Hi everyone, so let's get started. Today's topic is 2D geometry. We mainly deal with length and area, and in the next lesson we will study angles. Today, Lief is busy with his concert, so there will only be me. Let's start with the warm-up question. A regular, the regular hexagon ABCDEF, shown in the picture, has a perimeter of 36 each vertex is the center of a circle. The radius of each circle is equal to half the length of one of the sides of the hexagon. So the question is, what is the perimeter of the shaded figure here? If you need a little bit of reminders, then the perimeter of a circle is 2 pi r, where r is the radius, the circle, so they say the perimeter is 36 right so that means um we take 36 divided by 6 that's the length of each side of the hexagon and then we need to divide by half and that we get r right so r is equal to 3 but usually I would it's good to have everything consented to the chat yes but usually I would not worry about the numerical values usually when you do geometry you just you know we sort of most of the time we use ratio proportion and then it's good just to keep the very variables to the end to avoid numerical error so each arc here because we know that it's 120 degrees so it's going to be a third of a circle right so each arc going to be um 2 pi r and then we need to multiply this with a third right and then um and then 2 pi r yeah and then we need to multiply by 6 because they're sick of the new symmetry and r is equal to 3 so actually this r and the 3 is going to cancel out and then we should have a 2 pi 6 which is 12 pi so the answer is b It's, as I said, we deal mainly with length and area in this class and then next lesson we're going to study angles. So what are the topics in 2D geometries? I know that at this level some of you have finished geometry and some of you have just started. So these are some of the topics that you may want to review. Parallel lines, angles, congruent triangles, similar triangles, and the special triangles, the right triangles, isosceles triangles, and the 30, 60, 90 degree triangles. And then the special lines in a triangle, the bisectors, median, the altitudes. We also work with circles. So I will come to the theory when it comes to the specific problems. And these are the formula for the areas of familiar shapes. I know you, I'm sure you know it by heart by now, but then we just put it here for you to review if you need to. In terms of problem solving strategies, it's a very good skill to be able to draw the picture as accurately as possible. As you know, the ancient Greeks, they don't use measure. You only draw geometry using a compass and a straight edge. And then actually, when you do the Euclidean geometry properly, we only need those two tools. And then lots of times, you need to draw additional points, additional line segments. And it's very often that we actually need to use algebra to describe the problem and actually solve for the variables. So we come to that when we come to problem solving. Let's start with the first one. P is the area shaded with vertical lines here, and S is the area shaded with horizontal lines. The diameters of the circles are 6, the biggest one, 4, 4, this 2, and 2, the smaller one. And then they ask us to compare P and S. So let me mark P, these vertical lines. So here, here, these three shaded parts are P, and S is this one, horizontal, yes. So what is the relationship between P and S? If you have an answer or a question, please send it through the chat. Oh, I just need, in case, to remind you the area of a triangle of a circle is pi r squared or pi d over 2 squared. So remember, there are two strategies to calculate area. The first one is that you break down into smaller, break down into manageable pieces, like the familiar one, not the funky shape, but regular shapes so that you can calculate the areas of each shape. And the second one is I use complementary areas. If the shape is funky, maybe we add in something to make it more regular. And that way, I would allow us to calculate the total area as we subtract away the parts that we add in. So in this case, it is kind of hard because you don't know where these intersect. And then the shaded areas, they all have very irregular shape. So one way to do that is that we can observe that if we add the vertical shade here with the y part, so I will just name this as 1 and 2 and 3. So if we add p and then we added the y part, 1 plus 2 plus 3, and what do we get? We get the area of the three smaller triangles. That would be pi and then 2 squared, first one, plus 2 squared, the second one, and 1 squared, the third one. So that would be 9p. And if we add s with the y parts, then we get the area of the biggest triangle, which is pi. And it's radiate is 3, so it's going to be 3 squared. Then that's 9pi. So actually, these are the same because when added with the same area, we get the same answer, right? So the answer is Cp equal to s. But another strategy is kind of not a very rigorous proof, but it's commonly used that when you see that the result is actually independent of the actual intersection