 Hi, I'm Zor. Welcome to Unisor Education. I'd like to spend some time solving very simple problems related to circles. By now, we have accumulated non-theoretical material to start solving problems, which is basically the main purpose of the whole exercise with website and all this educational material which we are providing on Unisor.com. Again, you remember that creativity is the ultimate goal of studying mathematics. So solving problem is the way how you basically create certain things. It's very important that you try to solve certain problems which you have not solved before, because that's exactly what's supposed to develop your creativity. That's the purpose, right? Okay. So construction problems in geometry are very much in line with this idea of creativity, how to construct something using whatever the tools you have and you know the tools are compass and straight edge ruler. All right. So if you construction problems related to circle, very simple by the way, I'm delaying with all the real complex problems until later when all this theoretical material will be completed. All right. So number one, divide in half and arc in a circle. All right. So if you have a circle and an arc from here to here, and you have to divide it in half. Well, obviously, you connect it with a cord and you drop the perpendicular from the center and continue. That's making a diameter actually. If you remember, diameter is the axis of symmetry of a circle, which means the left part is symmetrical to right part. Now, all the points on the left side of this arc are symmetrical to corresponding points on the right. Why? Because this one, this point has exactly the same distance from the diameter as this point, as we have proven many times before. Well, just in case, if you don't remember, you can always connect the ends of the cord of this cord and these two triangles have common side and equal in lengths. Hey, partner moves this because there are walls, radius is on the circle. So triangles are congruent, which means that these particular legs are equal in lengths. So for every other pair of points which you can obtain by drawing a perpendicular anywhere in this particular area, which means that the entire left part of the arc is symmetrical to the right part, and that basically proves that we have divided by this point, this arc into congruent parts. In this case, congruency is basically established by reflection relative to the axis. Okay. Next, find a center of a circle. I didn't mention actually this problem many times. If you have a circle, just pick up a couple of points and you know that the center is supposed to be on the perpendicular bisector between these two points. Now, have another pair of points, let's say this and this. Again, draw a cord and draw another perpendicular bisector. On the crossing, you will have the center of the circle. That's it. By the way, it doesn't really depend on what kind of points we have chosen, as long as it's just three different points. Construct a tangent to a given circle, the given point of their intersection. So if you have a tangent, which has one and only one by definition, single point of intersection with a circle. Now, how to construct this particular tangent if you know only the point? Well, obviously, if you remember, since tangent has one and only one point of intersection, that's the shortest distance from a center. Because on a circle, you have all the different points which are at the same distance, basically from the center, the ranges. Everything outside has the distance from the center greater than the radius. So this particular radius to the point where our tangent intersects with the circle is the shortest point from this line. So the way to construct it is if you have only the point, you have to draw a radius to this point from the center, and then the perpendicular line to this radius at this particular point is a tangent. And by the way, I don't want to stop on how to construct the perpendicular to a line at certain point. We already covered that in previous lectures. Next, construct the tangent to a given circle from a given point outside of a circle. Okay, so again, you have a circle and you have to construct a tangent if you have this point. So the tangent is not there. Tangent is the one which is supposed to be constructed. You have a point and you have a circle and a center obviously. Now, this is a little bit more tricky. Yes, we still know that this particular point should be found where the radius is perpendicular to the line. So we have to find out somehow how to draw this line if you have only one point. We don't have any other points. We don't have the tangent. But let's recall that since this is a perpendicular, now if you remember, if you have an inscribed angle which is supported by a diameter, it's always 90 degrees. Why? Because it's always half of a central angle and central angle in this case, which is supported by the same R is 180 degrees. So any inscribed angle which is supported by the diameter is the right angle. Now, this is also a right angle, right? And if we will draw a circle with this as a diameter and we do know this particular segment because we know the circle with its center and we know the point the tangent is supposed to pass through. So we have this, so we can actually build a circle on this segment as a diameter. Well, how? Just divide it in half and then use this as a radius. Now, since this is a circle with this as a diameter, then any right angle which is using this as a hypotenuse has to be on this circle. So basically this is a locus of all the vertices of a right triangle with this as a hypotenuse. Back to this particular drawing. If you have a diameter, then the circle around this diameter is a locus of all the vertices of all the different right triangles which have this diameter as a hypotenuse. Now, this particular triangle is the one which has this as a hypotenuse, which means that now we can find this point as an intersection between this circle which we know since we know it's diameter and the main circle from which we have started. Well, actually there are two points, right? So this point will serve as well. We can draw two different tangents from one point to this particular circle using this technique. So again, we use the segment between these two points, the center and the given point as a diameter, draw a circle and every point on that circle has the property that a triangle is the right triangle, which means one of these is the one which we are looking for. So this particular point should have the property being on that circle and at the same time being on that circle. That's why it's on the cross into these two circles. So we have two points, two points of intersection and that means that we know the point where our tangent is supposed to touch the circle. Well, since we know the point, we just draw a line here and draw a line here and these are two tangents which are two