 Hi, I'm Zor. Welcome to Unizor Education. If you have not spent time before watching this lecture on these problems, which I'm going to present to you, construction problems, just pause right now, go back to the website Unizor.com, go to the geometry circles, find that particular lecture notes contain all these problems, and try to do it yourself first. You really have to spend some time. If it works immediately without any problems, great. Then just, you know, wish my lecture just to make sure that you're right. If you cannot solve the problem, try to spend at least half an hour to an hour before giving it up. So it's very, very important because the purpose of the whole course of these lectures and exercises, etc., is to teach you to think creatively, to analytically approach the problem, and to solve problems, basically. Again, the purpose is to solve problems, not to find out what kind of theorems exist and the proportions and the properties, etc., etc. Solving problems is the goal. So try to do it yourself first before you listen to me. Now, without further ado, let me just go through these construction problems and I will present my solutions, not necessarily the best ones, but that's what I came up with. By the way, some of them might be a little bit more difficult. Quite honestly, I had spent some time actually solving these problems myself. Some of them were quick, some of them were not. So it requires thinking and spending time. That's what actually you will achieve by doing this thing. The creative mind needs exercise, right? So that's what it is about. Okay, find the point on a given line, such that the given segment is viewed from this point at a given angle. So you have a line and you have some kind of a segment. And you have an angle. So you have to find a point from which this segment is viewed at this particular angle. Okay, it's actually a very easy task. But to solve it, I think I would like to introduce some general concept first, which will be used in some other problems as well. So the problem is, find a locus of all points from which a given segment is viewed at a given angle. Now, if you have a segment and you have certain angle like this one, and let's say this is a point from which this particular segment is viewed at this angle, and this is also a point from which the same segment is viewed. And this one. Now, you know that all angles inscribed into a circle are actually congruent to each other if they are supported by the same arc. So if I will have one particular point and then I have these three points, this is triangle, so I will circumvent it with a circle, so it will be something like this. Then all inscribed angles, which are supported by the same arc, which is this one, would have the same measure. So these are congruent angles, because each one of them is measured by half of the central angle, which is supported by the same arc. Now, obviously, this represents a locus of all points which have the property of having this angle congruent to this one. Now, why a point which is not lying on this circle would not have this angle congruent to this one? Well, let's take this one outside. Now, as you know from one of the lectures, one of the theorems that this angle actually is measured as half a sum of this central angle which is supported by this arc and by this arc, because it cuts two different arcs from the same circle. And the theory is that if the point is outside of a circle, then this angle is measured as half the sum of the corresponding central angles which are supported by this arc and this arc. So it's definitely greater. Actually, in this case it's half of the difference, but in case it's inside, then it's half of the sum. So it's either the angle will be smaller than half of the central angle supported by this arc in case the point is outside, because there is a difference between this arc and this arc, or it will be greater, because there is a sum of this same arc and this arc. So basically, the answer to this problem is the locus of all points from which a given segment is viewed at a given angle is a circle. Well, more precisely, part of the circle, an arc of the circle. And all these points have this property. Now, you can definitely point out another arc in another circle which is symmetrical to this one. If it's symmetrical relative to this line, obviously this arc is exactly the same as this arc and all these angles also will be congruent to this one. So the locus is actually two arcs. Now, how to build these arcs? Well, very simply, take one particular triangle which has this particular segment as a base and an angle at the top congruent to this one. For instance, a right triangle. You can always build the right triangle with this angle at the top and this casualties equal to a given segment. So having these elements, the segment and the segment in an angle, you build the triangle and then you can just do a circle around it and you can build a symmetrical triangle actually this way and have another circle. So basically what I would like to say is that the problem of constructing a locus of points from which a given segment is viewed at a given angle is solved by this. And I will consider this as given and I will use this particular construction as given. So whenever I'm saying something like, okay, let's construct a locus of points from which this particular segment is viewed at this particular angle, that's exactly the technique which I'm going to use. Now, back to our problem. Well, our problem now is solved very easily because if I would like to find a point on that line from which a given segment is viewed at a given angle, all I have to do is let's construct a locus of all points from which this particular segment is viewed at this particular angle. Now, this will look like something like this maybe and if this line has any intersection with one of these two arcs, that's the solution. So in this case, I have two solutions, this and this. Well, again, depending on where the line is and what the angle is and what the segment is, there might be no solutions at all. Might be one solution if the line is tangent to these things or, well, for instance, line can be tangent to both of these so it will be this angle and this angle. So also two solutions. So it all depends on their mutual correspondence and