of the shapes over here, what we can do is that you can consider a very special case. So in this case, you actually consider the case when all the circles are disjoint, right? So you have s in the middle, the shaded, and then p are here. You can actually consider this case, right? And p and s. So in this case, p is just the total area of the three smallest circles, and s is the area of the big circle. And the intersection here is absolutely 0. And then you can sort of in the exam, you know that the answer is C. But of course, for a rigorous proof, then you have to go with this way. But this is kind of a very useful strategy to get started in many situations. OK, let's go to the next one. So it's something that's very familiar, right? The Pythagorean theorem, when the square of the hypotenuse is equal to the sum of the square of the two sides of a triangle. One thing I just want to remind you is that remember a few Pythagorean tuples, like 3, 4, 5, 6, 8, 10, a couple more of them. So before the exam, you just remember all of them, a couple of them, when the sides are integers. So let's do one question with Pythagorean theorem. OK, this one. In a convex quadrilateral ABCD, the length of the sides are expressed with natural numbers. And the perimeter is equal to 16. The interior angles C and D are right angles. And angle B is obtuse in the picture here. What is the length of BC? So remember, what is it? Perimeter is 16. And a very strict constraint here is that all the side lengths are natural numbers. They are all integers. So take a moment to see what we can do with this piece of information. We are asked to calculate this one. Actually, you know what? We can calculate the side lengths of all of them. So let's see if we can do that. Yeah, I already got a correct answer, so I'll just give you a little bit of hint. The perimeter is 16, right, so we can have many different ways to have four numbers sum up to 16. But the constraint here is that the angle here has to be 90. So, you know, we know that the Pythagorean theorem would be useful in this case, right, and the way to use the theorem is that you will have to create a right triangle, so this is one way to do that. You can draw a line here, bh, bh perpendicular to dc, so that means hb is parallel to dc, right. And with that, you can look at the triangle ahb, that is a right triangle, and that means we we can use the Pythagorean triplets that we talked about before, it could be 3, 4, 5, or in this case could be 6, 8, 10, right. But if you use 6, 8, 10, then right away we see that they are too big, and then that would exist 16, so the only ways for the perimeter to be 16 is 3, 4, 5, and then you took, you could use 3 here, 4 here, 5 here, but that means this one's going to be 4, right, the sum is going to be 3, 5, 4, that's 12, and we have a 16, so we have 4 left divided equally between these two, so we have a 2 and 2 here. But it actually, it really doesn't matter, you could have 3 here, and 4 here, that one's d5, and the sum of these are d, these three numbers d12, and then these two is d2, so that's why the question only asks you for BC, because it has a unique answer of 2. Okay, so let's go to the next one, the congruent triangles. We have three tests for congruency, right, psi, psi, psi, if the three pairs of psi are equal, then the two triangles are congruent, or psi, angle, psi, we only need two psi that are equal like this, and the angle between them are equal, then the two triangles are congruent, or angle, side, angle, when the side is sandwiched between the two angles, then we also can conclude they are congruent, right, so these three tests I mean we can prove, but let's remember it by heart. So let's do this one. The center of a square with dimensions 2 times 2 overlaps with the center of another square which also has dimensions of 2 by 2 here. What is the area of the common part of these two squares? This one. You don't have to print out, you can just quickly draw, again it's a very important skill that we can quickly draw the problem as accurately as possible. Supposing that E is the center of another square E here. Oh, I'm sorry. I think there's a typo in the question. Thank you. What a student pointed out, it's not the center of square. It is a vertex of a square. So this is a lot of the time for geometry questions, the picture is kind of self-explanatory. So I'm sorry, this is not a center. This is a vertex. Oh, that is a big, big typo. Yeah, it's an error. So we know that E is the center of ABCD. How to calculate the overlap. So I can give a small hint. Remember what we did before for the problem in which the three circles are intersecting, the four circles intersecting each other? When the shape is weird, kind of irregular, you could use complementary shapes. You could consider special cases. Because sometimes when you consider the special cases, that gives you an idea of the answer. And from there, you can go back and prove the result. But at least it gives you a rough idea. In this case, the question doesn't tell us how they intersect. They do not tell us exactly where