 Hi, I'm Zor. Welcome to a new Zor education. Right now we are starting a series of lectures, which I believe is the purpose of the whole course. Solving problems, of course. Well, in this case we will solve a few problems in solid geometry. Problems primarily related to pyramids. Obviously we will have other series of lectures for other aspects of solid geometry, but right now we are talking about pyramids, and we will solve three or four different problems. Obviously I suggest you very strongly to try to solve these problems by yourself. They are all presented on Unizor.com website. That's where the whole course actually is. It's much better actually to go to this website rather than watching from YouTube, because the site has comments and there is certain functionality in the site, which allows you to take exams, etc. Alright, so right now we are talking about a very simple problem. Well, it's as simple as solving a quadratic equation basically. However, what's interesting about this problem is that it's not solved just using absolutely direct method. Like you have a formula, for instance, for a volume of pyramid, and then you just apply this formula and everything goes smoothly. Now, in this particular problem you have to just make one little trick, if you wish. And that's very important actually. I don't really like the problems where you just go with some kind of calculations according to existing algorithms or formulas, etc. Problem must contain something interesting in itself which forces you to think about it. And I think this is an example of a simple problem which, after this little trick is applied, is really simple. Before, if you are attempting to do it a little bit more straightforward, that would be much more difficult. Let's put it this way. Alright, so what's the problem? As follows. You have a triangular pyramid which has an interesting property. Actually, this is a solid line. I see it. Alright, triangular pyramid. S, A, B, C. Now, what's known about this pyramid is the following. First of all, all angles are straight at the top. So, angles A, S, B equals angle B, S, C equals angle C, S, A equals 90 degrees. So, all these are right angles at the apex for every side phase. So, every side phase is at the right triangle where side edges are the cavity of these triangles. So, that's known. Another known thing is that three side edges represent their length actually, represent arithmetic progression. So, we know one of them, A. We don't know this and that, but we do know that they represent an arithmetic progression, which means S, B is equal to A plus X, where X is a difference, and S, C is equal to S, B plus X. Or if you wish, S, A plus, I mean, A plus 2X. So, that's also known. Now, is that sufficient to basically define our pyramid? Well, not exactly. We have one more condition. We have the volume of the pyramid equals E. So, we know A, we know B, and we know that these angles are right angles. Now, what we have to define, we have to determine actually is all these three edges. Well, we know one of them. So, basically, we have to define, we have to determine the X, the difference in this arithmetic progression by how much this edge is longer than this and this is longer than that. Okay, now, obviously, it would be nice to have some kind of equation relative to X, and the equation probably would be something like, well, if you can express the area of the base, A, B, C, in terms of A and X, which probably is possible. I just don't know how to do it easily, but I'm sure there is a way. You obviously can determine, since you know A and you know A plus X, and this is the right triangle, you know hypotenuse. So, you know hypotenuse, which is A, B, same thing. You have this, same thing. You have A, C. So, you know all three sides of this triangle in terms of A and X, or A plus X and A plus 2X. So, you know this using the theorem, the Pythagorean theorem. Now, since you know three sides, you basically know the triangle, which means you can determine its area that would be the area of the base of this pyramid. You can use the Heron's formula or whatever else. Now, a little bit more difficult would be to determine the altitude of this pyramid. Not impossible, obviously, but a little bit difficult. So, it will be lots and lots of calculations. And I think that it would be so much that you would probably give up, because it's just, you know, too tedious and boring to tell you the truth. We must think about something else, and that's exactly what I meant when I was talking that there is a little trick which would allow you to simplify the problem very, very significantly. So, what's the trick? Well, look, this is a triangular pyramid. Now, in the triangular pyramid, any vertex can serve as an apex of the pyramid, and the opposite face could be a base. In this case, I said that S is an apex and ABC is the base of this pyramid. But let's turn it on one of the sides. Let's say, let's consider ABS as a base and C as an apex. Let me just redraw this pyramid. So, C would be on the top, and then would be going down, would be CS. And from the point S, at the right angle, I have 1 and 2. So, I have A and B. So, this would be my base. In this case, this is invisible. And what's interesting is that this angle is the right angle, this angle is the right angle, and this angle is the right angle. All three angles at this particular vertex are right. Now, what does it buy us? Well, it buys us a lot, because number one, we know that since CS is perpendicular to SB and SA, CS is perpendicular to the ABS, which is the base. So, CS is an altitude. Now, this is A, this is A plus X, and this is A plus 2X. So, we already know that the altitude is very simply expressed in terms of A and X. In this case, altitude would be very complex in terms of A and X. In this case, it's much easier because it's already perpendicular. So, all we needed to do is to turn it on a side. Now, as far as the area of ABS, the area of the base of this pyramid, it's also simple because it's the right triangle with two characters A and A plus X, which means the area of the base is equal to A times A plus X divided by two. And altitude is equal to A plus 2X. So, very easy. Volume V, this one, is equal to one-third of the area of the altitude, A A plus X divided by two, and A plus 2X. And this is an equation for X, and this is a quadratic equation, obviously. So, all we have to do is just to solve the quadratic equation. And that would give us S, X, and knowing X, we know all these sides, because one is A, another is A plus X, and another is A plus 2X. So, let's just solve this quadratic equation and that will be the end of it. It's much simpler than to fight the original pyramid. I mean, it's still a regional pyramid, but consider it slightly differently because in triangular pyramid case, any apex can be... any vertex can be an apex and any face can be a base. All right, so let's just solve this quadratic equation. This is a plain exercise. All I have to do is make it right. Equals. So, 6 goes to D, right? So, here I have this, but every one of them is multiplied by A. So, X squared would be what? 2X squared and A. So, it's 2AX squared. Now, X would be AX and 2AX. So, it's 3AX and A. So, it's 3A squared X. And finally, the free one would be A squared and A. So, it's A cubed. So, that's my quadratic equation. I will do this way. I will put minus 6V equals to 0. All right? That's a canonical form of quadratic equation. So, solution to this R. In the denominator, I have double first coefficient, which is 4A. Now, here I have minus the second coefficient minus 3A squared plus minus square root of square of this, which is 9A to the fourth, minus 4 times this and this. 4 times this is 8A to the fourth. And this is 8A6 plus. So, it's 4G8A. All right, X equals. First of all, forget about this minus, because we are talking about the lengths. Excuse me. We are talking about the lengths of the side. And so, it should be positive. With 2 minuses, it will be negative. I mean, it's not the lengths, actually. It's the difference between the lengths. We have basically set the smallest one as A, and then A plus X and A plus 2X. So, we are assuming that X is positive. All right? So, the solution would be for A minus 3A squared plus square root of 9 minus 8A to the fourth plus 48AV. Okay, that's the solution for X, which means that we have determined all the side edges. This was given to us as B was equal to A plus X, which is A plus this expression. And SC was A plus 2X, which is A plus 2 times this same expression. So, that's the solution. That's the end of it. We have determined all three sides. So, what I suggest to you right now is to go to the notes of this lecture, read again the problem. Well, it has an answer, by the way. This formula is an answer. And try to do it just by yourself. Turn it on the side, draw a picture, and just write down this is the base and this is the area of the base. This is an altitude. That's its value. And then the value is equal to, etc. So, I do suggest you to do it yourself. Very helpful exercise. Well, that's it. Thank you very much and good luck.