution. Because that's the whole purpose of this website, just to teach you to learn, to study, to think about creatively, to basically develop your analytical thinking and stuff like this. Because none of the information which you can gain from these problems would really be useful in your practical life. It's all brain exercises, nothing more than that. So consider this in exactly the same fashion as you exercise your muscles when you go to gym. Right. Let's go. Given the circle and a sector in it, bordered by two radiuses and an arc. So you have a circle and you have a sector, bordered by two radiuses and an arc. Okay. Inscribe a circle into this sector tangential to both radiuses and an arc. So you have to build a circle which is tangent to these two, and to an arc which is basically for a circle itself. Now, in many cases, I actually was trying to make this point. If you can reduce the problem to a simpler one, well, do it by all means. You can replace a more complicated geometrical object with less complicated. Now, straight line is simpler than a circle. Well, I don't know how to define the word simpler, but it looks like to solve problems with straight lines is easier than with circle. So if I will be able to replace this problem with a simpler problem, where instead of circle, I have a straight line, well, that would be basically half of the solution if you wish. Now, this is the center, and this is the line which connects the center of a given circle with this one. This would be a point of tangency. And if I draw a line perpendicular to this, that would be actually the tangent to both circles, right? So now, consider this particular triangle. A, this is A, this is B, this is C, this is D. Consider triangle OCD. Well, this circle which we are looking for is inscribed into this triangle. Which means what? Which means it's on the crossing of its bisectors and the bisectors, right? So all we need to do right now is to build this triangle, and then basically we can reduce the problem to the one which we know how to solve, how to inscribe a circle into a triangle. Now, how can I get this? Well, very simply, if I have a circle and I have a sector, well, obviously this is a bisector of the angle. So you got the point, then you draw a tangent, and basically expand these two radiuses until they intersect with the tangent. So you got the triangle, and now reduce the problem to inscribing your circle into this triangle which means crossing of bisectors, angle bisectors. That will be a center. That's it. Always think about how to simplify the problem by reducing complex elements into more, into a simpler one. Inscribe three circles of equal radiuses into a given circle, such that each circle is tangential to all others. All right, so each tangential to all others. So you have three circles of equal radiuses, and you have to inscribe it into a circle which is given. So this big circle is given, three sectors are those which you have to find out. All right, now what you do know is they all have the same radius. I'll do exactly the same thing. I will simplify my problem by basically doing the following analysis. If I will draw tangents here, here, and here, just for pure considerations of symmetry, which definitely can be proven very rigorously, but I'm not going to do this. And it's obvious that this triangle is equilateral, because all these radiuses are the same, and there are many different ways to prove that this is an equilateral triangle. I'll leave it basically to you. So number one, what you can see is that all these circles are tangential, not to this circle which is given, but to a triangle which is around it. Now, how to build the triangle around a given circle, if this is an equilateral triangle? Well, that's simple. And again, since these are equilateral, these are all 60 degrees angles. So you figure it out how to build a triangle. That's easy. Now, number two, let's draw millions or bisectors, whatever it is. They're all the same. Now, as you see, this particular circle is supposed to be tangential to this line. And this one as well between these two. And this between these two. Again, from pure considerations of symmetry, this line, which is a bisector of this angle, cannot have more than one point of intersection with any of these two. Because if it has two on this side, it does not have any common points with this side, which is against the considerations of symmetry. In this case, when I'm talking about symmetry, I'm talking about basically symmetry relative to a rotation by 120 degrees, if you wish, because this is a 120 degrees angle. So the picture can be turned by 120 degrees, and it's not supposed to change. That's why these are tangents to both circles. So whenever you have this triangle and you have these three angle bisectors, you have right now every one of these three circles basically inscribed into correspondingly this triangle, this triangle, and this triangle. So what's the sequence of steps you have to do? If you have a circle which is given to you, circumscribe it with a triangle. If you circumscribe a good word for this, anyway, build a triangle into which this one is inscribed. And this triangle should be equated, pick the collateral. Then, well, by the way, how to do it? Well, pick any point, let's say this one, draw a perpendicular, then draw a line at 60 degrees until it touches this one. And draw this line 60 degrees. I mean, it's really very easy how to do it. I don't want to go into the details. OK, once you've done that, once you have circumscribed a circle with a triangle, which is not the right word, once you inscribe a circle into a