this intersection point, where is it on the side BC. So great, I have an answer coming in. Yes. So if I were in a hurry, I would say, oh, I don't know where this point is. That means the answer should not depend on the exact positions of EFBC. So imagine you stay here. And you can just kind of rotate this center of rotation. You can just rotate this EFGH around ABC. So you can put it in a very special position. If I rotate it such that EF is actually coincide with EB. And in that case, EH going to coincide with EC. And the shaded area is going to be BEC. Does it make sense? And EBC, you know right away, is 1 because the total area of ABCD is 2 squared, which is 4. So it's a quarter of that exactly 1. But that gives an idea because after I consider this special case, it actually gives me an idea of how to prove the result. Because then it's obvious, right? A lot of time, we actually work backward. Like, oh, I think that this is correct. And I work backward to prove that. So in this case, it actually gives me an idea that the area, this little area, EBM must be equal to ENC, right? They have to be equal, no other way around. And how to prove that EBM. And we look at that. It's not just the areas are the same. But these two triangles, they actually are congruent, right? And how do we prove that they are congruent? We have three different tests, right? So which one can we use in this case? Because these are actually congruent. So we know that EB is equal to EC, right? Actually, you don't need to write here. So a lot of time, it's just mark whatever length and angles that are equal around the picture. So these two are the same because the EB and EC is the center. So one side is the same. That is great because that's the biggest problem. The side have one side the same. This angle is 45 degrees, right? And this angle is also 45 degrees. So it's good. So I'm very close. I have one angle here. I have one side here. So the only thing missing is this angle. This is alpha. I say this alpha. And alpha, we notice alpha plus beta. This angle is beta, right? So we know that alpha plus beta is 90 degrees. But beta plus this little angle is also 90 degrees because it's the center of the circle, of the triangle. So that means this angle must also be alpha, right? And these two triangles, they are congruent according to the test angle, side angle, right? And then we go backward. And we notice that the area here has to be equal to the area of EBC, which is 1. And a lot of time, you're going to get the answer if you draw the picture accurately. And you can get the correct answer and go back and prove it. OK, so that is congruent triangles. And the next topic is similar triangles when the angles have the same shape, but score dilation. The sides are not the same. The ratio is not 1. So we have a side-side-side if the ratio of three pairs of corresponding lengths are the same and the two triangles are similar. And that means all these angles are the same. This one is going to be equal to this. This equal to this. And this equal to this. Or we have a single side angle, side just like the other case, the ratio of BA to B prime, A prime is equal to the ratio of BC over B prime, C prime. And this angle is the same. Then the two are similar. We also have an angle-angle. We do not say side because if this angle is equal to this angle and this equal to this, then automatically this angle is going to be equal to this angle. And right away, we have the ratio of length is the same. And one result that we use very common, we use a lot in similar triangles that if the two triangles are similar, then the ratios of area is going to be equal to the square of the ratio of length. And that's obvious, right, because area is length times the base times the height, half of it. And if the two are similar, then the ratio of the length here is going to also be equal to the ratio of the size. And that's why we need to square them. And one way to get similar triangles that we have a triangle and we draw lines that are perpendicular to the base here, then we have this angle is equal to this angle, this equal to this, and then right away, we have a pair of similar triangles. OK, so let's go to this one. What is the ratio of the gray area to the area of the whole rectangle? So this big thing is a rectangle. And yeah, everything is clear from the picture here. M and N are the midpoint. Yeah, so if you don't have the handout, you can quickly draw the picture. M and N are the midpoint of AB and CD. This side is 1, this side is 2. And we need to find the ratio of the gray area to the area of the whole rectangle. If you want, you can actually calculate the gray area. Of course, we should be able to, right? Because they asked about the ratio and we know the area of the whole rectangle. But you don't have to. My point is that you don't have to if you use ratio. But if you want, you can calculate. So that's the first thing I would give you a little bit more time for this. Oh, that's great. Okay, so you got the correct answer. So often you're going to see that you have more than one way to solve a