 Hi, I'm Zor. Welcome to Unizor Education. I continue solving construction problems in circles department. Well, some of them are related to circles, some of them are just using circles. As I told you many times before, construction problems are excellent for developing your creativity and analytical skills. The best thing would be if you solve all these problems yourself and then listen to my lecture. In any case, after you completed this lecture, you understand everything. Go through the problems again yourself and try to prove or construct or whatever is required yourself without any kind of additional information. Everything is on the Unizor.com website, so please use it as much as you can. And again, self-study is extremely important. Don't just listen to me. Try to do it yourself before and after you listen to the lecture if you really want to develop that thing. Alright, so now the problems, construction problems here, some of them are simple, some of them are a little bit more complex. So try to concentrate and well, let's just go on. Construct a quadrangle that can be inscribed into a circle which is not given by its given three sides in a diagonal. So we know that we can inscribe this quadrangle into a circle, but the circle itself is not given. We just know that it's possible. What is given, three sides and diagonal. Okay, now how to solve this problem just using these four elements and the information about the fact that this particular quadrangle can be inscribed. Well, if you remember, quadrangles which can be inscribed into a circle have a very interesting property. Some of the opposite angles is 180 degrees. Why are these two because this angle is supported by this arc and this is supported by the complementary arc making some is basically the complete circle, which means some of these two angles is 180 degrees, which is because they are inscribed. Now, we will use this particular property and here is how. First of all, considering you have these three elements, you can build this triangle by three sides. So the only thing which we don't really know is position of this point, right? But now having these three points actually is equivalent to having a circle because with a triangle we can always circumscribe a circle around it. So not only we can construct this triangle, we also can construct the circle. So now, how can we actually find this point? Well, obviously, because since you know this distance, just use a compass and have this distance as a radius from this point and you get the intersection with this one. So that's the fourth point. Construct a rhombus by its side and a radius of an inscribed circle. Alright, let's have a rhombus. I'll position it this way. Now, rhombus, as you know, is a parallelogram with all four sides equal in lengths. Now, as you also know, the inscribed circle will touch all four sides and what's important is that this is perpendicular and this is perpendicular which means that the diameter of this circle is actually the altitude of a rhombus. So what we know about this rhombus is not only the lengths of the side which is actually lengths of the all four sides since they are equal in lengths. You also know the altitude. So how to do the construction? Have the line, this one, have the side and another line on the distance equal to double the radius, the diameter. Now here, using this as a radius, you just find this point and similarly this point and that's your rhombus. So in the rhombus, the diameter of the inscribed circle is the altitude. Construct a right isosceles triangle circumscribe, a right isosceles triangle around the given circle. So you have a circle and what you have to do is to build the right isosceles triangle, something like this. So this angle is 90 degrees and these two sides are equal to each other. So given is a circle, we have to build this. But let's just think about it. If you draw this line which is perpendicular from the top to the base, this is obviously an altitude and an angle bisector and the medium of this, I mean, all these things. And in addition, these are angles of 45 degrees each. Because the sum of them is 90 degrees. So what we also know is that this is 90 degrees, right? Which means that this angle is also 45 degrees. So if you have a circle, the simplest thing to do would be to have this radius into any point, doesn't matter. And at 45 degrees draw a line here and tangent at this point. And that's basically it and tangent at this point. So three tangents, one at this point, one from this point, which is an intersection of a perpendicular of this line, which we initially started, and then the third one, tangent from this point. We're just explicitly using the fact that this is 45 degrees. Construct an isosceles triangle by its base and a radius of an inscribed circle. So you have an isosceles triangle and you have an inscribed circle. So what you know is a base, which is AC, and you know the radius of an inscribed circle, OG. So let's think about it. Base and the radius of an inscribed circle. Now, all right. So first what we have to do is obviously have a circle from the base. So you draw a perpendicular with the radius OG from the middle of this point. So that's how you get this and now you can draw a circle. Now, having points A and C and a circle, let me just make it a little bit further. All you have to do basically is draw two tangents. Well, don't pay attention to my drawing. These are tangents to the circle. So since D is the midpoint, so AD and DC are equal, it's very easy to prove that these two lines will be congruent to each other. So that's basically how you construct it. That's it. That's all easy stuff. More difficult will be next. No, not next. Construct a triangle by its side and two midgons originating from the end points of this side. So if you have a triangle, you have a side and two midgons originating from this side. So you have a C, you have AE and you have CG. These are congruent segments because these are midgons, right? Now, as you know, midgons are divided by the point of their intersection in the ratio of one to two. So this is one-third and this is two-thirds of the midgen. Now, what it means basically is that if you will take two-thirds of AE, you will have AM. Multiply by two-thirds equals to AM. And if you have CG multiplied by two-thirds, it will give you CM. Now, how can you multiply a particular segment by two-thirds? Well, you divide it by three and then multiply by two, right? That's what it means. Now, I'm sure you remember how to divide a segment in three parts. You just take any other line, have three equal segments. It doesn't matter what's the length of