location and sizes, et cetera. But anyway, this is the way. You construct the locus of all points from which that given segment is viewed at a given angle and if there are any intersections with this line, these are your answers. Now, in the future, I will not actually go into the details about how to construct the locus of points from which a segment is viewed at a certain angle. I'll just use it. Okay, next. Construct a triangle by a side. So you have a side. Let's say AC is given. Opposite angle, which is this one. This is angle ABC. It's also given. And an altitude falling onto this side from an opposite vertex. And this BD is also given. So that's what's given. The base, the altitude, and the angle at the top. Well, immediately what should we do? Obviously, build the locus, construct the locus of all points which this given segment is viewed at this particular angle. Now, that would be something like this and this. Right? Now, we have to find a point B on this or this arc with the distance BD actually given. How to do it? Well, from this line, AC, draw a parallel line at the distance equal to BD. Now, wherever this line crosses the circles, you have solutions. So you have, in this case, one solution, another solution, third solution, and fourth solution. So you have four different triangles. All of them have the same altitude because it's the distance between two parallel lines. The same angle because these and these points all lie on the locus of points from which the given segment is viewed at given angle. So that's the solution. What was important here to gain is the method of construction is to have one locus and then another locus. And then wherever they intersect, that's what we actually need. The first locus was all the points from which a given segment is viewed at a given angle. Another locus is all the points of the vertices of the triangles with the given altitude. So on the intersection of these two, you have all the solutions. And solutions can be, well, numerous actually, right? It can be no solutions, maybe the altitude is too big or it can be two solutions on both sides or it can be four solutions. I don't think there is a three or one solution. Yeah, that's right. It's either no solutions or two solutions if vertices are, if altitude is equal to the distance to the farther part of the circle. Or no solutions, yeah. Zero, two or four solutions, right? Okay. Given the circle and a sector in it, bordered by two radiuses and an arc, construct a tangent to an arc, construct a tangent to an arc such that these radiuses, the continuation of these radiuses, cut the segment of a given length. So mn is a given segment, basically, the lengths of the mn. And the sector is given. And now we have to construct such a tangent which has this segment mn having a given length. All right, actually this problem is exactly the same as the previous one. Why? Because if you will draw radius to this point, think about the triangle mon. What do we have in this triangle? What's given, basically, or what we can determine? Obviously, the base, which is mn. Obviously, the angle, which is given, that's the sector, which is given. And the altitude, which is the radius, right? Remember that the radius to the point of tangency is perpendicular to the tangent line. So this is an altitude. Okay is altitude in the triangle of mn. So, basically, the whole problem is to build a triangle which we have this base. We have the angle, and we have an altitude from this vertex to the base, which is the same as the previous one. So we built that triangle, and all we need is just position it in this exact place where our sector is located. So if you build this triangle somewhere else, whatever the triangle is, mn, o, you just take the om and put it on this radius and take om and put it on this radius. And these are two points which you are looking for. So that's basically the same as the previous, just formulate it a little bit differently. Just to confuse us. Obviously. Construct a triangle by a side, opposite angle. You have a side, a c. You have opposite angle, which is angle a b c. And a median falling onto this side from an opposite vertex. Median. Seems to be easy as well. So what do we do first? Since we have a segment, a c, and an angle at the top, first what we do, we draw a locus of all points from which this particular segment is viewed at this particular angle. So this is the locus of points. Right? That's what we have to do first. Now, we do have this median b d. Now, where are the locus of all points b, such that triangle a b c has a median equal to a given locus b d? Well, you put a compass in the middle of a c, and using this radius, you basically draw a circle. And wherever this new circle intersects two previous arcs, that's actually the points which have both properties. Number one, this angle is given. And number two, the distance from the b to d is exactly the segment which we need. Since d is chosen as a middle of a c, that's the median. So that's basically the construction. A couple of arcs first, using the locus of the points from which a segment is viewed at a given angle. And then another circle from the middle of the a c with a given radius. And all the intersections, wherever the intersections are, can be one or two or whatever. Probably one cannot be. Now, one cannot be. So it's either two or four. Yeah, two or four. All right, next. Given two segments on a plane by lengths and positions. So we have lengths and positions of two segments. Find a point on a plane such that the first segment is visible from this point at one given angle. So you have this angle given and you have this angle given. So you have these two segments and you have two angles. So you have to find the point from which this segment is viewed at this angle and this segment is viewed at this angle. Well, again, that's actually the simplest thing of all. Because first you do the locus of all points from which this particular segment is viewed at this particular angle and you will get something like this and this. And then based on this angle you construct the locus of points. This particular segment is viewed at this angle. So it will be something like this. This and this. Now, wherever these four arcs you have wherever they intersect each other these are the points which combine both