 Hi, I'm Zor. Welcome to a new Zor education. Well, let's solve another problem on pyramids. It's a very simple problem. However, it involves certain stereometric vision, if you wish, and it also based on some problems which we have solved before, it's related to similarity in three-dimensional space. Alright, so here is the problem. Let's say we have a regular triangular pyramid. S is the apex and A, B, C are the vertices at the base. Now, it's regular pyramid, which means basically two things. Number one, base is an equilateral triangle. So all sides are equal, A, B, C, and A, C. Also, regular pyramids have the property of the apex to be above the center of the regular polygon at the base. So if you have a regular polygon, it has certain center, which is usually the center of circumscribed or inscribed circle. And in case of a triangle, like in this case, if you have equilateral triangle, then the center is basically where all the medians, all the altitudes, all the angle bisectors are intersecting. And this is also a center of circumscribed circle and inscribed circle. I mean, it's center of everything if it's the regular equilateral triangle. So what we can say is that S, the apex, projects at the center of this regular polygon, regular equilateral triangle. So let me just draw this altitude. Okay, this is an altitude of the pyramid. And this is, let's call it O, this is the center of equilateral triangle, which in particular means it's the center of intersection of medians or altitudes, etc. Now, what we do now next is we have a plane which we draw from apex S perpendicularly to the median AM through this altitude. So it actually intersects, this plane will intersect side faces, ABS and ACS. Now, since it's perpendicular to the plane and it's perpendicular to AM, and AM is perpendicular to BC, so it should be, the intersection should be something like this. This intersection between this section should be parallel to BC, let's call it PQ. And obviously it will cut something like here, the front face, and on the back it would be something like this. So we have a triangle SPQ, which is within the plane perpendicular to the base, goes through this altitude SO, and it's perpendicular to AM in particular. So, plane PSQ is perpendicular to AM. Now, it's a triangle and it has certain area, so area of PSQ equals, let's call it S. That's not what we have to determine. Now, I will do a slightly different section. I will make a cut parallel to this plane, but a little bit closer, exactly at one half. So let's call this H, and now I will draw a plane parallel to PSQ, which is perpendicular to the same AM, through this point H. So it will be something like this. So it's another triangle, a little bit closer to the vertex A. It has a different area, so let's call it P' and Q'. So what I'm talking about, and this is S'. I need the area of this triangle in terms of the S, where S is the area of the bigger triangle. So we are cutting our pyramid with two planes, which are both perpendicular to the base and in particular both are perpendicular to AM, the medium, and one is through the altitude of the whole pyramid and another is slightly closer to the A, it cuts AM in half. So H is the midpoint of AM, Q is not. So my question is, what's the area of this smaller triangle? Alright, now let's think about it. We all remember what similarity in 3D is. I'm talking right now about, why am I talking about similarity? Because obviously you can consider triangles P' S' Q'. So triangle P' S' Q', similar to triangle PSQ with A centre of scaling. So A is a centre and basically what I'm talking right now about is that we can consider this as a scaling. Now, why can we do this? Well, primarily because all the plane triangles like AP' S' and APS are obviously similar because two parallel planes, PSQ and P' S' AQ are intersecting the plane ABS which means that the lines of intersection are parallel to each other. Same thing on that side, ACS. Both lines S' Q' and S' Q' are parallel to each other. So there is obviously a similarity between these two triangles and on the back two triangles as well. And the coefficient, the factor would be exactly the same. And here is how we can actually think about this. The coefficient between the scaling factor between, let's say, SP and S' P' would be the same as between AP and AP' because these triangles are similar. In turn, AP to AP' would probably be related as AO to AH for instance because these are also parallel lines. Similarly on that side, S' Q and S' Q' would be related the same as AQ to AQ' which are in turn would be related again to AO to AH. So they are all proportional to each other. And since they are all proportional to each other, we can talk about similarity of the entire triangles SPQ and S' P' Q' with a scaling based on this center and the factor which I was just talking about. Now, next thing is to remember that in two-dimensional world scaling increases or decreases all the lengths by the same factor. And in three-dimensional case, we can talk about errors. Now, errors are always increasing or decreasing in the square of the factor, right? So if we know the factor of the scaling, then we can definitely say that the area of SPQ and P' S' P' are related as a square of the factor. So the only question remaining is what's the factor? Alright, that's easy actually. Let's consider this triangle in two-dimensional world. This is an equilateral triangle. Now, AM is a medium. AM is a medium, alright? And point all is that's where all the medians are intersecting, right? That's the center of circumference, for instance, inscribe or circumscribe. Now, remember a very old and very trivial theorem in plane geometry that medians are always intersecting at the point which cuts each of them at the ratio of 1 to 2. 1 to 2, 1 to 2. So this piece is half of that piece. So one-third of the median is the short end and the two-third is the bigger end. Well, if you don't remember it, go to the plane geometry course and triangles. That's very well explained over there. So what I can say is that AO is equal to two-third of AM. Now, point AH is the center of AM. That's H, which means AH is equal to one-half of AM, of the median. So what's my ratio? AO divided by AH is equal to two-third divided by one-half, right? Because AM and AM would be reduced, which is equal to what? Four-third. Now I know the factor because AH and AO are related to each other basically as a factor. So if I divide AO by AH, I would have the scaling factor of increase. Now, that means that the area of the whole triangle, PSQ, is greater than the area of P prime, S prime, Q prime by the square of this factor. So if my area is x, then what I can say is, let's do it this way, that the area of the big triangle, which is S, relates to the area of the smaller triangle as square of this, which is 16 9s, from which x is equal to 9 16s S. So this area of this triangle is equal to 9 16s of the area of bigger triangle. So what's important to remember here? Well, similarity in the 3G. So all you have to do is you have to know the factor, and then the areas will be proportional to square of this factor. That's important. Well, everything else is basically relatively trivial. I mean, to prove that this is really similarity is really based on trivial properties of parallel lines or parallel planes, et cetera, as they intersect with other planes and lines. Well, that's it. I would suggest you actually to try to do the same calculations just yourself. Go to Unisor.com. The conditions of this problem are there. And that's it for today. Thank you very much and good luck.