get started. So last week we studied 2D geometry and the focus was on triangles and today's topic is on circles. So let's get started with this warm-up question. In the picture point O is the center of a circle with a radius r. The length of the segment BC is equal to r. Alpha is this angle DOA and beta is this angle OCB and we are asked to find the relation between alpha and beta, specifically what is the ratio of alpha over beta. So last week we talked about standard step in solving geometry questions right and you know the first step is just you know label as much as you can. The nice thing with circle is that we have a lot of isosceles triangles because any triangles that has two sides being the radii of a circle that triangle is is automatically a isosceles triangle. So in this case we have OA equal to OB so we want to just indicate it OB and then in addition we have a BC also equal to r so we have another isosceles triangles right that's pretty much what we have in this picture and using that information we just do angle chasing right indicate all the angles that you have in the picture in terms of alpha and beta and from that hopefully we can find the relation between the two variables. I can start by labelling what angles that we know, so this one is obviously beta, right? So last week we talked about the exterior angle theorem. So these angles, it's going to be, you know, exterior angle at B, so it's going to be sum of the interior angles at O and C. Of course, we can go through the intermediate step of 180 minus 2 beta, but, I mean, this works if beta is a number, but when it's not a number then usually we don't do this step and it's much easier to invoke the theorem and using, like, this is 2 beta right away. And then, because OB equal to OA, another isosceles triangle, so we have another angle of 2 beta over here, right? And this one, if we want, we can label as 180 minus 4 beta, I mean, there's no harm to do that, it's a step that we can skip, but, you know, choose this one, we can put as 180 minus 4 beta, right? So very standard step of what we talked last time, label the pictures, identify all special triangles, and then, because we are dealing with angles, so obviously we have to use a triangle sum theorem so that this is a straight angle, so again we have alpha plus 180 minus 4 beta plus beta, and that is a straight angle, so that's 180, right? And from that we would have, this is alpha, so alpha minus 3 beta equals to 0, so the ratio between alpha and beta is C, alpha equal to 3 beta. So this is pretty much like what we did last time, but, yeah, the only thing that's new is that we have, you know, right away an isosceles triangle because this is a circle, so we're going to do a lot of angles and perimeters area for today's class. This is number 9 circle, we have another one on 3D geometry. So this is a part of the circle with a quick review, this is the tangents, the lines that meet circle at only one point, the segment, this is a sector, right, part of the circle going from the center, this is an arc, this is a quad, so arc, we have to calculate this curve and quad is just the straight segment here, the secant, the circumference, and we have two formulas that we use a lot, area is pi r squared, circumference 2 pi r, this is the dimensions of area and this is linear dimension, right, so just make sure we don't confuse the two. Okay, let's do the first one. A coin with a diameter of 1 centimeter rolls around the outside of a regular hexagon with an edge length of 1, so this length is 1, as shown in the picture, what is the length in centimeters of the path traced out by the center of the coin? So we imagine the coin just rolls around here, make a complete loop back to here, and we're asked to find the length of the path it traces out. So either you have a handout or you can quickly draw these pictures, and that involves a little bit of, I don't know, physics or what, but we've got to draw, right, how the path looks like. I'll give you a few minutes to see. So there's old, oh, so I launched a new one. so obviously if it's like a wheels right obviously if on you know on the straight line and such surface like this it just follows a straight line right and you know at a constant height from this and probably what we're worried most of what happens at the corners right so at least I know that it should be like this it leads from here to here right and but what happens is a corner when it turns that what happened and over here if you know it's only engaged on this side then it also just trace out a path parallel to the edge of the hexagon so something like this right and the same thing in all the other edges but what happened at this point Yeah, since the, so the contact point is here at the edge, right, it takes some time to sort of like rotate to change the direction, but the point is that the center is always a distance r, the radius one here, from that point, so that means the center has to trace out an arc of a circle, right, there's no other way, it cannot be a straight line because then the distance from here to this point is not r, right, so that means