 Hi, I'm Zor. Welcome to Unisor education. Today is an interesting lecture. I call it not so easy problems about cylinders. Well, it's a part of the advanced course of mathematics for high school presented on Unisor.com and I suggest you to watch this lecture from this website because there is a very detailed there are detailed notes about everything. Whatever the lecture is, notes basically are like your textbook. So right now we have covered theoretical portion of many different solid objects in geometry and one of them is obviously cylinders. I would like to present the problem which might actually seem a little difficult from the very beginning. However, what I would like to say about this problem is number one, it's not a tricky problem. I mean, it's relatively straightforward, which involves some calculations, but no more than Pythagorean theorem as far as your theoretical background requirements are. So what also is interesting is this problem, you might not actually know how to approach it without really thinking about the way, the steps, and I actually was thinking about this problem myself in certain order analyzing how I can approach this and I have structured this lecture in exactly the same fashion. I will analyze the problem first. How can it be solved or calculated or whatever? And then after this plan is already in my mind, I was able to do it just step by step and every step wasn't really very difficult at all. What was most important in this particular problem was to create this set of steps, more or less easier steps, which combined together constitute a complete solution to the problem. And that's how probably, you know, every relatively complex problem should be solved. First, you have to plan, you have to analyze it. What can be the way to solve this problem and then implement step by step? Okay, back to business. Here is the problem. You have a cube. I tried to present this picture to the best of my drawing capabilities, which are not really very high anyway. But anyway, try to imagine bear with me. So there is a cube. Now, ABCD is bottom and A prime, B prime, C prime, D prime is the top face of this cube. Main diagonal is from this corner on the bottom a two opposite corner on the top, C prime. This is the main diagonal of the cube, the longest possible segment among two different vertices in the cube. Okay, so this is the main diagonal. Now, we have a cylinder which has the main axis, the central line between two bases of the cylinder, coinciding with this main diagonal of the cube. Now, obviously, it's supposed to hit the three faces of the cube on that side and three faces of the cube on this side, wherever two edges of the main diagonal are. So they are basically like, inscribed. If you wish this base, this circle is inscribed into a trihedral angle on this, and that's around this particular vertex. And on this, the base of the cylinder, this circle is inscribed into trihedral angle formed by these three faces, which has the common vertex A. Now, what else is known about this particular cylinder is that its diameter is equal to its height. Well, you obviously understand that if you will take a fatter cylinder, which has a greater diameter of the base, it will not go as far into the corner as if the base is a small circle, right? So as you are increasing the circle, the base, the height should actually decrease of the cylinder, right? The height is decreasing and the base is increasing. If we want it to be inscribed into this particular cube in this particular way, right? So bases are tangential to trihedral angles. All right, so there is one particular dimension of this cylinder when the diameter of the base is equal to the height. So it's not as I have draw that in this particular case. It should be fatter and shorter. So the diameter would be greater and the height would be smaller and then when it will be equal to each other. So we are considering the case when they are equal. Now in this particular case, and that's the problem right now, you have to determine the ratio between the volume of the cylinder to the volume of the cube. So there are no dimensions given at all, but we do have to determine the ratio between the volume of the cylinder and the volume of the cube using whatever this information I have just provided about how this particular cylinder is inscribed into into a cube. Now, first of all, does it make sense if we don't give any dimensions, but we still require some quantitative characteristic? Well, actually let's think about this way. All the cubes are similar to each other with a special with some kind of a scaling, right? Now, if you have a cylinder which has a diameter equal to the height, all these cylinders also obviously are similar to each other. So if you have a one particular cube and the cylinder inscribed into it of that particular type, the cylinder with diameter equal to the height, and then you will blow it out completely using some kind of scaling. Well, the result will be similar, but as you know with any scaling the volume grows as the scaling factor in cube. Which means that those volume of the cylinder and volume of the cube will grow by the same factor, so the ratio is constant. So that's why the problem makes sense. Now, how to approach it? Well, obviously the best way to do is let's just assign some number D to the size of the edge of the cube and then calculate based on this diameter of the cube, which is easy. It's D to the third degree, right? D times D times D. And now, using the same D, we have to evaluate the volume of the cylinder, and then we will divide it one by another and hopefully D will be reducible. So we will have just the number. So that's the plan. All right, plan is great. And again, the volume of the cube is easy. Now, how can we calculate the volume of the cylinder? Well, as we know, volume of the cylinder is the product of the area of the base times height. So we need the base and we need the height. Now, the only thing which we know about the base and the height is that the diameter of the base is equal to the height. And we have to determine these characteristics