 Hi, I'm Zor. Welcome to Unisore Education. I love these construction problems which I'm going to talk about today. I think they actually bring up the best of creativity in a young student. Everything is completely artificial. Everything is only the brain exercise, no more than that. And they develop this brain creativity to a very, very high degree. So I do recommend to solve these problems in as large quantities as you can. And that includes obviously all these problems which I present on this website. I hope you did try to solve all these problems before listening to this lecture. It's really very, very important for you to do it yourself. So in any case, I will present solutions to all these problems. I have many of them, by the way. This is just lecture number three. There were two others before it and there will be others after it. They're all dedicated to all different kind of construction problems which playing geometry really can present you as a challenge. Alright, so let me start. Given a circle and a point inside it. Okay, given a circle, this is the center, and a point inside it. Construct a chord that contains this point and is divided in half by it. Well, this is simple. You know that radius which is dividing a chord in half actually is perpendicular to this chord. Which means that in this case to construct this particular chord which will be divided by this point in half is just to draw a radius through this point and in this point draw a perpendicular chord. Now, since it's perpendicular, these angles are both right. Obviously these right triangles are congruent to each other because they have common characters and hypotenuses are ridges and they are equal in length. So we have hypotenuse and the characters and that's why the triangles, this one and this one are congruent and that's why these two segments are congruent to each other. Now, basically every problem which has a chord most likely will require the radius perpendicular to this chord which divides this particular chord in half. We will use this particular property of a perpendicular to a chord which divides this in half in many different problems which we will try to solve today. So this is an easy part. Okay. Now, next, divide an arc in two, four, eight, etc. congruent parts. Any kind of a power of two. And then we divide an arc. Let's say you have an arc from here to here. This is the center. How to divide it in half? Well, draw a chord and if you draw a perpendicular it will divide it in two congruent parts and I think I did this actually before, the proof of that I did before using the symmetry because every point in this arc is symmetrical to this relative to this as a diameter. Why? Because they are on the same perpendicular to this diameter and on equal distance from the diameter as I just proved before, second to go. So these two points are always symmetrical relative to the diameter which means the whole arc is symmetrical to this one and that's what actually makes these two arcs congruent to each other. Now, if we can divide it in two we obviously can divide it in four parts, equal parts how? Divide it in two and then divide into each part which means again draw a chord and perpendicular to the chord here and here. So I have four pieces, how to divide in eight? Well, again, obviously every piece you divide in two halves. So this is the next. Construct two arcs in a circle if their sum and difference are given. Okay, that's an interesting thing. So you have a circle. Now, there are two arcs which you don't really know what they are but what you do know is you know their sum let's call this arc x and this arc we will call y. So you know their sum which is equal to some arc like this. Okay, so this is x plus y and you know their difference somewhere here. Let's say this is x minus y. Alright, so you have these two arcs this one and this one but you have to construct x and y. Alright, so how can we do it? Well, let's call this arc A and this arc B. So x plus y is equal to A x minus y is equal to B. Now, let's just think about this from an algebraic standpoint. These are actually two equations with two unknown. A and B are known, x and y are unknown. Now, how can we solve the system of two equations with two unknown? Well, in this case it's very easy. Let's sum them up and you will get two x equals A plus B and x equals half of A plus B, right? Now, if you subtract them then you will have two y equals A minus B so y is equal to A minus B, one half. So, let's just think about this. These are solutions to our two equations. So if knowing the arcs A and B we can construct half of their sum and half of their difference that actually would be our x and y which we are looking for, right? So, let's forget about the equation now and let's concentrate on building the solution. Now, how to build one half of A plus B if you know A and B? Okay, let me just draw another circle equal to this one in all respects. Now, this is A and this is B. Well, let's start from one point then have a compass, measure this particular chord and basically put it here, the same distance. So, this chord is equal to this chord. This is also A arc. Now, from this point, I take this distance in the compass and put it here. This is an arc B and whenever I put them one after another and let's tag them together, they represent A plus B. So, this chord is corresponding to arc A plus B and all I need right now is to divide it in half and we know how to do it. We just draw a perpendicular and this is the half which we need. This is an x. Now, similarly, A minus B is, again, it's very simple. Basically, the same thing, first you do A then you have to subtract arc B which means you take this distance and instead of going further along the circle you go backwards, so this would be your B which means that this would be an A minus B and perpendicular to A minus B divides it in half and this is the Y which we're looking for. That's it. We found x and y, we found both arcs and that's what was required. Next, given a circle and a point anywhere outside it. Okay. A