from here to here it's going to have to trace out an arc of a circle, and then we got to calculate this arc, this, so this we know how to calculate, but that we do not know, okay, so this distance we know that's one, right, one, and the diameter of the circle r is a half, so this distance here is a half, that's why, because we have a diameter equal to one and the edge length equal to one, so this is a half, and then now we have to know like how much, what fraction of a circle this one is, so for that we have to use angles, right, so this one is 90 degrees straight from here, this is 90 degrees, and this is a regular hexagon, so regular hexagon, remember we can, we can like, you know, divide them into six congruent equilateral triangles, so that means this one's going to be 120 degrees for that, and 120, 90, 90, that means this is 60 degrees, and a full circle is, a full circle is 360 degrees, so that means this is a sixth of a circle, but we have a one, two, three, four, five, six, so we actually have a full circle, right, very good, you already started to get the correct answer, we have a full circle, we have a full circle, and plus six edges of the hexagon, and the full circle, that's going to be two pi r, right, two pi r, that's the circumference of a circle, but r is equal to a half, so this is going to be two pi and a half, so that's the circumference of the circle, exactly pi, and the six edges here, it's going to be six because it's one, one, six, so the answer is b. So I think for this question, the hardest step is this, and maybe, I don't know, whether it's a physics flavor or not, Andrew, but yeah, this could be a tricky thing, but when you are in DAO, which is, oh, I'm sorry, we're not supposed to turn on the camera, sorry, but because of the setting, Andrew, could you please change the setting, okay, good, so that is the first question that we make use of the formula because of circumference. So the next one is with area, Andrew, could you please do this question? So the diagram shows three large circles with equal radius, and four small circles with equal radius, where the centers of all circles and all points of contact lie on one straight line. The radius of each circle is one. What is the area of the shaded regions? So we have a radius, it's one, Yeah I just basically summarized the techniques that we discussed last week about how to calculate area and the circle is also a regular shape right because we don't have a formula to calculate area directly. Do you see the poll, Andrew? Okay, good. Okay, so to solve this question, we should use indirect calculations of error. So if we look at the shape, you can see that we know that all four of the small circles are equal, and all three of the large circles are also equal, and that all the centers of the circle are collinear, so they're all lying on one point. So because we know that, we know that this area here is actually equivalent to the sum of these two areas here, and that this area here is equivalent to the sum of this area as well. So basically what this question is asking us is for the area of the larger circle minus the area of the two smaller circles. So in order to calculate that, we can take the radius of the larger circle, which we know is 2. So the area is 2 squared times pi, and then we subtract the two small circles of 2. So we have 2 times 1 squared times pi, and this is equal to 2 pi. So our answer is B, which most people got. Yeah, thank you, Andrew. Yeah, so for this last time we used complementary areas, this one we have to move around a little bit, and there's a lot of symmetry, so that is also something that we also use again in today's class. Move to the next one, inscriber angle and central angle. So you have a central angle here, like central from this defined as an angle, you know, standing from the, from the center of the circle. Okay, so this is like one of the major theorem in circles. We have a few questions on this, not a whole lot, but basically the theorem states that an inscriber angle subtended by the same arc side equal and equal to the half of the central angles. So whether you, if you have A and B here, and these are angles subtended by arc AB, and it doesn't matter where C is, right? This is A and B, we have arc AB. It doesn't matter where C is, C, C prime, C, C prime, this angle is always the same, and that is equal to half of the angles, the central angle here. So this is like a very beautiful theorem, but we don't have a lot of problem with that, but this is sort of a special case that we use a lot. So if we have angles in a semicircle, right, then if A and B is diameter of the triangles, I'm sorry, of the circles, then the angle C here, ACB is 90 degrees. And actually, this is something that you can easily prove, right? Because if we have a O here, then we can connect O and C, so we have a COB and COA, both of them are isosceles triangle, if this is x, this is x, this is y, this is y, then we have a 2x plus 2y equal to 180 degrees, and that's why x plus y has to be 90 degrees, right? And just like another way to define a circle is that it's the locus of the points