 Hi, I'm Zor. Welcome to a new Zor education. I'd like to solve a couple of problems in solid geometry today. It's about pyramids. We have already solved certain number of problems, some of them with little tricks maybe, and this one actually there are two problems here. Originally, I wanted to introduce only the second one, but then I decided that the first one would be really a good introduction into the second. Well, obviously as usually I encourage you to try to solve these problems just by yourself. They're all presented on unizor.com website and that's where I suggest you to which to select you from because there are notes and notes in this particular case is basically the conditions of the problem itself with an answer. Answer is very important for you if you want to just solve it yourself. So, you will check your answer for correctness. So, I will tell you why I have introduced the first one when I will start the second problem in this lecture. So, let me start from the first one, which is not really very difficult at all. Okay. What do we have? We have right rectangular prism. Okay. A, B, C, D is lower base. A prime, B prime, C prime, D prime is upper base and the sides. So, if it's a right rectangular prism, I need three dimensions. The base is a rectangle. So, it should be X and Y, widths and lengths or depths, whatever, and height would be Z. So, X, Y and Z are three numbers, which basically completely define right rectangular pyramid. So, that's the beginning. Next is what I will do is I will connect certain vertices to get a new object inside of this pyramid. I will do this. I will connect B and B prime and D prime, B prime and C, A and C, and B prime and A, and A and D prime. Is that it? No? One more. Okay. So, what I would like to do is there are eight different vertices in this right rectangular prism. I choose four, this, this, on the top base, and this and this on the bottom. And I will cut basically all other corners. Corner A prime, corner C prime, corner B and corner D, which is behind. So, I have a triangular pyramid left. I hope you see it. Let's consider B prime as its top. A, D prime, C as its base. And you have three edges, side edges, B prime, A, B prime, D and B prime, C. Now, what I would like to know is the volume of this pyramid, which is inside this prism in terms of X, Y and Z. Now, if I will try to find out its volume, just based on its own dimensions, and basically I know all the dimensions. For instance, AC is a diagonal of a rectangle X by Y, which means I know the length of AC, which is square root of X square plus Y square by the theorem, by the Pythagorean theorem. And same thing every other. So, every edge of this pyramid is basically a diagonal in the corresponding rectangle. So, I know basically all the dimensions. But it would be very difficult for me to find out its volume because I have to find out somehow the area of the base, which is AC, D prime. Well, I can do it because I know three dimensions of the sides and I can use the heron's formula for the area of triangle. But to find out the altitude of this pyramid, which is perpendicular from B prime to a plane defined by A, D prime and C would be kind of difficult, I would say. So, it's difficult. But what's easier, well, easier is to take the volume of the entire pyramid and subtract these four Pyramid, which I cut out because I'm cutting out also Pyramid. Like, for instance, from the C prime, I'm cutting the Pyramid, which has apex at C prime and the base at C B prime D, this triangle. Why is it easier? Well, because the volume of this Pyramid is much easier to calculate. Look, I know that this is also X, right? This is X. This is X. This is X. This is Y. This is Y. And this is Y. This is Z. This is Z. This is Z. And this is Z. So, this is a Pyramid, which has a right triangle as a base. And I know the altitude because this C C prime is perpendicular to the base, which is C prime B prime D prime. So, it's much easier to calculate the volume of this piece. And same thing with every other. Now, let's just think about every other. Well, let's start from this one and then we'll go to two others. What's the, for instance, volume of this Pyramid? It's one third, right? Area of the base, which is X by Y, two category of the right triangle. So, it's X, Y divided by two times altitude, which is Z, which is one sixth, well, X, Y, Z divided by six. Now, let's take any other Pyramid. Let's say the Pyramid, which has an apex B, which I'm cutting out B and the base A, C, B prime. No, sorry, A, C, D prime. Am I right? No, no, no, no. A, C, B prime. I was first the first time. So, the B is apex. And I'm cutting off the Pyramid, which has this apex and the base A, C, B prime. Same thing. The base is X by Y, right triangle. So, it's one third X, Y divided by two. And the altitude is again Z. So, again, we have X, Y, Z divided by six. So, one, two. Then this Pyramid, which I'm cutting out, A prime is apex and A, B prime, D prime is the base. Same thing. I can use the A, D prime, B prime as the base and A prime A as an apex, as an altitude. And I will have again X by Y over two. This is Z and one third. It will be exactly the same thing. So, all four Pyramids, which I'm cutting out, have this particular volume. So, there are four of them. One has an apex at A prime, C prime, B and D, which is from that side, from behind. I can consider D as an apex and then the A, D prime, C as a base. But again, I will turn it on the side. So, my base would actually be A, D, C and apex would be D, D prime. Same thing with everyone. So, all of them, there are four of them. So, it's four times XYZ divided by six, which is two-third XYZ. Two XYZ divided by three. That's the volume, which I'm cutting out. Okay, great. So, what's the volume of the original prism? Well, it's obviously X times Y, which is the base times Z. So, I have to subtract from XYZ. I have to subtract to third XYZ and the result would be XYZ divided by three. So, the