 Hi, I'm Zor, welcome to Unizor education. I would like to present a couple of more problems which I can conditionally call final problems on 3d geometry, solid geometry. They're not very difficult at all, one of them seems to be a little bit more difficult, but it just seems to be, it's not really. However, I would like also to pay some attention to be a little bit more rigorous in whatever the calculations are required for these particular problems. Okay, the problem number one, which looks actually very easy, you have a cone, circular cone which means the base is a circle and the apex projects to the center of this circle. So it's on the perpendicular to the plane where the base is and the perpendicular goes through the center of the circle. Now we have a sphere which is inscribed into this cone. Now, considering you know the parameters of the cone, radius of the base and the altitude, height of the cone, you have to determine the radius of inscribed sphere. Now, first of all, let me say something very simple which will lead us to a solution and then we will think about how to make it a little bit more rigorous. So the simple thing is that everything seems to be quite symmetrical, so if I will cut with a plane from the apex of the cone through the axis, through this altitude of the cone, in the section I will have, well the columns section would be a triangle which goes through the center of this circle as the base and the section of the sphere would be a circle which is inscribed into this triangle. Now, what do I know about this circle? Well, I have to determine its radius but I do know that the height is h, now this is isosceles triangle and this half of the base of this triangle is r, that's what I know. How can I find the radius of an inscribed circle in this case? That's called r. Well, there are probably many ways I personally prefer the following. I will connect the center of an inscribed circle with vertices of a triangle, now I can consider the triangles area being on one hand half of the base times h which is rh, right? On another hand, I can consider this triangle as consisting of three triangles, this one, this one and this one. Now each of these triangles have the altitude equal to radius of an inscribed circle, so the area of this triangle is the product of this length times r, then this triangle is this length times r and this triangle has this length times r, so all of them have the same altitude, so if I will add them together I will have the perimeter which should be divided by 2 because triangle area is half of the base times altitude, so it should be equal to one half of the perimeter of this triangle times r. Question is what's the perimeter? Well, perimeter is simple because if this is r, this is h, so this is the Pythagorean theorem square root of r square plus h square and there are two of them, this side and this side and then there is a 2r, so 2 this plus 2 this that's a perimeter. Now if you will substitute it here and this is one half so this two will go, so you will have rh equals to 2r, right? Equals to r plus square root of r square plus h square times r, so r is equal to rh divided by r plus square root of r square plus h square, so that's the answer and it's relatively simple to get. Now what have I missed? Well obviously I missed whatever I did from the very beginning like I said, okay it's obvious if we will make a section we will get something like this, so that actually should be proven and that's also part of the problem. So how can I prove that all these manipulations are lawful and I indeed will have this picture if I will cut with a plane my column along its altitude. Okay first of all again let me repeat that this is a right circular column which means this is a circle and the apex is on the perpendicular to the plane through its center. Alright now how can I prove that if I will just cut the plane I will get exactly this? Well first of all let's prove that the center of a sphere is on this altitude. How can I prove that? Well let's just take the center of a sphere somewhere here. Now first of all I was talking about this during the previous discussion of the previous problem, previous lecture. What is a tangential, what is the tangency between the sphere and a surface of a cone? Well I have defined it as the following. If my sphere is tangential to every generator which connects apex of a cone with a point on a circle which is its base so the circle so the sphere should be tangential to each point which means it should have only one common point with each other. So the tangential if you wish all the points where each particular generator is touching the sphere that's what it is. And I was actually talking about this being a circle and all these points are lying in the same plane so let me just repeat a little bit what the logic was behind it. So if this is the center what I can say so let's just take any two generators. So if I will connect my center to this and to this now I was also talking during the previous lecture that the radius from the center of a sphere to a point of tangency with a line is perpendicular to this line because it's the shortest distance everything else is outside of the sphere and only this point is on the sphere right so this is the shortest so it's a perpendicular. If it's a perpendicular then if I will take these two points of tangency to two different generators is obviously I will have triangle SOA and SOB congruent to each other because these two are radiuses OA and OB and OA is equal to OB because it's radius of a sphere and hypotenuse SO is common between these two triangles right so they are congruent which means the angles are the same. So it looks like the angle between any generatrix and this particular line which connects apex with the center of a sphere is the same right well if that is true then it should be obvious but the same the same property actually has the altitude of a cone so the angle between the altitude and any generatrix is the same. Now from these two circumstances that angle between SO and any generatrix as well as angle between the altitude of the cone with any generatrix these angles are the same actually follows and it's probably very easy to do I just don't want to spend time on this that all should belong to this altitude so the center of a sphere is supposed to be on the altitude just because SO has equal angles with each generatrix right okay well I'm just wondering how it can be done actually it's very easy if you connect S and O doesn't really matter whether it goes to the center or not but look at this consider these two points C and D so the same two generatrices I was using before SB goes to SC and SA goes to SD right now all generatrices of the cone are the same lengths that's obvious because again you can consider right triangle so any generatrix has the same lengths because all these are the same and this gadget is common for all these triangles and obviously they are all right triangles because the altitude is perpendicular to the cone right so all generatrices have the same lengths right now so SC and SB sorry SD are the same lengths now angles between generatrix let's say SC and the line which goes through the center we have already proven that this is the same right doesn't really depend on the generatrix so we have a side we have an angle and another side so the triangles are equal which means if we will put this letter M where SO hits the plane at the base so CM and DM should be the same because triangles SCM and SDM are congruent by side angle side so this is the center so that's how it probably completes the proof that my sphere is centered on the altitude and since it's centered on the altitude and it's also touching the plane of the base which means the radius is perpendicular but all M is perpendicular because it belongs to altitude so the point of tendency between the sphere and the base