that, you know, that, you know, if you take C over here, I'm sorry, it's the locus of the points that make ACB a 90 degree angle, so like all those points, you also get a better circle. So just pay special attention to it because it is a theorem that we use a lot. And then we have that, then of course, we can use Pythagorean theorems, so let's go, I think these are very beautiful questions here. Suppose there are two intersecting circles of different radii, big one and smaller one, the center of the larger circle with radius r lies on the smaller circle, and the points at which the two circles intersect are the endpoints of the diameter of the smaller circle, right? So what is the area of the shaded region? So basically first we draw the big circle, and then this is a very special point A and B, such that when we take the center of AB and we draw a circle with the radius from O to A here, then S, the center of the big circle, it's going to lie exactly in that small circle. So A and B are not arbitrary points, they are at a very special locations, since that's I'd like to think about it, this is the center of the smaller circle here, and we are asked to calculate this area. So again, a very irregular shape, right, little progressions in this case, and it's hard to calculate directly, we don't know how to break it down into smaller areas, so in this case it looks like the only thing we could do is to use complementary area. And you see like if you add this one to another area and see if we can calculate the total area and then subtract away the complementary area. So if you have the handout, you can use the handout, if not, try to draw these pictures and really make it as accurately as possible. you may need to draw some actual lines because AB is the diameter of the big circle so try to make use of that information and R here is the radius of the big circle So naturally I would, I would draw AS, right? Because first of all, you know, we have to indicate in which we have to use R and R in this case equal to AS. And I will also draw, draw SB. And remember the theorem we were talking about just now? Because AB is the diameter of the small, of the small circle. So that means this angle ASB is 90 degrees. So that would be the first hint. And I think that most of you probably already recognized. And now we've got to choose what is the complementary area, right? So in this case, I mean, which one? Because we got to add to this. So that is actually, we're going to have to use, oh, this looks like the most natural choice. AB, I mean, not natural in sense that we still don't know how to calculate this one, but at least because we know R. Okay, so I think maybe we do a little bit more work here. We know this is R, right? This is R, this is R. So right away, it's not just right triangle, but it's right isosceles triangle. So we know that AB is equal to, AB is actually equal to root 2 of R, right? Remember the Pythagorean theorem. And that means we know the radius of the small circles, which is AB divided by 2. So it's going to be 1 over root 2 of R. And if we know the radius of the circle, that means we can calculate the area, half of this area. So that means if we know the area of this crescent, then we must also know the area of this, right? So it's kind of the other way around, how to calculate this complementary area. I will just label this as P, how to calculate this area here, the dotted one. Yeah, in a sense, it's equivalent to calculating the area of the crescent, but maybe there's another way to do it. While we cannot calculate the area of the crescent, maybe we can calculate the area of the dotted area. And I call this area P. So how to calculate P? So I will write a text here that we're going to choose this complementary area for the crescent to be P. But again, P is also irregular, right? So P is also an irregular shape. So maybe I've got also need to calculate a complementary area for P, right? And then with that, the next natural choice for the complementary area for P is this one. Yeah, let's call it Q, right? Because again, we know, why is that so? Because, you know, because this is 90 degrees. So that means we know this sector and we know the radius of the big circle. That means we can calculate the area here, SB, the whole sector here. And that will be exactly a quarter, right, of the area of the big circle. And this one is an isosceles triangle. We know R, so its area is known. And that means from that, we can calculate P. From that, we can calculate the crescent. So that would be like a lot of intermediate step. But I think that that's the way we should do it. Like we think about like the big picture of what step we need to do before we actually start doing the calculations. So maybe from that, we can get started together. You can calculate together. I already had some answers, but the rest of the class, we can just do it together. So let me call this area that we are interested in, right now, I think we call it T. It's not natural, but just call it T, right. So we have a T, we have a T plus P is a half of the area of the semicircle, of the small circle. So it's going