of the cylinder using only one parameter D. So the question is how can we do it? Here's what I suggest. Now, since the main diagonal of the cube coincides with the central line between the bases of the cylinder and if you have a cylinder, the central line is obviously perpendicular to both bases. So I can say that the central line, AC, or the main diagonal of the cube base, AC prime, is supposed to be perpendicular to the plane which contains the base. Now, in this particular case, this triangle is the intersection between this plane, so the plane where the base of the cylinder is, and three faces of the cube. So since this particular circle is inscribed into this trihedral angle, so if you will cut the plane where this particular base is located, so the plane also will cut obviously these three faces. Now, how the plane can cut these three faces? Well, it will cut each edge A, A prime, A, B, and A, D at certain points, P, Q, and R. Now, from the considerations of symmetry, you should feel that if you have a cube and you have a plane perpendicular to the main diagonal of the cube, it should cut equal segments around this particular vertex from which the diagonal is originated. Now, why is it so? Well, consider it this way. So the plane is perpendicular to the main diagonal, right? So what happens if I will turn the whole picture around this main diagonal? Now, if you have a line and you have a plane which is perpendicular to this line and you turn the whole picture around this line, the plane turns into itself, right? So if you will turn this particular cube around the main diagonal, then this plane which intersects these three edges which are emitted from the vertex A will actually turn into itself. It will be the same plane. The image after the transformation of the rotation of the plane will be the same. Now, point D will turn into A prime, A prime into B, and B will be turned into D, right? If I will turn it this way, the whole cube. So that's why this point of intersection should be transformed into this point of intersection. So that's why, since the plane remains the same, the diagonal, the axis of rotation remains the same, and this point becomes this point, so this should become this. So it feels like it should be the same distance, primarily from these considerations of symmetry. Okay, we have to prove it. And I think this should be our first step. So if I will have a plane which cuts our cube, and the plane is the plane where the base of the cylinder is located, then it's perpendicular to the axis, obviously, because the base is perpendicular to the axis, and that's why it should actually cut equal sides. Now, if I have proven that, and I will, it will be our first step. What does it buy us? It buys us the following. If you consider this triangle, PQR, which consists, which has vertices of intersection of this plane, where the base of the cylinder is located, and three edges of the cube, then the base as a circle, what I'm just stating right now, and it should be relatively obvious, this base, which is a circle, within the plane, because the plane PQR is in the same plane as the circle, because it's the intersection of the plane where the base is located with these edges of the cube. So the PQR triangle and this particular circle, they are very much related to each other. Triangle and the circle are in the same plane, so we are talking about the same plane. And what I'm stating right now is that the circle is supposed to be inscribed into this triangle. Now, why is that? Well, let's just think about it. I mean, this is the line where our base plane intersects with this particular face. Now, if my circle is touching the face, which means there is only one common point, and the intersection between the plane of the base and the face is PQ, then PQ should be actually intersecting the circle in one point. And one point between circle and the segment is actually only when the circle is tangential to a segment. So it feels like it should be inscribed circle. So that will be my second step. So after I have proven that these three segments are equal, I can probably calculate the radius of the inscribed circle, and that would give me the radius of the base. Now, I know radius of the base. Now, how can I get the height to equate, to get some kind of equation or whatever? Well, let's just think about it. How can I get the height of the cylinder? Height of the cylinder is equal to the height, to the length of this main diagonal, minus heights of two pyramids, where A is a vertex and PQR is a base of the pyramid. So turn it this way. And from that side, C prime is the apex, and P prime, Q prime, R prime is the base. So this pyramid must be the same as this one. That's also something which should be very briefly discussed. It's obvious. And then knowing the height of the pyramid, I can subtract two heights of the pyramids from the diagonal, and that will be my height of the cylinder. Well, knowing the height of the cylinder and knowing the radius of the inscribed circle, I can equate them into each other. And that's how I will find my dimensions of the cylinder based on one parameter, G, which is the H of the cube. So that's the plan. All right, let's try to implement it. Okay, so the problem number one is, I would like to prove that these three segments, A, P, and AQ and AR, are equal to each other. If PQR is a triangle, which is formed by intersection of the plane perpendicular to the main diagonal. Okay, how can I prove that? Well, I'll do it the following way. If I will prove that PQ is parallel to the diagonal BD, PQ is parallel to the diagonal BD, what happens? Now, this is a square, right? A, B, C, D is a square. Diagonals are perpendicular to each other. So if my PQ is parallel to BD, obviously, since these are equal to each other, triangles are similar, since this is a parallel, right? So all angles are the same. From equality of these two, follows equality of these two. So that would be easy. All right, so how can I prove that PQ is parallel to BD, or which is the same? Instead of parallel to BD, I can prove that it's perpendicular to