properties. One segment is viewed at one angle and another segment is viewed at another angle. So that's your construction. By the way, if you have noticed I'm kind of doing two things together. First I'm analyzing the problem and then I'm constructing. So analyzing I mean that I know that this point is supposed to belong to two locuses. Now the construction itself is okay, draw the first locus, draw the second locus and then choose the intersection point. And it's always like this. First you analyze the problem and then once you've done that you propose a solution. Okay, find a point inside a triangle such that all three its sides are visible from this point at equal angles. Okay, now we have three different segments combined into a triangle and what we need is to find a point inside that all three angles are supposed to be the same congruent to each other. Well, if they are congruent to each other it means that each one of them is equal to obviously 120 degrees, right? Each one of them. Because the whole thing is 360. Now, knowing that the solution is obvious you build the locus of points from which this particular segment is viewed at an angle of 120 degrees which is this arc. And then same thing for this segment which is this arc and wherever these arcs intersect each other that's the point from which this segment is 120. And this segment is viewed at 120 which remains actually this segment also to be equal to 120 because the whole sum is supposed to be 360. Notice that all these problems which I am just talking about they are all based on the same approach. If you have a segment and an angle this segment is supposed to be viewed from some point. First you build the locus of all points from which this segment is viewed at this angle. That's what I have started the whole lecture with. Next. Construct a triangle by an angle at some vertex and originating from this vertex medium and altitude. So you have triangle you have altitude and medium. So what you have is you have angle A, B, C you have an altitude B, D and you have medium B, E. This is the right angle. And A, E is equal to B, C obviously because B, E is immediate. And you have this angle. Actually, this is the problem which I have spent a little bit more time thinking about. And it's not solved in one shot. This is the first step which I am making and everything is obvious. I have to do certain additional construction before even I start analyzing the problem. The question is what this construction is supposed to be. How can I help myself? Now, my first thought was look at the triangle B, G, E. Now, we know the catatous B, G and the hypotenuse B, E which means we can actually construct it. So, okay, having these two elements I can construct this triangle somewhere. Okay, fine, that's done. This is B, G, E. Okay, now I know that A and C are somewhere here on this line. So, I have to find a point C and on equal distance from E on this side it will be A such that this angle is equal to the given. And I don't know really how to do it quite frankly. Now, if instead of this angle I was given this angle then I can do it very simply. That's the beginning of giving how I have started thinking about this problem. It's not this angle which kind of makes the whole thing easy. It's this angle because if this angle is given that means that B, G, E the D, B, C is 90 degrees minus this angle so I know this angle so all I do is after I constructed the triangle B, G, E I just draw a line at this angle and that's it at 90 degrees minus whatever is given but this is not given angle so I can't really do this but something actually triggered in my mind that I have to like shift something from this angle to some other angle and here's what I have come up with let's oops, sorry my medium at the same lengths here so B, E and E, M are congruent segments now think about A, B, C, M this is parallelogram why? because these are two diagonals which are dividing each other in the middle in the midpoint let's call the theorem about this so this is a parallelogram well if this is parallelogram then what we have is obviously we have this angle equal to this angle I mean not equal sorry 100 h-degree minus this angle so if this one is given then this one is 100 h-degree minus this one so we know it knowing this angle we know this angle now what do we know we know B, M because it's twice the B, E and that's why using just analysis I can say that point C is supposed to be on the locus of all points from which this segment B, M is viewed at this particular angle which I have calculated in some kind of circle here on which all the different points C are like and knowing this line I have intersection between the circle and the line so now how can I construct the whole thing now using this analysis now let me wipe it out analysis is finished and let's just go to construction so first using B, G and B, E I am constructing the right triangle B, B, E now knowing angle A, B, C I am calculating angle B, C, M equal to 100 h-degree minus angle A, B, C now so I have this angle now B, E I continue gauging the point M and using a segment B, M and this angle I am constructing an arc where all angles where all points from which B, M is viewed at this particular angle are located and the intersection between this arc and the continuation of D, E gives me the point C now having the point C going to the left from the E by the same distance well it's supposed to be a little bit differently on my drawing it's supposed to be a little further somewhere there so having this distance on this side and that's where I get point A and that's my frame now point C is chosen as intersection between the line where it's supposed to be and the circle from which the extended double median is viewed at this particular calculated angle and that's the end of the construction so again I actually did give it a thought I was thinking about this problem for some time and it's not like you know, automatically the solution however, I would like actually to point out the more problems you solve the more approaches you kind of develop to certain situations so you know that in a similar situation you acted this way to help you now if we are talking about medians in many cases solutions are related to extending median to the same lengths and converting a triangle into parallelogram so if