volume of this particular pyramid inside is XYZ divided by three. Let's just remember this and this is the end of the first problem. And now I'm ready to go to the second problem. Now, but before that, let me just mention one thing in this particular case. Look at this interesting pyramid, which is, let's say, having D prime as an apex and A, C, D prime as a base. Well, it's a tetrahedron, right? So, let's consider, for instance, sides, it ages B prime, D prime and A, C. Now, B prime, D prime is a diagonal in this rectangle. A, C is also a diagonal in a rectangle, which is the lower base. Now, these two rectangles are the same and since they are rectangles, their diagonals are the same. I mean, they look different because there's different projections, but if you just look from the top, it will be a rectangle and two diagonals are equal to each other, right? So, what I have here, I have an interesting property of this particular pyramid. Opposite sides, C, D and B prime, D prime are equal in lengths. Same thing with any other two opposite sides, which belong to two opposite rectangles of the prism. For instance, side B prime, A and compare it with C, D prime. Now, B prime, A is on the front face and C, D prime is on the back face of the prism. Now, these two faces, D prime, C prime, C D is one rectangle and A prime, B prime, B A is another rectangle. Rectangles are the same. So, diagonals are the same. They're just two different diagonals. This is this way and this is and this is this way. So, they are the same. So, these opposite sides in this particular tetrahedron, C D prime and A B prime exactly have the same length. So, that's another pair and obviously the third pair is obviously on the same level. It's A G prime comparing to C B prime. So, this is on the right face and this is on the left face. Two faces are the same rectangle so their diagonals are the same. So, this particular pyramid has an interesting property that every pair of opposite sides and opposite means they're not really related to each other. They're on opposite sides of this prism. They're all equal in lengths. This to this, this to that and this to this. And that's the end of the first problem with this commentary and now let me go to the second problem. Second problem is the following. Given a pyramid which has this particular property that the opposite sides are the same. So, I have pair A, B and C are three different sizes. So, for instance, A B prime equals to A and so is C D prime. Now, let's say A C is equal to B and so is B prime D prime. And finally A D prime equals C. It's on the left and C B prime on the right. So, let's say we have this particular pyramid and I would like to find its volume in terms of A, B and C. Well, when I was trying to solve this problem, forget about the first problem for a second with this prism, etc., I just have this pyramid. I was trying to solve it and I was actually thinking for quite some time. I tried different ways. Something like direct approach. I mean, if I know these six edges of this pyramid, I can somehow calculate basically the area of one of the bases, for instance, again using the Heron's formula. And then to evaluate the altitude, I really have to do a lot of work. I mean, I have to draw a perpendicular to a side within the face. For instance, within the face A, B prime, D, I have to find out what's the height of this, what's the altitude of this triangle, then have a B projected down. I mean, it's really complicated calculations. I wasn't able to get through. So, I was basically at the loss. I didn't know what to do. I was thinking maybe I should really extend this particular pyramid to a prism, but not right rectangular prism, like in this particular case, but just a prism when you can do this parallel line. So, let's say if you have this pyramid, you can draw parallel line here. So, it will be like entire new prism. Something here, here and here. Right? And here. Something like this. So, I was thinking maybe it would be easier to solve the problem if I will just create this prism. I will triple the volume, obviously, right? Because the base is the same and the height is the same, but now this is the prism. So, the pyramid would be one-third of this. Now, is it a little easier? Maybe a little bit easier, but still I was not able to go through. And then, I was just doing something else and the problem, which is not this one, but similar problem, actually caught my eyes. And then I realized that in this particular problem with the prism and these vertices connected, the pyramid which came to be has this property of having opposite sides equal to each other, the same as in my problem. So, I was thinking that maybe if I can somehow convert my pyramid into this prism, then I can use basically something which I have just introduced in the problem number one as the calculation method. So, I think it's just interesting for anybody to see how somebody else is thinking. I mean, I was thinking using some direct approach, maybe some a little bit indirect, but still not exactly sufficient. And then they're completely different, which doesn't really come to mind easily. So, if you don't know anything about the first problem, if you just have the second problem, so you have a pyramid with opposite sides equal to each other like this, how can you get the volume? I mean, without knowing that it's possible to build a prism, which basically makes the whole job much easier to calculate the volume. It's difficult to come up. So, it's only when you are doing something else, your associative thinking actually comes to play. And well, it might just play in such a way that you basically realize that it can be used. And that's what happened with me quite frankly. I realized I can use. So, now the question is, can I, knowing just the pyramid, let's say B' is an apex and ACD' is the base, if I know just this pyramid, can I construct a prism which basically looks like my drawing right now? Well, I was