AC, right? Perpendicular to AC, because it's a square, right? All right, so let's just think about it. Consider triangle A, C, C prime. A, C, C prime. It's basically a vertical plane catching through our cube. Now, what do we know about this? We know two things. Number one, C, C prime is perpendicular to this triangle AC, C prime, right? Therefore, no, not this one. C, C prime is perpendicular to the base, to the ABCD. C, C prime is perpendicular to the base, right? This is the cube. So the edges are, side edges perpendicular to the base. Now, what follows from this is that C, C prime should be perpendicular to any plane, any line on this plane, right? So if the plane is ABCD and C, C prime is perpendicular to the whole plane, then C, C prime is perpendicular to any line on this plane, right? To this, to this, it's all perpendicular. And even outside, right? Because you can always draw a parallel line. It will be perpendicular. Therefore, these are perpendicular. That's the definition of the perpendicularity between the skewed lines. All right, so. Therefore, C, C prime is perpendicular to PQ, right? Since C, C prime is perpendicular to entire plane, it's perpendicular to any line on the plane, including PQ. All right, what's next? We also know that AC prime, AC prime is perpendicular to the plane which cuts our cube, which is triangle PQR. Because that's how we actually draw the plane. It's perpendicular to the main diagonal. The plane is the plane where the base of the cylinder is and we have stated that the axis of cylinder, the central line of the cylinder coincides with the diagonal, right? And obviously the central line of the cylinder is perpendicular to its base. So PQR as the triangle within the base, within the plane which contains the circle base, is supposed to be perpendicular to AC prime. And therefore, again, AC prime is perpendicular to any line on this plane PQR, on the triangle PQR, including the line PQ. So again, if the line is perpendicular to a plane, it's perpendicular to any line on that plane. So line PQ actually has two properties. It belongs to two planes. ABCD and PQR. It's an intersection between these two planes. So as part of the ABCD is perpendicular to AC prime. As part of the PQR, it's perpendicular to AC prime. So it looks like PQ is perpendicular to AC and is perpendicular to AC prime, which means what follows is that PQ is perpendicular to triangle ACC prime, ACC prime. Again, since PQ is perpendicular to CC prime, because it's part of this plane, and PQ is perpendicular to AC prime, because it's part of the PQR, therefore it's perpendicular to entire triangle ACC prime. And therefore it's perpendicular to any line on this triangle, on this plane, which encompasses this triangle, including line AC. So PQ is perpendicular to AC. PQ is perpendicular to AC. And again, within the square, ABCD, if PQ is perpendicular to AC, it's parallel to BD. So these triangles are similar. Now these are equal, which means these are equal as well. So AP is equal to AQ. AP is equal to AQ. AP is equal to AQ. So we have two of these three segments equal to each other. Now you obviously understand that I can do exactly the same logic with another plane. And I will have, because it's all symmetrical, right? So if I found how AP and AQ are equal to each other, using different, instead of AC prime C, I can use something like BD prime D, that particular plane. And I will see that another pair of lines, pair of segments would be equal to each other. So all these PQ and R are equidistant from A along the edges of the cube. So that's my first theorem. Now, what follows from it is, let's introduce another variable, call it X. That's X and this is X. They're all Xs, right? So using this X, I will express the triangle dimensions of the triangle PQ and R. Then I will use it to calculate the radius of the inscribed circle. Then I will try to use it to calculate the height of the pyramid with apex A and base PQ R, this from the A to the center of this circle. And that would give me another one and then I will have an equation. If I will equate the diameter to the height, I will find an equation where I will get the X from, all in terms of D, all right? Next. So I have proven that AP is equal to AQ equal to AR is equal to X. Now, let's consider triangle PQR. Now, what is PQ, for instance? Well, this is X and this is X. This is X. And obviously APQ is the right triangle. Two characters are equal to X and X. So hypotenuse, which is PQ is equal to X square root of two. Same thing with PQR, X square root of two and QR. So it's equilateral triangle with a side equal to X square root of two. Now it's a very simple thing to calculate what's the radius of the inscribed circle. This is the base of our cylinder in terms of X, obviously. So what is this? Well, you know that in the equilateral triangle center of medians, altitudes and angle bisectors, they're all intersecting in this center, right? Now, since this is X square root of two, this is half of it. So entire height is equal to X square root of two, square minus X square root of two, two square, square. This is entire altitude. Now the radius is, you remember, the old medians are always intersecting in the ratio of one to two. So it's one-third of the median. So I have to divide it by three and this is my radius of the inscribed circle. Okay, let's simplify it. One-third square root of two X square minus X square root of two, right? Two-four, yes. So this is three over two, three seconds, right? So it's one-third square root of three, seconds, and X goes outside. Well, let's not have radicals in the denominator. I'll multiply it by square root of two. I will have X square root of three, square root of two divided by six. So this is my R. Let's write it down. Radius of inscribed circle is equal to X square root of three times two divided by six. Okay. Now, so I know the radius of the circle. That's a lot, but we now have to determine the height. And the height I will determine from the fact that the height is equal to the diameter of the circle. On one hand, on the other hand, the height of the cylinder