you have a triangle and a median in many cases you just double the median and have this parallelogram and something might actually come up as a solution using elements of this parallelogram it's again, it's something which is developed with experience and the more problems you solve the more standard approaches you might develop okay construct a triangle by a side an angle with side forms with another side so you have a side AC an angle BAC and and an angle on the third side is made, makes with a median dropped from the angle okay, so you have a median here and you have this angle AC so you have these elements alright well obviously if you have AC and you have an angle you can start from this this is your AC and this is your line where the point B is supposed to be now, how to find point B well let's just think about it now if I draw a line parallel to AB this would be the middle point since these are congruent segments these will also be if these lines are parallel alright so since I know the line AB I know the line EG because these angles are the same BAC and CEC are the same angle so from the middle of AC I can construct another line which is parallel to this and now again I'm using the fact that since I have this angle and I have a segment AC I can always construct an arc where all the different points from which AC is viewed at given are located so here if I will do this arc the crossing would be my D and this would be my point B so that's the construction this angle is equal to given because it lies on this arc where all the points from which AC is viewed at this angle and now since these lines are parallel and these segments are congruent because I chose midpoint of AC that's why these two also would be equal which means a D would be immediate construct a parallelogram with two diagonals and one angle so you have a parallelogram you have two diagonals and let's say this angle ABCD so you have AC you have BD and angle B these three elements alright this is actually easy because again you start from BD as a segment which is given you draw a locus of points from which BD is viewed at this angle so that would be some kind of an arc right now diagonals in the parallelogram are intersecting in the middle which means if you divide AC in half and take this particular radius from the middle and using this radius you draw a circle wherever it crossing our locus of points from which BD is actually viewed at this particular angle you get the proper angle right so this is your angle and since the AM is equal to half of AC this is your parallelogram so again BD is given you draw an arc where all the different As are supposed to be located and at the same time since AM is basically half of AC centering at the midpoint of BDM you can draw another circle of this radius and wherever the circles are intersecting this is the point A the last one you have to triangle by side opposite angle and sum or difference of two other sides ok so you have ABC you have AC you have opposite angle and let's say start from the sum AB plus BC well whenever you have a sum obviously you have to basically continue that thing so BD is congruent to BC alright AC is given so angle B is given ABC which means you can calculate this angle which is 180 minus ABC and then considering these two segments are congruent these angles are congruent so knowing this angle you calculate this angle which is 180 minus this and then you calculate these which are actually half of this or another one this is exterior angle and these are two interior which are supposed to sum up to this one but since they are equal to half of this so you know this side because this is the sum of AB plus BC and you know this side and you know an angle at the top so you can construct this thing starting from AD which is equal to the sum then having this line at this particular angle which is half of the ABC half of this angle and then having AC as a radius just draw a circle and it will cross into different places alright so each one of them can be point C for C prime to have your point B all you have to do is this is point D all you have to do is draw a perpendicular bisector of CG now this is equal to this so this is point B so either ABC would be your triangle because this angle obviously would be twice as big as this one and the sum of these two would be equal to AD so everything seems to be fine now this solution would be as good as well now this is sum now what if the difference instead of a sum basically it's practically the same thing so you have a difference so you have ABC since it's a difference instead of going that way and extending AB we go this way and put the point G here so that BG is equal to BC we know AD and we know this angle why? because since we know this angle ABC and DBC is isosceles triangle these triangles are congruent to each other so basically we know these angles which is a 100-degree minus given angle and divided in half and that's why we know this angle which is a 100-degree line is this one and now the situation is exactly the same as in the previous case you know AD you know an angle and you know AC so you do this first this is your AD then at an angle which you have calculated you draw a line and then using AC as a radius you draw a B now to get to the B you draw this and you draw a perpendicular bisector to CG since it's isosceles triangle right? DBC and that's where your point B is okay that's it construction problems are very interesting and primarily because you have to really think about what additional drawing I have to I have to do to make my life easier so it's very important that you let me repeat myself solve these problems yourself now even if you could not solve it yourself and you just listen to me after you have finished go back to the website unizor.com go back to this lecture to the notes of this lecture which contains all these problems and try to solve them yourself again and remember the technique so next time it will be easier and parents and supervisors and teachers please use the unizor.com to control educational process of your students because you can enroll students they can go through exams and you can examine basically their scores and you can mark a particular course as completed or not completed ask them to repeat it until the score will be perfect that's it for today