triangle, that's probably a better way. You draw bisectors, angle bisectors, or median or altitude, they're all the same, right? Because it's an equilateral triangle. And into each of these three smaller triangles, you inscribe a circle. And that's your solution. I think it would be a very interesting exercise for you to prove that this is exactly the right solution. It's easy proof, so that's why I don't want to stop on these details. Given a circle and a point inside it, construct a chord containing this point such that the difference between two segments, the given point divides the chord, is equal to the given segment by c. So you have a circle. You have a point. So you have to build a chord, mm, in such a way that am minus am equals given segment a. All right. How to do it? Well, actually, in this case, the best way is draw a circle with the same center and with a radius OA. It will be concentric. Now, let's think about point b. Now, since these are concentric circles, ma is equal to bm. This piece equals to this piece. How to prove it? Well, easily, actually. You can draw a perpendicular here, connect this, connect this, connect this, and connect this. And obviously, this triangle is equal to this one, this triangle from the center k. k and o is equal to km o. And that's why kn is equal to km. So from kn, we subtracted kb. From kn, we subtracted ka. ka and knb and kb are equal to each other again. So that's why we have equal pieces. Now, so what is ab in this particular case? Well, obviously, ab is equal to an minus m minus bm. an minus bm. That's what ab is. But since ma and bm are the same, this is equal to an minus an. So whenever you draw a concentric circle, this chord must have this length given to us. So how to construct this? Well, very easily. Again, you have a regional circle. You have a point, a, draw a circle concentric with a radius oa. And now, within this inside circle, you just have to build a chord of the given length, which means you just take the compass, point one leg into the a, and these are two solutions. So this would be a chord looking for, and this would be the chord you're looking for, two solutions. Given two intersecting circles, construct a second that contains one of the intersection points between these circles such that its part inside a union of two given circles is congruent to a given segment. You have two circles, and you have to build a second, which is passing through one of the intersection points in such a way that a, c is equal to a given segment a. This is a, c. So given a segment and two circles crossing each other, so point of intersection is the point which you have to use to build a second in such a way that the part inside the union of these two circles is given. OK, so how can we do this? Well, as usually, we can have two perpendiculars, which divide each chord a, b, and b, c in half. Now these are two center lines, centers. So now obviously, the lengths between k and l is equal to half of a, am I right? Since this is half of this and this is half of that, then together they are half of the whole thing, which means using the hypotenuse and mp is parallel to kl. Now these are perpendicular, so this is the rectangle, which means mp is also the same thing. So triangle mpn can be constructed, because you know the hypotenuse, which is given, it's two centers, and you know the catatose, mp, which is half of the given lengths. So the way how you approach this construction problem is as follows. You built this triangle using catatose and hypotenuse, and once it's built, you basically use, let me just draw it again. So you have one circle, you have another circle, you have these points, and the first you do is, this is b, the first you do is you construct a triangle by this as a hypotenuse and this as catatose. So you have the point b. Now you continue the point b further, and from this point b, it's a perpendicular, right? So you drop a perpendicular here. It goes parallel to this. Or basically, you don't have to go to perpendicular, you can go parallel to mp. That's not really meant to. And then your combined lengths of these two chords will be twice as big as mp, which used to be half of the a. All right, so that's the construction. So what was important here, just to realize that again, by the way, whenever you have chords, you probably have to build the perpendicular to the chord from the center, which divides it in half, and then you're using that this is half. Given a straight line, point on it and point outside it. Straight line, point on it and point outside it. Construct a circle tangential to a given line at a given point, and containing the other given point. So you have to build a circle like this. Well, this is really easy. And in the center, it's supposed to be on the perpendicular to the given line in the given point. And also, it's supposed to be equidistant from these two points, which means it's supposed to be on the perpendicular by sector. That's easy. That's not even worth the time. OK, inscribe a circle of a given radius into a given angle. That probably is easy as well. So you have to inscribe into an angle if you're given a radius. Well, obviously, the bigger the radius, the further it is from the vertex, right? Now, but if you are given the radius, it means that your center is on this radius distance from this leg, which means if you draw a parallel line on this distance, it will be a locus of all points, of all circles of this radius, which are tangent to this line. And same thing for this particular leg of the angle. You draw a line on the same distance from this leg. And the crossing of these two lines is your