to be a half, right, pi. And this one is going to be R over root 2 square, right. Right, so that T plus P is equal to this pi R square over 4, right. So it means if I can get P, then I can find T. So the next step is that how to calculate P plus Q. So P plus Q is a quarter, because it's 90 degrees, is a quarter of 60 degrees, 360. So it can be a quarter of the area of the big circle is R square, right. And then, so this is 1, this is 2. Did I make mistakes somewhere? Okay, good. Okay, so from here, I mean, in principle, we can calculate like Q, okay. So we also know Q, and then from there, we subtract Q to get P from here, because it just like happens that from 1 and 2, we know that T equal to Q right away. We don't have to go through the intermediate step, because the two right-hand sides are the same. So if T equal to Q, then I only need to calculate area of Q, right. But Q is right isosceles triangle, so that one's going to be R square divided by 2. Oh, is it? Okay, good. So it's just really half of R square. And you're fine. Okay, good. So I have a question. Great. I mean, I have a question, please feel free to get from the chat. So Q over here, so we got the theorem that if AB is a diameter of a circle, right, then the angle subtended by this AB is 90 degrees, because S, we have to look at S not as a center of the big circle, but S is a point on a small circle, where AB is diameter. So ASB is 90 degrees, right? Hope you agree with that. And then we have to look at S is the center of the big circle, so SA is a radius, so it's equal to R, and SB is also a radius, so it's also equal to R. So we have ASB as an isosceles, right, isosceles triangles, so that's why it's going to be R square divided by 2. Does it make sense? Okay, good. So from that, we know that the area here is actually C. Q over here. So actually, it turns out to be that the area of the crescent exactly equal to the area of this triangle. Of course, we can calculate P, but it's unnecessary in this case. P is going to be, this is pi R square divided by 4 minus half R square, but it's really unnecessary in this case. So I have to say that, okay, is it half times pi due to half the radius? Oh, the Q is the area of the, the Q is the area of this triangle ABS. So that is just R square divided by 2, there's no pi in the area of ABS. Does it answer the question? If you want to take the whole area T plus P over here, for T with P, it's a half of the small triangles, so we have a half here, and the radius of small triangles are divided by root 2, and that's what we have over here, pi R square divided by 4, but that, because this is half of the circle. But for Q, it is just the area of this right-angled sausage triangle, and there's no, there's no pi in it. Anyone still has a question about this? Oh, no, in this case, we're going to have a question with 360, 90 degrees rule, but in this question, there's really no 360, 90 degrees rule. I just want to go, come back to this, because we, you know, if you, this kind of seem too exotic, you can ignore this, but this is something that really need to know. If you have an arc here, which is a quad here, which diameter of the circles, and this angle is always 90 degrees, and that's what we use over here, AB is the diameter of the small triangle, so SASB is 90 degrees, and there's no 360, 90 degrees triangles here. It really doesn't matter where S is. I can have S over here, and this ASB is still 90 degrees. I can have S on this side, and, I mean, S can be here, and ASB is still 90 degrees. Okay, that, does it make sense to you? Square root of two, yeah, square, oh, square root of two again, square root of two, because ASB is an isosceles right triangle, okay, now I got it, right? The square root of two over here is, because this ASB satisfies isosceles triangle, so I put it here, okay? So I put it here, okay, good. I mean, this is a very beautiful question, and I'll explain to you why it's a difficult question in the next, in the next slide. So this one have ABS is right isosceles triangle, and that's why we have this relationship, okay, so I hope that's clear up this part. Okay, now why do you get P plus Q equal quarter? This one is because the sector ASB is a quarter of, we call it the disc or the circle, the big circle, right, because the full circle is 360 degrees, and because this one is 90 degrees, so if you can extend it over here, it's really exactly a quarter of the big circle, and the area of the big circle is, it's me again, it has to be pi, okay. Okay, thank you everyone for correcting me, we have a pi here, because that is area of the big circle, Andrew, when I make mistake, please feel free to correct me, yeah, it's a half pi square, okay, thank you for pointing it out, because the area of big circle is pi r square, and then that's why we have a T equal to Q, but for this, we do not have pi, okay, I hope I answer all the questions, but if not, then we can come back, okay. Did I answer all the questions? I mean, this question is so important, because it really sort of encapsulate and has all the elements that we've been talking, I mean, important for the