is exactly the center of a base okay now when everything is done what we do is we have any diameter SD and point S and then we cut with a plane SCG now we have this now since this plane goes through this point which is common point actually for perpendicular from S to the plane and from S to the center of a sphere because center lies on the on the altitude right so this plane contains the center of a sphere so point all here now SC and SD are to Generatrices as we have already known are equal lengths so it's a nice also this triangle now these actually would be points B and C and obviously this circle is supposed to be tangential to any line because if it's not then basically there is no common points between sphere and these A and B so it should be at least one but it cannot be more than one because if the circle contains two points of intersection then the sphere would contain two points of intersection with a generator this on this and this is not the case because it's tangential so that's why we have a circle which is inscribed into a triangle so all these words are actually necessary if you want to do the calculations which I did very quickly in the beginning more or less you know lawful and based on some real logic rather than just intuition and intuition was perfectly fine that's actually how things are done in many cases in science first you intuitively feel that it should be something like this and then your calculations are easy and then you think about how to somehow to support your calculations with this proof that this is really correct next problem okay next problem okay you have a cylinder and you have a triangle which is kind of inscribed into this cylinder in the following way two points of this triangle are on the base and one point on a side so let's connect A, B and C now what I know about this triangle that number one is isosceles A, B is equal to A, C number two I know that B, C is equal to six now this is a perpendicular this is isosceles triangle and this is the perpendicular to the base A, D which is median and angle bisector and an altitude rate so A, D is equal to two I also know that angle between the plane A, B, C and the base is 30 degrees that's what I know now what do I have to define I have to find out the radius of a cylinder have to find out the radius of a cylinder okay now here is the center of a cylinder now the base is a circle right so if from a center I drop a perpendicular to a chord B, C it will divide this chord in half right so D is since this isosceles triangle point G is also middle of the B, C since A, G is perpendicular to B, C so this OG is perpendicular to B, C right now so let's continue it further let's call it E now what I am actually staging right now that angle A, D, E is actually this angle beta which is 30 degrees why well if you remember how I defined the angle between two planes I actually defined it as an angle between two perpendiculars within each plane to a line of their intersection so now I know actually a lot about this now what happens if I drop a perpendicular you know what I didn't really write it correctly let me just make my picture slightly more plausible the point would be somewhere here on the surface that's A that's on the front part of the cylinder what I would like to say is the following that if I will drop a perpendicular from point A now this is a right circular cylinder I didn't say but I can presume it so it would be a generator right and it would be down to I'm staging that this will hit exactly the point E but why let's just think about it how can I prove it well let's just leave it alone and continue with calculations and then we will return to this point the same way as I did in the previous problem it's kind of obvious that since these two are equal AB and AC then if I put the perpendicular down it should be in the middle of B and C middle of the projections right because this is an isosceles triangle forget about cylinder forget about everything else it's just between the plane ABC and the base plane if you have this type of an isosceles triangle and I drop down perpendicular it should actually hit the perpendicular from C from from E to to this intersection should actually hit the middle of it but we will prove it later on now considering I have proven that what I can say right now about this particular problem well here it is I have a circle let's just consider the bottom of this a circle what do I know about this I know that there is a chord BC and they know its lengths what I don't know is well I have to determine radius right but I probably can can do it by calculating separately this particular distance OD and DE separately how can I do it well what do I know about this well I know that this angle is 30 degrees and I know AD now ADE is the right triangle I know hypotenuse AD which is equal to 2 so what exactly is my DE we're just thinking about so if I have a hypotenuse and it's 2 and this is 30 degrees well it doesn't look like 30 degrees I'll make it 30 degrees this is 30 degrees this is 2 now what is this length well remember this is 1 right because it's across the 30 degrees so it's half of the hypotenuse so this is square root of 3 right 4 minus 1 yes so I know DE so this piece is square root of 3 now BC is 6 so this is 3 and this is 3 right so what can I say about the radius of this particular circle well it's kind of okay if you will connect these you consider this is R right and OE is also R so on one hand OG is equal square root of R square minus R square minus 3 square right minus 9 on another hand OG is equal R minus square root of 3 so here we have an equation R square minus 9 is equal to R minus square root of 3 square right I squared both sides now which is which is what R square minus 9 is equal to R square minus 2 R R square root of 3 plus 3 right R is out so it's 2 R square root of 3 equals 12 R square root of 3 is equal to 6 R is equal to 6 divided by square root of 3 well let's simplify it multiply by square root of 3 both sides now square root of 3 times square root of 3 is 3 6 divided by 3 is 2 so it's 2 square root of 3 alright so that's the answer we have found what R actually radius actually is but let's go back and find out how to make this assumption that A projects to this E which lies in the same radius as OG so that's the case all right so let's just forget about the cylinder we don't really need it well what we do need is the following so if I will connect this and this and consider triangles A C E and A B E now what can I say about these triangles both are right triangles because A E is perpendicular to the plane right and that's why it's perpendicular to any one of them now the A E characters is common now A B and AC are equal by condition of this problem I said the triangle ABC is isosceles right so I have a hypotenuse and the characters which means that these two triangles are congruent which means that BE equals CE which means that BCE is the right the isosceles triangle right and since it's isosceles triangle a perpendicular from E to BC also goes exactly into the center so OG is perpendicular goes to a center of BC and ED also perpendicular and it's also goes to a center which means it's one line so the radius from A through G hits exactly the point of projection of A so that basically completes the the proof of whatever we have done is correct because otherwise we can count only on our intuition and again I would like to point your attention to this particular fact that first we kind of based on intuition try to solve the problem and then I have returned and intuitively obvious things proved relatively rigorously all right as usually I suggest you to go through all these proofs and calculations just by yourself go to theunisor.com and everything is in there these problems are presented with answers so do it and you will feel satisfaction all right thank you very much and good luck