solutions to our problem. What was actually the creative moment in this particular little problem? The creative moment was to realize that there is certain property characteristic for this particular picture, the property that this is the right triangle and the locals of all the right triangles can be built using this as a diameter. So this is a creative part. So you have to think about how to find this particular point because since you don't have it, you have to find it using certain technique. So this is a creative moment. You have to realize that this point is supposed to belong to two different curves. One is it's obviously on our initial circle and secondly, it should belong to the locals of all the points which have the right angle at the top and supported by this as the end. We have to construct a right triangle, sorry, inscribe a right triangle into a circle given the length of one half of this. All right, so we have to inscribe the right triangle and as we know many times, repeated all the right triangles are supported by the diameter. So if we know the circle, obviously we know it's diameter and then any point on this circle can be used as a vertex of the right triangle, this, this, et cetera. Now, we are interested in the right triangle with a given categories. Well, that's actually simple. You take this length, use this as a center, mark this, you can actually mark this too. So you have two triangles with this particular size of the categories. You can actually do exactly the same thing from another end of the diameter and have this and this two points again, so you have again two different solutions. So we have four different solutions for any given, well, not for any given. Obviously, if this is too big, then you will just go outside of the circle. So obviously there are certain restrictions on the size, but if the size is less than a diameter, then the crossing will always be found. So again, we have a circle, we have a diameter, any diameter, and from both ends, you can use this as a radius of your compass and then have all these four points will be vertices of the right triangle which we are looking for. That's simple. Construct a circle of a given radius tangent to both rays of a given angle. So you have an angle and you have to inscribe a circle so both rays are tangent, but you are given the radius of this circle now obviously these are too perpendicular since these are tangents. So the lines, the segments from the center to the point of touching the tangent with the circle are perpendicular. So these are perpendicular. Now, radiuses are given, right? Okay, but this is now a really simple thing. Since radius is given, it means that this point is on this distance from this line. Now, what is the locus of all the points on a certain distance from the line? You just draw a parallel line which has the distance between these parallel lines a given segment, right? So we don't know this point, but we do know that it's supposed to be located on this line which is parallel to this line and on the distance which is our radius. Now similarly, this same point is supposed to be on the same distance from this line, which means again we draw a parallel line on this distance and the crossing of these two parallel lines is the point where the center of the circle is supposed to be located. Since this point belongs to this line, this perpendicular is equal to our R and since this point is on this line, the lengths of this perpendicular is always R, which means that if you will use this as a center and draw a circle of radius R, it will touch both rays in one particular point for each. So that would be tangents. All right, that was easy too. Inscribe a circle into a triangle. All right, so we have a triangle and we have to inscribe a circle so it's tangent to each one, to each side of the circle. Well, let's think again. If you will use the center and draw all three perpendiculars, let's think again where exactly this center is supposed to be located. If you have a triangle, well, obviously if you draw this particular line, it will bisect this angle. Why? Because these two triangles, this one and this one, are supposed to be congruent. Now why? Well, because these are two radiuses, right? And this is a common hypotenuse. So these two triangles are congruent, which makes these two angles congruent to each other. So it looks like this particular point, the center of an inscribed circle into a triangle is supposed to lie on the bisector of this angle. And similarly, it's supposed to lie on the bisector of this angle and this angle. So you take two bisectors of the triangle, angle bisectors of a triangle and you will get the inscribed center of an inscribed circle. And obviously the third one is supposed to be crossing the first couple of pairs, the pair of bisectors in exactly the same point that can be very easily shown. So let's just repeat again. If you have a triangle and you want to inscribe a circle into it, you have to use angle bisectors. If you want to circumscribe a triangle, which we did before actually, now if it's a circumscribed circle, then as you know, the center is supposed to be equally distanced from all three vertices, which means it's supposed to lie on a perpendicular bisector of each side. So inscribed with the bisectors, circumscribed circle is on the crossing of perpendicular bisectors of all sides. As you see, all these problems are really very, very simple. Construct a line tangent to two given circles. All right. So you have two given circles, one and another. And we have to draw a tangent to both of them. Well, obviously we can draw another one on this side. All right. How can we do it? Here is what we can say. Now these are two perpendiculars to the same line, right? So the radius to the point of touching a tangent is perpendicular to the tangent, which means they are parallel to each other. Now, let's bring the whole line, the tangent line down by this distance. What happens? If you have this r and this r and you do this and you will have a circle here with a radius of r minus r, now this line relative to this circle is shifted down. And as you see, this circle of the radius big r minus small r would be actually would be actually touched by this line, which we shifted down, and the line would be this line would be also tangent to this. Now, why it's tangent? Well, because since we shift parallel down, which means that this used to be a perpendicular and this would be a perpendicular, right? So we have a line which is perpendicular to the radius and it's on this particular distance which is equal to the radius. So we have subtracted from r because big r was subtracted small r, we have this r minus r. So basically what I'm saying is that the new line, which is dotted, is tangential to a smaller circle of the radius equal to the difference between the radiuses of big and small circle. Now, can we build this line, the dotted line? Yes, we can because