resonance circle, so I want to get it through, but if not, we can, you know, stay after class as usual, and then I just quickly go, it takes one minute, but I just want to tell you why this question is difficult, so this is actually a question solved by Hippocrates, an ancient Greek mathematician, like, you know, in around like 280 BC, and the question that the high, the Greek mathematician had at this point, like, can you quadrature any arbitrary figure, and by quadrature, it means that we're going to create, given an arbitrary shape, we want to create a square that has the same area of that shape, and, but the restriction here that we can, we are only allowed to use compass and straight edge, that means no ruler or protractor to measure distance and angles, so this is a restriction, so they could do it, you can square a rectangle, you can square a triangle, and then you can square the quadrature, you can, you know, quadrature the loon, exactly the problem that we, we solve this now, because after we prove that the area of the loon is exactly equal to the area of this triangle, and they already know how to square the triangle, then that, this question is solved, it's a long-standing question, it's like a huge achievement in ancient, you know, Greek math, and they didn't know like three, two thousand years later, it appeared in math can grew, but after they did this, and they thought that they could actually quadrature the circles, but then in the end, I think at the end of 19th century, when they proved that, you know, pi is a transcendental numbers, and no, we cannot quadrature the circles, so just want to point out that we actually were solving a very famous question. Okay, so the next thing, the next topic we are doing going to be tension, so if we have a line tension to the circle, so always, you know, draw this radius, if the question say that a line is tension to circle, always draw this radius from the center to the tension point, and that is 90 degrees, if we have a two circles that are tension to each other, then by all means, you wanted to draw the lines that connects the two centers, and the passing through the point of tension point, and the distance between the two centers exactly equal to the sum of the two radii. So, this is a direct application of what we just talked just now, let a and b represent the length of a right triangle's legs, okay, so the left means this one and this one are a and b, so if d, small d is diameter of a circle inscribed in the triangle, that means this circle is touching each side at this, at exactly one point, that is small d, and big D is diameter of a circle circumscribed the triangle, like on the outside, and we are asked to calculate small d plus big D in terms of a and b, right, so a and b are given to us, so if I put this one is like a and b and c, a and b and c, then by convention, we call this side b, a, this side b, here from here to here b, and we're asked to find d. So, take a moment, draw this picture in your, either you can use a handout or, you know, quickly draw the picture, but again, please try to draw accurately. This 90 degrees, so you know what is special about a and d, right, take care of the fake, that fake enjoying the pictures. because we are given that it's a right triangle, right? So that means it's 90 degrees. So, I mean, always indicate all the information you have in the picture itself. And because of that, we know that, because of this information, we know that AD, again, we use that fact again and again, AB is a diameter of the big circle. So that is the first thing we know. That means this one is gonna be 2D, from here to here is diameter. So this is big D, but from here to here, this is B, this is A. And now we also have to indicate, right? So that takes care of the fact that D is a, the big circle circumscribes the triangle on the triangle. And now when you have a small circle inscribed the triangle, you also want it to translate all that information on the picture itself, right? And the first thing that you have a circle that we want to indicate the center, right? And also we want to draw some special radii. So if I have a circle, then I definitely think I need to indicate where its center is, call it O here. And to translate the fact that this angle is tangent to all the sides, what did we do in the last slide? We're gonna have to draw all the radii, right? All the radii. So I draw the first one here, the second one here, try to draw as accurately as you can. And the third one here. And all of them are 90 degrees. And that was D, this is D over two. Okay, so I think we are so close. I give you, I already have some answer. Just give it a minute to figure this one out. And the next step you try to, so draw, indicate all equal length in the pictures. So for example, over here, I would have, Yes, so I purposely draw, like, indicate, like, A opposite A, B opposite B, and C opposite C is sort of, like, the way to sort of preserve symmetry when you write equation, and a lot of you who have taken, you know, trigonometry, you probably realize it about sine, cosine, and tangent. But let me tell you, I've