each one of them. Connect the end points and draw parallel. Since these are equal, these will be equal to each other. We have proven this theorem before. So we divide it by three using this technique and multiply by two, which means just attach one to another. So this is how you get AM. Now, since you have AM and similarly we have CM, then triangle AMC can be built, right? So that's the beginning of your construction. You have AC, then you have AM and CM. This is your point M. Now, all you have to do right now is half of this, well, you already actually had one-third somewhere. So you just add another third and another third of this segment and here is your triangle. That's it. So you're explicitly using the fact that mediums are intersecting at a point which breaks each medium in the ratio of one to two. Okay. Construct a triangle by three mediums. Okay. One medium, two medium, three medium. A, B, C, D, E, F. We will do exactly the same thing. Explicitly use the fact that mediums are divided in this one to two ratio because look at this. If you will continue BE by the same length which is equal to this one and connected to this, so this will be M and this will be M. Let's see. Now, let's think about this. Now, AM is known. That's two-thirds of this particular medium. Now, M and is also known. Since this is one-third, this is two-thirds of this medium. And as it is very easy to prove, AMC and is a parallelogram because its diagonals are intersecting in the point which divides each one of them in half because these two segments have the same length since E is midpoint and these two segments have the same length by construction. So AM is also equal to two-thirds of the third medium. So this triangle has three sides, each of them equal to two-thirds of the corresponding medium which means we can start construction from building this particular triangle. Having three mediums, we just take two-thirds of each of these three mediums and start our construction from building this. So this is how it will be. This is our first step. Now, now it's easy. Point E is midpoint of MM, right? So this is E. We draw a line and continue by the same length and that's our point C. Now, if you continue this by the same sides, we will get B. This segment Bm is the same as Bm. This is two-thirds. This is one-third plus another third. This is also two-thirds of this medium. So that's our B. And here is our triangle ABC. By the way, this is something which I would say... I don't know. I do have sometimes some aesthetical feelings about mathematical problems. This problem seems to be has a certain aesthetic appeal, at least to me. I hope to you too. It's kind of very symmetrical. And it's just, you know, I don't know. It's just a statical appeal. That's all I can say. So we have built a triangle by three mediums. Given the circle and three points on it, construct a triangle such that it's three-angle bisectors intersect the circle and given points. Okay, that's interesting actually. So it might actually be a little bit involved. I do remember that it's not a trivial point. So let's say you have your triangle and you have three-angle bisectors. One-angle bisector, another-angle bisector, and the third-angle bisector. You need to build a triangle using these three points. So these three points are given. Okay. Now, we have to find these three points. So these are points of the original triangle and these are points where angle bisectors intersect the circumscribing circle. Well, let's think about this way. Since all these are angle bisectors, the corresponding arcs must be congruent to each other because arcs are angles, angles are the same. That's why arcs are the same because the central angles are the same. By the way, if I will use something like, okay, the measure of this arc is something and something, I actually mean the measure of the central angle which is supported by this arc is something and something. So if arcs are equal to each other or congruent to each other, it means central angles are equal or congruent to each other. That's obvious, right? So now let's think about the arcs. So since this is angle bisector, this arc is equal to this arc. In lengths and in the central angle, if I will draw a central angle which I don't want to do, it's too complex a drawing will be. Same thing here. This particular arc and this particular arc are equal because these angles are the same, are congruent. And finally, these three, these two will also be equal to each other. Okay? Fine. This is good. But now let's think about this in the following way. Let me put letters maybe so it will be easier actually. Let me just put this is x and this is x and this is y and this is y and this is z and this is... Oops. I'm sorry. This is y and this is y and this is z and this is z. All right? Now when I'm talking about x, y and z I mean angular measurement of these arcs which means measure of the central angle which is supported by this arc. So what I can say is that 2x plus 2y plus 2z equals 360. 2x plus 2y plus 2z equals 360 is a full circle which means the sum of these central angles is equal to 360, right? Now what also is important is the following. Okay. Let's connect these guys. Now points f, g and e are actually given. Now what does it mean? It means that we basically have expression for this arc from f to e which is z plus y. So z plus y, y plus z is known and this is equal to let's say a. Whatever the measurement of the central angle which is supported by this arc is that z plus y. Similarly, x plus z is also given and let's say it's b and similarly x plus y is also given and let's say it's c. So sum of this is one thing, sum of this is another, sum of this is the third one and we also know that a plus b plus c is also equal to 360. Well, to tell you the truth it's enough to algebraically solve these equations. Let's think about this. From this you get x plus y plus z is equal to 180 and since y plus z is a you have x is equal to 180 minus a. Similarly, from here we get y is equal to we subtract from this, this and we get 180 minus b and similarly z is equal to 180 minus c. Basically this is a solution because you know what to do. So you take the angle a which is I think it's this one y plus z is a central angle from let's say f o e. I don't want to put one, actually I can use another color. Okay, this is our a and this is our b and this is our c. So what you do is you take a hundred and h minus a which means you have an a and have the complementary to 180 angle. How is it called? No, supplementary. Supplementary, yes, supplementary angle. And supplementary to a would be x which is this one. Then y and z