since we know both circles, we know both radiuses, that's why we know the difference between them so we can draw a circle concentric to the big one with the radius of r minus r. So this circle can be constructed very easily. Now, this is a point now, this is a center of a smaller circle, which means from a point, we know how to draw a tangent to a smaller circle. So that's how we build the whole thing. We draw a smaller circle inside the big one with the radius of big r minus small r and from a center of a small circle, we draw a tangent. So since we know how to do this, we know this point. As soon as we know this point, we just continue this particular line and then we continue and then we draw this line parallel to this line. So we get this point and we get this point. And these are two points which are supposed to be where our tangent, which we have to construct, is touching two circles, two given circles. All right, so again, first we build the difference between two reduces, this minus this, subtract it, use the difference as a radius concentric to the big r and then from the center of a small circle, we draw a tangent at this point. Now, how to draw a tangent from a point to a given circle? We already know in the previous problem which we have solved and then just along this particular line, we find one point and this is a parallel find another point. It's a little less trivial, I would say. This fact that we have to shift the whole thing down by this radius is again a guess which you really have to make to simplify your problem and not just to simplify but to reduce it to the one that you have already solved before. Actually, that's how everything is solved. You reduce your big complex problem to a smaller one, which you have already solved before. And then go back. Inscribe a square into a given circle. Okay, so you have a circle and you have to find four points which make a square. Well, how can you do it? This is actually easy. Obviously, if you connect this to this and this to this, you know we will get the center of a square, right? So, the way how you build this particular square is the following. You have a circle, you have a diameter, put a perpendicular diameter through the center and basically connect. Now, how to prove this is a circle? Well, very easily. I mean, all these triangles are congruent to each other. They are all right and equilateral and the, sorry, antisocialists. And so, you have all these angles 45 degrees. This is 90 degrees. So, all these angles are 90 degrees and all these sides are equal to each other. So, two perpendicular diameters will give you a square. Okay. You know how to inscribe a square into a circle, okay? How to inscribe a regular hexagon into a circle? Well, hexagon, it might be a little more difficult. This is something like this. Well, you can say that this is a diameter. Well, yes, it is. But now you have to divide this piece into three equal parts, which is not easy, right? Well, but actually it is quite easy. Let's think about it. Since this is a regular hexagon, since this is a regular hexagon, now all these are equal to the radius, right? And all these angles are equal to 360 divided by six, which is 60. All of them are 60, which makes these also 60. Which makes every triangle equilateral, which means if this is r, then this is also r, r. So, these chords are equal to the radiuses. All chords and all radiuses, they are all the same. Now, how to build the hexagon? Very simply, just take a point and using this radius, just put all these chords and that will give you the regular hexagon because this is equal to the radius. This side is equal to the radius. So, that's how you inscribe hexagon. That's the simplest, it's probably even simpler than to inscribe a square because there you have to draw a perpendicular. Here you just don't have to draw anything. Just put the points one after another, all six points which are vertices of the hexagon. Again, let me just stop a little bit. What was the constructive moment in this particular case is just to do this little research and basically find out that the lengths of the side of the hexagon is exactly the same as the radius. As soon as you know that, the construction itself is simple. So, you have to really have some kind of research first to how to approach the problem, but once you know, to solve it is really easy. And the last one, inscribe an equilateral triangle. So, equilateral triangle inscribed into a circle. Well, that might not be actually as easy as hexagon, or is it? Well, the way how I suggest you to solve this problem is very easy. First, inscribe hexagon, and you know how to do that. And then connect every other point. Just skip one. One, skip two, three, skip four, five, skip six, game one. And this would be your regular equilateral triangle. Again, a little guess that instead of just thinking about what's the lengths of this particular side, it's complicated. I don't know what the lengths of this side, just off the top of my head. I have to do some kind of calculations, some kind of maybe additional constructions. It's difficult. But if I have just guessed that providing you know how to do the hexagon, the regular triangle, the equilateral triangle is actually can be obtained by connecting points, vertices of the hexagon, just skipping every other one, well then the problem is easy. All right, so I just want to emphasize how important it is to think about how to approach the problem. Construction problems in geometry are very, very much like this. You just think about this and then the situation actually clarifies. And the construction itself is simple. You just have to know how to do it. And how to do it is actually requiring certain level of thinking, creativity, and guessing if you wish or something. And the more problems you solve, by the way, the more proficient you will be in this particular art of solving these problems. Because actually, if you think about it, the number of methods of solving different problems is really finite. So the more you solve, the more you enrich yourself in this methodology and you will always be able to basically choose from one which you have already done before as a method to at least approach the new problems. Every new problem might actually be solved by certain combination of the methods which you have already used before solving other problems. So the more you solve, the better you will be off. And all the problems in the world we are planning to put on unisor.com, special attention for parents who would like to supervise their children's study and mathematics. It allows you to enroll your children in certain programs. Then exams actually can be taken by the students and US parents can actually examine the results and mark the particular topic as completed. Or ask again, do it again, do the lectures again, do the exercises again, and do the exam again until it's perfect. All right, that's it for today. Thank you very much.