seen a lot of math questions for level 9 and 10. You can solve it with trigonometry, but you really can solve them with just classical Euclidean geometry. So when you see trigonometry, if you can solve it, it's fine, you can use trigonometry. But don't assume, don't assume things, like, don't skip steps and, like, things that is familiar and poor in the result, because it may be not correct. So try to solve it from, you know, from the beginning. So for example, over here, yeah, we don't have a question like that, but instead, we have a couple of questions when, you know, I was solving it in trigonometry, and some students solve it using classical Euclidean geometry in a very elegant way. Okay, so in this case, we're gonna just rely on our, the tools that we have in hands, which is no trigonometry. So this is M, I just want to indicate this is M, N, P, right? So for example, if you look at this, what, what segment are the same? If you, for example, we connect O, A and O, then we know that the two triangle AON and AOP, they are actually congruent, right? Because OP equal to ON, and then they share the same length here, and they try, these angles are the same, right? So actually, they are congruent, or you can say that they are congruent because AO is the same, ON and OPs are the same, so using, using Pythagorean theorem, AN and AP are also the same, right? And that's what applies to all these tangent lines. So if I indicate this one is X, and this one is X, it's also X here, but it uses different colors, it's gonna be X. Okay, I think finally we got some correct answer, this is X, right? And if this is Y, this is Y, this is also Y, they are the same, this is Z, remember this result we're gonna use a lot in today's class as well. And then if we add everything together, we actually have a D, okay, D equal to X plus Z, right? We already know one thing, but then what about Y, this CPOM is actually a square, so Y is actually equal to half the radius of the small triangle, of the small circle, okay, and that, so Y is actually equal to D over D, over 2. So if we add D plus D, so if we add D plus D, we're gonna have actually X plus Z, and plus Y over 2, right? But if we look at it in a different way, that's 2Y, I'm sorry, because Y is a half of D, D over 2, so D is gonna be 2Y, but you look at it a different way, you can regroup this, we're gonna have equal to X plus Y, and plus Z plus Y, and that turns out to be X plus Y equal to B, and Z plus Y equal to A, so actually we just have a B plus A, right? And you know, this, oh, how did we know Y's radius, that's a good question, because we look at this, this point here, so we have, okay, I can call the small radius is R, right, knowing that R equal to D over 2, so we have this P, O, M, C as a rectangle, right, because all angles are 90 degrees, but because, you know, we have two adjacent sides are the same, then it has to be a square, so that means Y equal to R. Okay, does it answer the question? Y equal to R, and that's why it's a radius, Y is a half of D over 2. So, again, I don't know how, if you do Euclidean geometry in real life, I always have to draw things with a compass and, you know, a ruler, because a lot of the time, your success really depends on how accurately you draw the pictures, and most of the time when you draw the picture accurately, you pretty much see the answer, you have to prove it, but you know what to look for. Another question with tangent lines, this one, I think I'm going to go through it quickly with you in the sense that it's kind of complicated, but it's very good, and it sort of illustrates the point that we're talking about now. So, we have the radius of a larger circle in the pictures here is three times the radius of the small circle, that's the only description we have. Oh, we also know that, you know, this length is 9, this little length is 1, so we're asked to calculate the length from here to here. So, if you don't have the pictures, then try to draw a circle, and this is definitely an isosceles triangle, right, draw a big one, and draw a tangent. So, a lot of things are unspoken in this picture, I mean, in the question itself, but you can see that you assume that this is actually a tangent line, this is another tangent line to both circles, this is another one, and this is another one. This is going to be ABC. I mean, we just do this together because it's more like a lesson than problem-solving sessions. So, as we said before, we've got to translate all the information, like, how do you indicate that it's a tangent line? I know it's tangent, but how do I actually show it in the picture, right? And I show it by, you know, drawing the center of the circle here, I have O, I have I, right, and then over here, remember, we make use of symmetry whenever we have symmetry, so that's definitely going to go straight like this. And then to indicate that this is a tangent to this line, I'm going to draw this one. So, all of this, I just want to convince you that these are the standard steps that we're going to have to follow. So, I'm going to draw this one, and then I'm going to draw this one, we're going