 Hi, I'm Zor. Welcome to a new Zor Education. I would like to continue talking about pyramids, and in this particular case I would apply our knowledge about pyramids to determining the volume of a very complicated figure. I was trying to draw it before starting the lecture with my very limited artistic abilities. Anyway, this figure is called cube octahedron, and I will explain what it is later on. So, what exactly this is all about? This is just one of the lectures on advanced mathematics for high school students. I presented it on Unizor.com website. I suggest actually to watch this lecture from the website, because it contains notes and exams actually for those who are signed in, etc. So, cube octahedron, what this beast is about. Alright, first of all, we have a cube A, B, C, D, A prime, B prime, C prime, D prime. And then what we do, from each vertex, we cut the corner of this cube. Now, the way how we do it is the following. We divide in half the side A, B and A, G and A, A prime. And connect these midpoints together, three points. I cut the plane through this, through these three points. And that cuts off this little pyramid. If these points are, let's say, P, Q and R, I cut this pyramid APQR from the cube. And I repeat exactly the same thing at every vertex, and there are eight of them of this cube. Whatever remains, and this is something which I was trying to present in this picture, is actually called cube octahedron. It's part of this which is related to the cube is basically, because we started with six faces of the cube, and octahedron because you have additional eight faces of this figure which are a result of the cutting. So whenever we cut off this corner, we left another face. So we have six faces remaining from the original cube, and eight additional faces which are the result of our section. The question is, what part of the volume of the cube, the initial cube, this cube octahedron actually retains? Well, obviously the easiest way is to calculate the volume of the pyramid. There are eight of them and they are absolutely equal to each other, and subtract it from the volume of the cube, which is very easy. What's tempting is to call the vertex itself of the pyramid. This, this, or this, call it an apex of the pyramid, and the triangle like PQR, a basis. However, it complicates the calculations. It's easier to basically consider this pyramid slightly differently. I will consider ARQ as a base and P as apex. Now, it's much easier because these are all right angles. So we know the AP would be the altitude, and the ARQ, which is the right triangle, would be the base. So it's very easy to calculate the area of this particular triangle and the volume of the pyramid. So let's do it. Let's consider our cube has the side, quote, the side lengths of A. Well, then obviously the volume of the cube equals A cube. Now, let's talk about the pyramid. Every pyramid out of these eight has this side equal to A divided by 2. And AR also and AP also, all of them are half the edge of the original cube. So let's just talk about this pyramid. So this is A, this is P, and we have this pyramid. So these are right angles. So this is A over 2, this is A over 2, and this is A over 2. Now this is the right triangle, and its area is equal to A over 2 times A over 2 times one-half. That's the area of the triangle. Now the altitude is also A over 2, right? So I have to multiply it by the A over 2 and by one-third because we're talking about pyramid, right? So one-third of the product of the area of the base and the altitude. So this is the volume of my pyramid. Equals to A cube divided by 48. Now, we have eight such pyramids. Now, since we have eight such pyramids, now the volume of eight pyramids is equal to A cube O over 48 times 8, which is A cube over 6. So the volume of our cube octahedron is equal to... Well, if A cube over 6 is the volume of all these pyramids, whatever remains is obviously 5A cube over 6. Because we have to subtract from A cube, we have to subtract A cube over 6, and that's what remains. So this is basically the answer. So the ratio is 5, 6. You can say that the cube octahedron takes 5, 6 of the volume of the cube. Now, incidentally, if we want to calculate the volume of this cube octahedron in terms of its own property, and its own property is its side, right? So let's call this side B. And this, and this, and this, and this. All of these sides are equal to B, right? Because they're all diagonals in this particular... Well, not diagonals, it's just... It's hypotenuse, if you wish, of the triangles with the side A over 2, right triangles. So, if my catatris is equal to A over 2, then hypotenuse B is equal to this, right? We're talking about Pythagorean theorem. A over 2 square plus A over 2 square is equal to B square, right? So this would be my B, right? Now, if this is B, then in terms of B, this formula would look how? Well, let's just resolve it for A and substitute into B. So it's 2B over square root of 2 is equal to A. Now, 2B over square root of 2 is actually B square root of 2. So instead of A cube, I can substitute A instead of A B square root of 2. And I will have 5B square root of 2 cube. And the parenthesis should be here. And divide it by 6, which is equal to 5B cube. Now, square root of 2 to the power of 3 is square root of 2 to the power of 2 and 1. So it's 2 square root of 2 over 6, which is equal to 5 third B cube square root of 2. That's another representation of the volume of the cube octahedron in terms of its own characteristic. And its own characteristic is the length of its edge B. And that's the final formula. And that's basically all I wanted to talk about in this today's lecture. I do suggest you to try to do all these calculations yourself, because it's kind of useful thing. And it's a very interesting figure, this cube octahedron. It has 6 square faces and each of these, which are 8 of them, 8 triangular faces. So if you will count how many faces and edges and vertices this figure has, it would be what? We have faces. Faces is equal to 6 squares from the original cube plus 8 triangles. So it's 14. Now, edges. Okay. So we have... It's not easy to count them, right? Well, I think we have 4 edges here and 4 here, here and here. So these are all edges of the squares. But triangles will not bring up any new edges. So it looks like I have 6 squares, each of them has 4 edges, so it's 24. It looks like it's 24. I think that's what it is. Because triangles do not bring any new edges, because each edge of the triangle is actually edge of a corresponding square. And squares between themselves do not have any common edges. And now let's talk about vertices. Okay. So what's about vertices? Let's think about how to count vertices in a nice and convenient way. Well, how about this? I have 6 squares, each square has 4 vertices. However, each vertex is counted twice as part of this and part of that. So it looks like I have 12. So vertices seems to be 12. Now let's count them differently. If I'm counting only by triangles, triangle has 3 vertices, right? Now we have 8 triangles, 3 by 8 is 24. And again, every vertex in this case also counted twice in this triangle and in this triangle. So it's again 12. It looks like I've right. And let me just go back to the Euler's formula about relationship between F, E and V. F plus V minus E equals 2. This is Euler's formula, which we will still attend at some time in the future, but it still checks here. 14 plus 12, 26 minus 24 is equal to 2. So that's it for today. Thank you very much. I do suggest you to calculate whatever I just did on the board yourself. And good luck!