corresponding. So you have basically all the angles. Now having all the angles it's very easy basically to do the job. All you have to do is for instance you found x. Now having od and having x you just draw a central angle which you have just calculated and that would be your point b, etc. So I have demonstrated something which is maybe a little unusual I'm using algebra to solve geometric equation. Fine, whatever works. By the way the same thing can be geometrically described to you but the drawing would be a little bit more complicated. So I think it's as good basically there's nothing wrong with the solution. If you can find a better one send me an email, that's fine. Okay, next. Given a circle and three points on it construct a triangle such that three altitudes intersect circle given points. So it's exactly the same kind of a problem but instead of angle by sectors I have altitudes. So I have altitude to this. I have this point. I have altitude to this. I have this point and I have altitude to this and I have this point. So these are right angles. Okay, here is what's very interesting about this particular problem. Consider these two angles. This one and this one. They're both right angles, right? So they're equal to each other. Also, I'm sure you remember that any angle which is inside which is formed by two chords inside a circle, let's say this angle is measured as half the sum of this arc plus this arc. When I'm saying half the sum of the arcs means same half the sum of central angles supported by these arcs. So it's half of this plus half of this. Now, consider these angles. They're both 90 degrees. They're both right so they're equal to each other. Also, they are supported by this. Now this angle is supported by this arc and this arc. Now this angle, which is congruent to this 90 degree, is supported also by two arcs. One of them is exactly the same. So what does it mean? It means that the other arc which supports this angle. So let me put letters A, B, C, D, E, F. So what I have actually said right now is that arc, arc, C, D should be congruent to arc, C, E. C, D and C, E. Why? Again, because this angle is supported by this plus this divided by two. This angle, which is equal to this one, also 90 degree, is also supported by this. But now the second component is this. So if angles are the same and one component is the same of the sum, then another component of the sum should be the same. So these are equal in lengths, in measurements or whatever else. That's very important because now these points are given to you, right? So how can I find C? Well, if this is point C divides this arc in half, all I have to do is draw a chord and perpendicular to the chord and that's how I get the C point. Same thing here. I'll draw a chord and perpendicular, that's how I got this. And same thing here. That's the construction. So it was very important actually to realize that these two should be the same. The last problem. You have a triangle. I mean, you don't have a triangle. You do have a circle which circumscribes this triangle and you have three points. Not good. Let me try again. This triangle, okay? Now the median from this point, the angle bisector and the altitude. So, median which I called M, angle bisector which I call S and altitude which I call H. So these are points of intersection of median, angle bisector and the altitude from the same top point B. You have a circle and you have three points on it which are intersections of median, angle bisector and altitude from the same vertex. By these three points on this circle, you have to reconstruct the triangle. Let's think about it. Bs is angle bisector. So these two angles are congruent which means arc ACAS should be congruent to arc CS. AS and CS they must be congruent to each other because this angle ABS is supported by AS and SBC is supported by CS. Since angles are the same, inscribed angles the arcs must be the same. Central angles which are supported by these arcs must be the same. S is a midpoint of the arc and therefore if you connect a center of a circle with the midpoint of an arc it must cross the cord which connects end point of this arc. It must intersect the cord in the middle and be perpendicular to it. Remember, let me just draw a different drawing. It would be more obvious. If this point divides the arc in half then the cord which connects end points of the arc would be perpendicular to radius of the midpoint of the arc. We have proven this many times before. What it means is that if I will connect the center of the circle with a point where angle bisector M, S and H. If I will connect this it will be perpendicular to this cord which is basically a triangle and it will divide it in half. But now, BH is also perpendicular to the same AC because this is the altitude which means that OS and BH are parallel to each other which means if I will draw this I will hit the point B. Hooray! We've got one point of our triangle. Now, if I connect B and M it will intersect this radius OS will intersect in the middle of AC. Because AC is the side and BM is medium. If I will connect this to this the intersection point is the midpoint of AC. Great! Now in this I will just draw a perpendicular which is perpendicular to our height of point A and C. So that's our triangle. That's it. Very easy, right? This is actually another construction problem which I consider not only mathematically so to speak but also aesthetically. When I first look at the problem you have these three points on the circle of the bi-sector and altitude it's kind of scary because you don't really know where to start looking for whatever the solutions are. But then if you think about this and you realize that the bi-sector divides the arc in half and that's why you can just draw the radius to this it will be perpendicular to the core. It unravels so to speak but in the beginning it's kind of interesting and a little bit scary for me it was. But then if you think about this you find the solution. That's exactly what the whole course is designed for. It's designed to push you towards solving problems which you just don't know how to solve. I did not know how to solve this problem until I realized I think about it and come up with a solution. Thank you very much. Please do all these problems again yourself. Go to the website and just look at the notes to this lecture. All of them are there and don't forget you can enroll into a specific course or your parents or teachers or supervisors can enroll you where you will have to go through exams, get some score and make sure you will get the maximum score available. Thank you very much. That's it for today and good luck.