circle and a point outside it. Construct a new circle with a center in that point that intersects a given circle in two points on opposite sides of the given circle's diameter. So basically you have to build a circle with this as a center and which actually crosses the given circle on opposite sides of the diameter. Well, again, it's very easy and it's using exactly the same technique. If you draw the line between these two centers now we know that diameter is divided by the center in two equal parts, each equal to a radius which means that from this circle's perspective this is a radius which divides this chord in two equal parts and it's perpendicular to it. Okay, this perpendicularity is all we need because how to construct it? How to construct this circle? Well, easy. You connect this center to this center and draw a perpendicular to this line, a diameter, because it's through the center. These two points are basically what we are looking for because the radius of this circle which we are looking for is the distance from here to here. So again, connect the centers, draw a perpendicular to this center line until it intersects the given circle and these two points are what you need to find out the radius of this circle, of a new circle. And again, as everywhere else we are using this theorem that if you connect a center of a circle with the midpoint of a chord it's perpendicular, or if you draw a perpendicular it will be a midpoint of a chord. Given a straight line and a circle not intersecting with it. A straight line and a circle. Find a point on the line closest to a circle. Well, intuitively it's obvious that the point which is closest to the center is if you connect the center with a perpendicular line to a given line. So this point is on the shortest distance from a circle. So this distance from A to B is shorter than any other distance from any other points, let's say C and D, which you can pick from a circle, one point in the circle, another on the line. Why? Simple reason. Connect this. You do know that, and you can actually put a tangent here as well. This is a tangent in this point A, which means it's perpendicular to the radius and that's why it's parallel to this one. Now, you know that all points which are lying on a tangent are outside of a circle, except one of them which is touching the circle, right? Now, also you know that A, B is a mutual perpendicular to these two parallel lines, which means A, B is shorter than let's call this point E. So A, B is shorter than E, D. You also know that all E is shorter than O, C plus C, E. Why? Well, primarily because all A is equal to O, C. These are two radiuses, and now we are adding something. So basically the whole road from O to C through E to G contains a couple of segments which are greater than the corresponding segments on the perpendicular line. This segment O, E is greater than A, B. Segment O, C is equal to O, A, but then we add yet another piece which is C, E. So no matter where you pick the point C, the overall distance from C to D will be greater by two actual increments. One is because this E, G is greater than E and then we add one more point because it's basically outside of this parallel line. So that actually makes the C, G a greater distance than A, B. So A, B is the shortest, and if you want to find the distance from the circle to the line, you just have to draw a perpendicular to that line, and wherever it intersects the circle and the line is the shortest distance between these two geometrical figures. After a while, actually, all these problems are very much like each other, and I mean, there are some more complicated problems, but they're not this type. All these are, whenever you see the chord, just draw a perpendicular to it. Whenever you see a tangent, draw a perpendicular to the line of tangency which will be perpendicular to the tangent itself. So these are typical approaches, typical techniques which the more you solve these problems, the more you almost automatically will use them. All right. Given a chord in a circle and an angle. Given a chord and an angle. Construct another chord in that circle that intersects a given chord at a given angle and is divided in half by a point of intersection. All right. So let's draw something like this. So this is equal to this. It's divided by a point of intersection in two equal parts. The angle is equal to the given angle. All right. Now, since we have a chord, we know what we have to do. We have to connect the midpoint with the center of a circle. Okay. Now, what do we know? So we know that this is a perpendicular because the perpendicular is the one which divides the chord in half. Now, we have another chord, the given chord. We can also put the perpendicular to this, right? So AB is given. MM is what we have to construct. We put two perpendicular to these chords and this one actually divides it K. MK is equal to KM. And the angle MKB is equal to, let's say, is alpha. Okay. Now, what can we say about this? Let me think how to construct it. All right. Well, obviously these are two perpendicular to these chords, which means this angle is also alpha. That's obvious, right? If two lines are making an angle alpha and you have two perpendicular to these lines, we actually went through this theorem a long time ago. The angle between the perpendicular is the same as the angle between the lines. All right. That's fine. Now, what does it mean? It means that we can build the point K by the following construction. First, you have an AB which is given to you. So you draw a perpendicular from O to a point L, which is midpoint, by the way, of AB. So you draw this perpendicular radius from the center to a given chord. It falls in the point L, which is in the middle. Now, you have this angle alpha. So you take this angle alpha and you build another line from O-L, you build O-K. And wherever it intersects, whenever this radius which you have built intersects AB is your point K. So let me start from this. So you have this AB. First, you draw a perpendicular, which is point