to have to start with, like, we draw all of these lines just to indicate these are tangential. We can indicate, like, with this line and with this line. So, do as much as we can, and also, I think it should go all the way straight here because of, you have to be tangent to this line as well, right, don't forget that. And after you do all of that, you can just indicate, and you can do P, Q, I mean, due to symmetry, you do not have to do everything, you can just, you know, do on one side, that should be enough. I probably do it K, not a very good color, then we can do H, we can do E, usually you want to, you know, label it in a systematic way, not like arbitrary use of letters, but, yeah, okay. So, now, after we do all of that, right, so now I'm sort of all set by, you know, representing that this circle is tangential, all of these are tangential lines, so the next step we do going to be length tracing because that's where we're asking for. So, this is 9, this is symmetry, so this part here, AE here is going to be 4.5, right, so I just indicate right in the picture, this is 4.5, and just like the question that we said just now, if this is a tangent source, then AE is going to be equal to AH, so AH is also 4.5, 4.5 here, right, and then I go from this line, this is 1, so this tiny little thing here is 0.5. If this is 0.5, then MM is also 0.5, like what we did just now, 0.5, and so the only thing we are missing here is DM, but DM and HD, but notice that because all of these are tangent lines, so DK equal to DM equal to DH, so we also have a, I could call it A, and this is A, and this is also A, right, so in, finally, we're going to have X equal to 4.5 plus 0.5 root of 5, and plus 2A. I sort of boil down to the only one length that I need to calculate is A, and remember, we sort of like avoid calculation as much as we can, and what we've been doing here really is just do length tracing in the figure itself, right. So now, what is the easiest way to calculate? You can draw a ton of pictures and, but I think when I look at it, the easiest way to calculate D is using congruence, so we have this AEOH, AEOH and DKIM, these are congruent. Of course, you can sort of split them into two triangles, right, two triangles and use all kind of argument to make them, but right away, you can see these are, I'm sorry, not congruent, are similar, they are different by a scaling factors, right, and the scaling factors is 3, because the area of the, I'm sorry, the radius of the big circle is 3 times the radius of the small circles, so from that, I mean, you don't have to go through the triangle, I think we can just get away with this, you know, quadrilaterals, and that means AE equal to 3 DK, and that means DK is equal to 4.5 divided by 3, which is 1.5, so you add together x equal to 5 plus 2 times 1.5 is equal to 8. Okay, good. So this is one question that, you know, you could make it really complicated. I've seen other solutions that are really complicated by adding actual lines, but pretty much you can get away after you draw all this, you know, in a sense kind of standard lines if you have a tension circles and tension lines. So I think this question is clear. The next question is actually a very beautiful one, because it also, it doesn't really require any special knowledge, but you will see. So A and B are on the circle with center M here. PB is tangent to the circle at B. The distance PA and MB are integers, and PB equal to PA plus 6. How many possible values are there for MB? So let me, Pb is tangent. So when you draw this, you draw a circle, right? And then you draw M as a center. You draw a line here and take a P as of some point, sort of, you know, a little bit further away from the circle and then draw Pb as a tangent. And how do you indicate that P is a tangent to the circle? Right, at this point, it should be clear how to translate that piece of information into the picture itself. And then this is very important, Pa and Mb are integers. And Pb equal to six plus Pa. So from here I could, you know, after you finish drawing the picture, remember what we've been talking about in the last two, three questions, always connect the circle, center of circle to the junction mark. So I would put 90 degrees here, and I indicate this is R, AM is also R. Okay, R is an integer, so that's a good thing. So we say R is an integer. Now as happens a lot with geometry is that after we set up the question, we have to use algebra, a lot of algebra, we use a lot of fraction proportional and, you know, the formulas that, the Pythagorean theorem that, you know, we use a lot of length, and we have an unknown, it makes sense to indicate that this PA equal to a variable A, and PB is going to be A plus 6. Because there's nothing else we can do at this point, we can't stop drawing, and if we want to move forward, we somehow have to, you know, find some relationship between R and this length, so I think it's just natural that we're going to assign an unknown to this length PA, and that also allows it to find PB, so all the lengths that we have in the picture can be expressed in terms of R and A, right? So radius