center. Easy. Also, not worth the time. We are much more sophisticated right now after all these problems, which we have solved, these are just peanuts, right? Construct a circle tangential to the two given circles with a given point of tangency on one of them. All right. And there are different cases, I guess. So you have two different circles, and you have a point of tangency. So it would be like this. So you are given A and B circles, and you are given a point M. Now, obviously, you should connect these. So this is a circle which you have to find out. You have to construct. And obviously, one locus is connect A to M, and that would be the line where your center is supposed to be located, right? But that's not enough. You need something more than that. All right. Now, the first thing which you can do is you can certainly simplify the job by blowing up this circle by this radius. Now, I'm converging, basically, the problem when this circle is tangential to this circle to another problem where the bigger circle is just passing through a given point, which is the center. So I'm reducing my one circle, which is given. Let's say this is bigger than this one. So I'm reducing the bigger by the radius of a smaller one. So this, call it C, let's say. So AC is equal to AM minus VM. So I'm expanding this circle, which this is analysis. So I'm presuming that this is already drawn. So I'm expanding the radius by this length, and I'm reducing this radius by the same length, and this radius by the same length. So this is transformed into a point, and this is transformed into a smaller circle. And now, quite frankly, the situation is, well, similar, but simpler. Now, similar in the way that also we have to find exactly the same point O, but it's simpler because now you can reduce the problem to a different one. You have to find a circle, which is tangential to a given circle in the given line at a given point and passing through a given point outside of the circle. Now, we already solved that problem because the way to do it is notice that O, C, and O, B are radiuses of this bigger circle, which means the point O is equidistant from B and C, which means it's on the perpendicular bisector between B and C. Now, after all these analysis, how to build the circle which we're looking for? Well, step number one, reduce the size of this circle by this radius so you get a smaller circle. Then you draw the line to a given point, M is given point, and C is just crossing with a smaller circle. So you have point C. Then B and C, you use it as two points from which you are making a perpendicular bisector to a segment connecting them. And these are two lines. You have original one, AM, and this perpendicular bisector gives you the center of the circle which you're looking for. Now, there might be some other cases which are probably more or less the same. For instance, you can probably think about this circle lying inside of this rather than outside. It's slightly different, but in theory, it's exactly the same thing. So if you just draw a picture, you will see that this is more or less the same. So you have a circle and a point on it where the tangential events supposed to occur. So you're looking for a circle which is tangential at this point and also a tangent to this. So this is given, given a circle with a center A, a circle with a center B inside it, and there is a point M on this circle. So we have to find this circle with a center O, which is tangent to this one and tangent to the bigger one at the given point. Now, here we do exactly the same thing. We expand this circle by this radius. So it will be like this. Now, if we expand now instead of reducing, now we will expand our given circle with a center A by the same radius. So it will be like this. And now we have done exactly the same thing. We need, this is also a known point, so we need to build a circle which is tangent to a given circle at a given point and passing through a given point inside it. Again, you have two points. You draw perpendicular bisect. Same thing. So basically, no matter how it's located, we can also do the same thing. Now, the very last situation is if you want an inside tangency between this and this, I don't know if it exists, actually. It might or it might not. The tangent comes on the situation. But anyway, if I want to do something like this, that is also possible, right? So this is A. This is B again. Now, what should we do in this case? Well, in this case, we should reduce the radius of the circle which we need by this radius and expand this one by the same. And again, we have actually reduced the problem to having two points equidistant from this one. The center B and this point, which is crossing through a bigger circle. And so again, perpendicular bisector gives you another focus. All right, so in many different combinations, you can consider this particular problem. And it's solved in exactly the same way. And this way is, let me just repeat, it's a simplification of the problem. You're reducing more complex geometrical object, which is a circle, with a simpler one, in this case, a point. And basically, you have reduced the problem to the one which you have already solved many times before. All right, that's it for this particular lecture, these problems. I'm not sure. Probably it's not the last one which I will do for circles. In any case, it's very important for you to spend time to do yourself all these problems. So please don't hesitate and pay attention to Unisor.com. For parents, for teachers, it will give you the control over your students' educational process, including exams, et cetera. Thank you very much.