is the diagonal length minus two heights of the pyramids, which I have on both opposite vertices, right? So let's draw this pyramid slightly differently. So I have A on the top, and then I have something like this, PQR. R. So PQR is equilateral triangle, and A is projecting to its center because everything is equal, right? AP is equal to AR equal to AQ, so all of them are equal to X. And this is X square root of two. This is X. So the question is, what's the AP? Well, what is the height? Let's call this point, let's say M. Now, obviously I can use again the theorem of Pythagorean theorem. All I need is this, from the vertex of the triangle to the center of triangle. Now, back to equilateral triangle. If this is the center, then this is the segment which I need. And it happened to be twice as big as this one, which is the radius of the inscribed circle, which we have already determined, right? So this is the radius, actually, of circumscribed circle. Radius of circumscribed circle around equilateral triangle is twice as big as inscribed circle, because this is one-third, and this is two-thirds of the median. So I can determine immediately the radius of the circumscribed circle, this Rm in this drawing is twice as big, divided by three. And now I can determine the altitude, AM, of the pyramid. It's a Pythagorean theorem, again. H is equal to, hypotenuse is X square, and the catatoules is square of this, which is X square times six divided by nine square root. Well, instead of six, nine, I can put two-thirds, right? So I will have one-third. So I will have what? X divided by square root of three, is that right? Right, X square root of three divided by three. So that's my pyramid height. Now, what's the height of the cylinder? Therefore, H is equal to the full diagonal of the cube, which is D square root of three, right? Well, if this is D, this is D, this is D, this is D, AC is equal to D square root of two, right? D square plus D square and square root of it. So that's D square root of two. And this is D, again. So CC prime is equal to D. So AC prime is equal to square of this, which is D square times two plus square of this, which is D square and square root. So it's D square root of three. So is equal to D square root of three minus two H. Minus two H, because there is a height of the pyramid here and height of the same pyramid there. I didn't spend some time actually proving that the pyramids are exactly the same. I think it's just a very trivial consideration if you will just consider a couple of triangles over there. So let's not spend time on this. So pyramids are the same, their heights are the same, so I subtract two heights and that's what I have. Now, what should I say that H is equal to two R? That's my equation, right? From which I can determine X. So let's put it here. H is D square root of three minus two H, which is two X square root of three divided by three, right? H minus D cube three minus two H, which is this. That's H, right? And two R is X square root of three square root of two divided by three, right? That's the equation for X. Okay, square root of three is happily reduced. We multiply by three, that would be 3D, from which X is equal to 3D divided by two plus square root of two, right? Okay, I don't like radicals in the denominator, I will multiply it by two minus square root of two and that would be four minus two, which is two, 3D, two minus square root of two, that's X. Okay, I have X now. Let's, so all we actually need right now is, if I have the X, I have the R, so I don't need anything more than that. I just need the R. So X is equal to 3D, two minus square root of two divided by two. So what's my R? It's X, 3D, two minus square root of two over two times square root of three, square root of two and divided by six. That's R, R, right? Okay, six and three can be reduced. What else? I think I will multiply this square root of two to here. I will have D square root of three divided by four, D square root of three divided by four and here I will have two times square root of two minus square root of two by square root of two, which is two, which is, so two outside and divided by four, it would be two square root of two, two minus one, I think that's what it is. Yeah. Okay, I've got R and if I have R, I have also H, which is equal to two R, right? So H is equal to D square root of three, square root of two minus one. Okay, now we can calculate the volume. The volume of the cylinder is equal to the area of the base times height, which is pi square root, which is D square three, square root of one, square divided by four and times H, which is this, which is another D and square root of three and square root of two minus one. So we have determined the volume. Volume is equal to three D cube square root of three and square root of two minus one to the third degree divided by four. Now, what's the volume of the cube? That's D cube. So what's V divided by V cube? What's the ratio? It's three D, sorry, no D. D is as I promised reduced. Oh, I forgot pi, I'm sorry, there is a pi here. Pi square root of two minus one to the third divided by four. So that's our ratio and that's the answer to the problem. Now, what I suggest you to do is read very carefully the notes for this lecture. It's very helpful. I was trying to use some kind of a reduced symbolic notation for the proof, whatever I was just proving. And then there are some other formulas. I don't present any calculations inside the notes. I just give the results, for instance, result what's R, what's D, et cetera. So I suggest you to do something yourself. Well, basically that's it for today. I do suggest you to use the website unison.com, not only for source of the lectures, but also as the website, which can set the educational process you can enroll into topics you can take exams and get the marks. And if you have a supervisor, which is actually watching over your marks, then supervisor can actually mark this particular topic as completed and enrolled into a new topic, et cetera. So the whole process can be arranged around the website, unison.com. Thanks very much and good luck.