 Hi, I'm Zor. Welcome to Unizor education. This lecture is part of the course of advanced mathematics for high school students and it's presented on Unizor.com Which contains not only the link to the lecture, but also some notes that so I recommend you to watch it Not directly from YouTube, but from this website By the way, the website contains lots of other things functionality of basically The whole educational process can be arranged through this website with enrollment exams Etc. Okay, so this lecture is Kind of a final theoretical lecture about symmetry in 3d space and Not that I will not Touch this issue again. Most likely I will in some problems But this lecture is a problem actually. It's also a problem which encompasses all three types of Symmetry in three-dimensional space Symmetry relative to the plane like a reflection Symmetry relative to the point. It's a central symmetry and Symmetry relative to axis which can be arranged as a rotation of the space So all these three types of symmetry Have combined into one problem Not very difficult at all, but I'm going to present the problem and its solution obviously and I do recommend you just to to go through this material as some kind of a Final Touch on your understanding of symmetry. All right, so what's the problem? Let's assume we have a plane It's called gamma we have one point on the plane We have one point Outside of the plane now But I'm going to do is I'm going to introduce a perpendicular To the plane at this point Let's call it D Next is I'm making two different Symmetrical transformation of point a one is reflection relative to the plane Which means I have to drop a perpendicular And the same distance Should be point B the same distance from this point this is the point where Let's call it Q Where this perpendicular hits the plane? gum at the same time I would like to make a reflection of this point a Relative to the point P. So it's a centrally symmetrical thing which means I have to connect them and continue By the same distance something like this, I guess Let's call it C and now what I am going to prove is that the points B and C are Symmetrical Relative to the axis D now, I Hope you understand that this is not the Flat surface of this board which B and C belong to We are in the 3d So it doesn't look like that. They are symmetrical relative to this line D But that's only if you consider B and C belonging to the plane which is this board in in theory. They are Slightly turned as well as these two are slightly turned So basically this is the right angle and What I'm going to do is to prove that three things I need to prove Which basically constitutes symmetry relative to the axis number one that this segment BC actually Intersects with G That this is perpendicular BC and D and the third one that they are on the same distance from intersection R So that's what I would like to prove so again, they Intersect G and BC They are perpendicular to each other and B and C are in the same distance from the intersection point Let me just write it down BC is supposed to intersect with G That's not empty set Now BC is supposed to be perpendicular to D and BR Should be equal to RC Where R is actually the intersection BC intersect So that's what we have to prove these three things It's called a B and C three conditions All right, so I was explaining for such a long time the conditions of this problem The proof is not really very difficult not very not not very long Okay, so let's just think about this way A B is supposed to be perpendicular to gamma and D is also perpendicular to gamma That's how we have constructed in the first place which means they are parallel So two lines perpendicular to the same plane are parallel to each other since they are parallel These two lines Belong to some plane right So let's call this plane delta. So a B belongs to Delta and D belongs to Delta Parallel lines are always belong to some Plane which encompasses both of them All right Now what does it mean? Well, since a B belongs to this plane Delta then both a and B belongs well and incidentally Q as well because Q is midpoint of the AB now a belongs and Since G belongs then this point B also belongs to this plane obviously and since a and P belong then the whole line AC belongs so what we have come up with is is AC is also belong to this point to this Delta So this belongs this belongs and this belongs so everything belongs to this plane Delta So what I'm going to do right now is not number one obviously The G and BC are not really going Passes each they're not passing each other without intersection because they all belong To the same plane right so D belongs to the same plane as BC Since B belongs to the plane and C belongs to the same plane Delta So BC belongs to the plane Delta so from this and this D belongs and BC belongs Follows this they must intersect All right That has been problem Now let's just consider triangle a B C Now Q is midpoint of the AB P is midpoint of AC right because this is the reflection relative to the plane Gamma and this is reflection relative to the point P So P Q is a midline of a triangle and we all know the properties of midline of triangle Well, not only it's equal to the half of the opposite side, but it's also parallel to the opposite side, right? So P Q is parallel to BC and equal to half of it now We know that this D is parallel to a B. So what is actually PR PR is also midline Because P is the middle of this and PR is parallel to a B and If you have a triangle And you have a line which intersects side at its middle and parallel to the base in this case a B is the base Then this line is midline and it cuts in half the Opposites the opposite side of a triangle which is BC From which follows the BR is equal to RC That's this What's left is perpendicularity Now it's also very simple now BC and P Q are parallel to each other now P Q is obviously perpendicular to a B because A B is perpendicular to the whole plane and that's why it's perpendicular to the P Q line Which means BC is also perpendicular to a B Because these are parallel. This is a midline in a BC Well and since BC is perpendicular to a B It's also perpendicular to D because D and a B are parallel That proves this Okay, so all these three properties Are proven which means that B and C are Symmetrical relative to the axis so here we have a composition of all three types of symmetry in this problem That's why I liked it actually as a kind of a final theoretical Polishing on the symmetry in 3g. We have the reflection a and B We have a central symmetry a and C and we have an axis symmetry between B and C here is kind of interesting thing All right, basically, that's it. I do recommend you to go through notes for this lecture at Unisor.com and Just go through The proof I was trying to use as much symbolics as possible because mathematics is actually much better if you Converse in in symbolics. It's shorter and it's more understandable So basically that's it for today. Thank you very much and good luck