R is an integer, PA equal to A is also an integer, so now we have, we have two variables, right, and we don't know how to solve for that, so no other information except that they are integer, so the next step, and that's naturally to ask, like, how are R, A, and R related to each other, right? How to establish a relationship between A and R, right, that would be, you know, just trying to convince you that the steps that we are taking are kind of standard and natural. Maybe some suggestions, Andrew, would you like to suggest to the students what they should use to find the relationship between A and R? Because this is a 90 degrees angle, right? So this is a Pythagorean theorem, right? Exactly, because, I mean, like, any time we have angles, like we talked about last time, we really don't have much tools, many tools, so we have a 90 degree angle, so we got from the fact that this tangent to the circle, so make use of that fact, and we have a length here, so the next natural step by Pythagorean theorem, right, just go through our, like, toolbox and see what we can use. So if I use a 90 Pythagorean theorem, I would have A plus 6 squared plus R squared, right, and that would be equal to A plus R squared. So that would be, if I expand it, I have A squared plus 12A, right, plus 36 plus R squared, that would be equal to A squared plus 2AR plus R squared, right? Sometimes we don't know where it leads to, but really that's the only thing I can do at this point, and there's nothing else I can do, so I just delete this one and this one, oh, okay, good. So actually we turn out to have, okay, we still have a one quadratic term is here, so we have, we come back to number theory, we have an equation with integers only, plus 36, right, equal to 2AR, and at this point we want to divide by 2 to make it simple, equal to 18 plus AR, and as we said before, when you have an equation like this with integer, it's much easier to work with products than working with sum, right, because product allows it to use, like, you know, divisibilities and things like that. So in this case, I mean, naturally I would take 18 equal to AR minus 6A, because I have a common factor of A here, so it's going to be 6 minus R. And really there's no restriction, I mean, A can be as small as possible, it can approach zero here, and A can be as large as it can go to infinity, so really there's no restriction on A and R here, except that R has to be smaller than 6. Is that correct? Yes, because it has to be positive numbers, so we have a 6 minus R, that number, we want it to be, what other? It should be the other way, R minus 6 is greater than zero. Oh, thank you, thank you. So this should be, like, 18, no, I think 18 is 6, okay, I thought it's the R minus 6, okay, thank you. Thank you, Andrew. So R minus 6 really can be anything, right, there's really no restriction that R has to be bigger than 6, okay, thank you. So R minus 6 is going to have to be, so both A and R minus 6 are positive, right, positive divisors of 18, and that means we have the choice of, once we have R minus 6, I mean, we can find A, uniquely determine A, so let's start with this because we have a restriction on this, so it's going to be 1, and it's going to be 2, 3, 6, 9, 18, right, and these are the possible divisors of 18, so we, and from there you can find R equal to 7, all the way to 24, and I mean, for each R you have a corresponding value of A, right, so the answer is 6. So I think this is a good question, it doesn't, I mean, some of the questions we had today, I have to admit that's pretty difficult in the exam, this is a good question, right, Andrew, like it didn't really require anything except for Pythagorean theorem, and then it has a combination of like a little bit of number theory, question number 29, so of course we have to expect that it's also not straightforward. Okay, so really to summarize what we did last week and this week, we have a pretty much I think last week we have a slide with step-by-step doing geometry questions. The first thing, draw the picture accurately and then indicate all the information given in words and into the pictures and do, find out special triangles, do angle tracing, do length tracing, use a couple of these that we have at our hands and, you know, try to manipulate them. So I think we pretty much cover what we need for the circles, right, Andrew? Yeah, a couple of techniques we have here using symmetry, using complementary, doing, using the area is getting a little bit more complicated when you, you know, involve circles. But apart from that, I think the technique and pretty much similar to what we did last time with triangles, the picture itself gets a little bit more complicated. And then a big part of that, for today's class that we're doing with tangent line, it may be new to you, but really if you draw all the tangent lines and all the radius, the relevant radii, most of the time we can get the answer, right? Okay, any other questions for today's class?

geometry

circles

angles

isosceles triangles

exterior angle theorem

arc lengths

circumference

Pythagorean theorem

tangents

congruent shapes