L. Then, from O-L radius, you draw a radius at your given angle alpha, and that's how you get the point K. Now, having the point K, now you again can build an angle alpha, and that will be the chord which you want to find out. That's it. And again, we're using the fact that perpendicular to a chord divides it in half. Or if the radius divides it in half, it's perpendicular in both ways. Okay. Given an angle, a point on one of its legs, angle, a point, and a segment, construct a circle with a center in a given point on the leg of an angle, like intersecting with a circle forms a chord, a chord congruent to... Okay, so this is the center. So we will make something like this. So this is our angle. This is the point. And we have to build a circle with this as a center, which cuts the piece from K to L with another leg of an angle which is congruent to this segment which we have. So this is the length which we know. All right, fine. So how can we do that? Well, first of all, again, if there is a chord, most likely this perpendicular would be very handy. Now, perpendicular, we can always draw from a point to a line. So basically, the perpendicular is also almost like known. It's very easy to construct. We don't need anything, just draw perpendicular to it. Got the point, M. Now, what do we know next? Well, we know that the perpendicular divides the chord in half, which means that Mk and ML both are equal to half of the KL. So what you do next is you divide your segment which you are given in half and use this half from the point M on both sides, and you get K and L. So basically, you have the radius now, which you have to use from the point O. Use this as a radius and you draw a circle. Simple is good. All problems which I am presenting here are simple, which means it takes probably a minute for you to be able to solve it. Now, this problem is very similar, but instead of a point on another leg, we have a radius. So given an angle, a segment which you have to cut from another leg, and instead of a point, you have a radius. So if you have a circle, so you don't know this point. What you do know is the radius, which is the same in these cases, right? And you know this length. This is 8. Okay, but now let's think about it. Again, draw a perpendicular which bisects the segment A in half. So you do know this side of this right triangle. And you know the hypotenuse, which is your radius, right? Which means you can build this triangle. A right triangle can be built by characters and by hypotenuse. So you can build this triangle. Okay, build it somewhere else. One side is people one half of A, which is this one, and hypotenuse is R. So you build this triangle, which means now you know another character, which is this. This is a distance from the line which you have been given to a center. Now, where are all the points which are on certain distance from this line? Well, it's a parallel line on this particular distance along the mutual perpendicular. So how to build this line? Well, you build a perpendicular anywhere to this leg of an angle. You use this length, this calculus length as the catching point on this perpendicular. You get this and draw a parallel line to this. And wherever this parallel line intersects the other leg of an angle, that's your center. Now you have the radius and the center and now you can draw a circle. Okay, give it a straight line and a point on it. Okay, straight line and point on it. Construct a circle of a given radius such that a given line is its tangent with a given point being a point of tangency. So you have to build a circle which is tangential to the line and this point which is given is a point of tangency. Now, we know that the radius which is connecting the point of tangency and the center is perpendicular to a tangent. Already discussed it many times before. So in as much as I was telling you that every chord in all these problems most likely requires the radius into its midpoint, tangents usually require the radius into the point of tangency which is perpendicular to the tangent. So, what do we know about this? The radius is given and the point of tangency is given which means how to construct it. You just take this point you draw a perpendicular and cut the length which is the radius and that's your center and this is your radius for a circle. That's it. This is the simplest. The last problem on this construction number three construct a tangent to a given circle parallel to a given straight line. Alright, so you have a circle you have to construct a tangent which is parallel to a given line. Very simple. Again, these are parallel which means this perpendicular is perpendicular to both. So all you need to do is draw a perpendicular to a given line and wherever it intersects the circle that's your point of tangency. So you got this point. Now how to draw a line? Again, it's perpendicular which you already have constructed because the tangent is always perpendicular to the radius in the point of tangency. So that's it basically. Perpendicular and another perpendicular. Okay, that concludes my ten problems for this particular construction number three. But I have others which I will just put into another lecture. So that's it for today. Don't forget that all these problems and the future ones, etc. try to solve it yourself. That's number one. Number two the website Unisor.com is for not only for students who can just listen to the lectures solve problems, etc. but also for parents, supervisors or teachers who would like to supervise a group of students and basically give them enroll them into one or another particular program which Unisor provides and that enables the students to go through exam and parents or supervisors can basically check the results of this exam maybe analyze if they want to if they don't just say the student until the exam is done completely, you are on it. And if everything is okay mark this particular program as completed and enroll into the next one. One by one you will cover the whole math. Okay, thanks very much and I'll see you the next time.