thinking that actually it's not really very difficult. Here is how we can do it. Well, let's say we would like to construct the base of this prism. Now, I know the diagonal and if I would take this diagonal, which I also know, this is part of the pyramid edge, right? And bring it down parallel to itself. So, what I will do is I will take this midpoint of this AC future diagonal. So, I know AC, right? AC is edge of my pyramid. So, I take the midpoint and then I take the line, I construct the line which is parallel to B' and D', something like this, through this midpoint and measure the same lengths on both sides as I have here on both sides because the diagonals are equal to each other. I will get points B and D, right? Same thing on the top. If I will take the midpoint and draw a line parallel to AC through this point and measure equal parts on both sides equal to these parts, I will have points A' and C'. And now I have all 8 points of my prism. Now, the fact that this is a rectangle is obvious, right? Because I took two lines of equal size, which are intersecting right in the middle. And obviously, everything is all sides are equal and angles are right. I mean, it's very easy to prove. If you want to prove it, by the way, it's a good exercise on plane geometry. Now, so, I consider this as an obvious fact, which everybody knows how to prove. So, the top and bottom are rectangles, right? Now, I think, again, it's a good exercise for anybody to prove that these sides are also rectangles. And I don't want to spend right now the time on this because it's really kind of a simple thing. Just use the plane theorems of Pythagorean theorems or whatever. And now, having built this particular prism, I know the volume of this pyramid inside in terms of x, y and z. What I do not know is, I do not know the volume in terms of A, B and C. But they are very close related to each other, right? I mean, I can always say that x square plus y square is equal to x square and y square is A, C square, which is what? B square. Now, x square plus z square is A, B prime, which is A square. And finally, y square and z square, y square plus z square is equal to C, B prime, C, B prime, and c square. So, this is a system of three unknown variables, x, y and z. System of three equations with three unknown variables. I obviously can find out what they are. Very easy. Well, let's first do it this way. If I will summarize all three of them, I will have twice x square, twice y square and twice z square, right? So, I will have x square plus y square plus z square times two. And that will be the sum of this. So, if I will divide it by two, I will have this. Now, knowing sum of three, I can subtract from this sum of three, let's say, sum of these two only and I will get the x square, right? So, x square would be equal to sum of all three, which is A square plus B square plus c square over two minus sum of these two, y square plus d square, which is c square. Now, y square would be same thing, sum of all three minus. To get y square, I have to subtract x square and z square, which is minus A square. And same thing with z square is equal to A square plus B square plus c square over two minus B square, right? So, now, since I know x square, y square and z square, I can have the volume. And the volume is equal to square root of, well, one-third, square root of x square, x square, y square, z square, right? That's what it is. Let me just do it a little bit more accurately. Again, volume is equal to divided by three, square root of x square times y square plus z square. Now, why do I need square root and square of these? Well, because I know what is x square, y square, and z square. I don't know what x, y, and z. So, I have to use the square root, right? So, which is, now, let me just, this expression, let me just use letter m for this. So, it would be square root of m minus A square, m minus B square, m minus C square divided by three, where m is this. So, everything is in terms of A, B, and C, the volume. Well, what lessons should we actually learn from this particular problem? Sometimes, a very unusual approach can really simplify the problem significantly, because if you will try to do it, like, directly with calculations, etc., I'm almost positive that you will not be able to do it. Even if you know how to do it, let's say, you have a system of three equations with three unknowns, but they are all, like, square roots, squares, cubes, whatever. It's basically unsolvable in a reasonable amount of time. But if you do something really artificial, if you just come up with an idea which will help you to simplify the problem, that would probably be the right solution. Now, how can you come up with this solution? Quite frankly, I wasn't able to until something else prompted me. So, what's very important in this case, as a lesson, is to know that the more problem you solve, the more different associations you actually produce. For n problem you solve, you have almost like n square, the order of n square, different associations between them, or even more than that. So, it all depends on how your fantasy is working. So, you always have to have, as your repertoire, a set of approaches. So, now you probably can think about it this way. Okay, if I have some pyramid, maybe I will convert it into another more palatable, more understandable object, like in this case, right rectangular prism, which would help me to achieve the goal which I want. So, that's very, very important as a lesson from this particular problem. I would suggest you to do these two problems again, just by yourself. And well, basically, try to put it in writing in very accurate thing, maybe with a nice drawing, much better than whatever I do this, maybe in two colors. I was trying to put blue color for the pyramid, but I'm not sure how well visibility is on the video. But in any case, that's what I do suggest you to do, and I hope you will succeed. So, thanks very much and good luck.