geometry question. So, for example, I can do it in two different ways. So, for example, if you want to calculate that and we also use, you know, symmetry and just moving the pieces around. So this one, this is like, I can name it like 1 and 2 and 3, right? So, you know, MB is parallel to DN and we see we have a parallelogram here. Due to symmetry, you can easily see that 2 is actually equal to 3. The two areas are the same. So the total area of the gray area is going to be actually equal to A, Q, and D, right? So if I can calculate A, Q, and D, I'm all set. But A, Q, and D, this area is harder to calculate than this complementary part C and Q. So ABCD is 2, so that means ACD is 1. So I can take 1 and subtract away NQC. That's one way to do that, right? If you want to calculate that, which I think is a good exercise and we do that, we can calculate, subtract away the area of NQC. And NQC, we already know that the base is 1, so I just need the height, right? So I put a height here, QH here. And this one, I can't quite calculate. Actually, I can because another way to look at that, that you have MP parallel to BQ, right? You see that, right? Because these are parallelogram. So that means P is going to be the midpoint of AP and PQ, midpoint of AQ. So you have AP is going to be equal to PQ. Similarly, because NB is perpendicular, if we look at this way, NQ is parallel to DP. Then, and N is midpoint of DC, that means Q is midpoint of PC. So right away, we have PQ also equal to QC, right? So if you look at this way, if you look at the triangle ADC, now I have to look at triangle ADC, you have a QH, QH is parallel to AD. And because the ratio of CQ over CA is a third, right? So that means QH going to be a third of AQ, AD, which is AD equal to 3, which is a third, right? So you know the height of NQC, you know the base 1. So the area of NQC is going to be half of that, which is 1 sixth. So we have a 5 sixth, 5 sixth at the area of the gray area. But then we need to take the ratio, so we take 5 sixth, and then we need to divide it by 2, the area of ABCD, and the ratio is 5 over 12, right? So that's one way to do that, and I think it's a very good exercise to practice with parallel lines and congruent triangles. But there's a faster way to do that, and I will show you here. Another way to look at that, we also start by observing that these three segments are the same, right? So that's very helpful. And then if you look at the triangle as we said earlier, if the ratio, so we have ABC and, for example, AEF here, these two triangles, they are congruent, and the ratio of AB over AE is a half, right? So if I want, I want to calculate the ratio of areas, I would have to square that. That means ABC is going to be a quarter of AEF, right? So usually in that case, I just name the smallest area is 1, right? I call it 1 unit, and that means the area of CBEF is going to be 4 minus 1, which is 3, right? So it's going to be 1 and 3, and over here, I asked about 1 over here, and now I need 3 over here, because these are, this is symmetry, they are the same units, right? And the only left over are this piece and this piece, which are the same. So how to do that? Now we look at this way, because FH is parallel to CM, right? And the ratio of, okay, sorry about the notation, OF over OC is a half. That means FH is going to be half of MC. Does it make sense? So we have MC, it's going to be equal to 2FH. And HF is actually equal to CB, also due to symmetry, so that means MC is going to be equal to 2CB. So if you look at the triangle ACB and AMC, they have the same height coming from here, right? I'm going to delete it, but they have the same altitude, and the base here is twice that base, that means the area of AMC is going to be twice the area of ACB. So I'm very close to the answer, because then I can indicate this area is 2, this area is 2 as well. And if I sum up everything, the gray area is going to be 5 and divided by the total, which is 1, 3, 2, so 6 multiplied by 2 is 12, and I get the answer right away. And I do that without having to do, I mean, doing a lot less computation. And the nice thing is that if you use this, so it's kind of very restrictive condition here in which this side is 1, this side is 2. But if you kind of look at this solution, a lot of time you can actually generalize it, and you can actually relax this condition, because the only thing I've been using as a ratio, so if I have a rectangle in which this side doesn't have to be 2, this side doesn't have to be 1. It can be anything, but as long as these 2s are midpoint, and I draw it this way, I draw it this way, then, okay, I'm sorry, since this one here to the midpoint, as long as these are the 2 midpoints, then you can see that everything that we did earlier, all the ratio remains the same, and that means the ratio of the shaded over the entire rectangle is defined, regardless of the relative ratio of the two sides of the rectangle. Okay, if any comment or questions, send to the chat, and I will try to answer. If not, then we're going to go to the next one. Okay, ABCD is a square. What is the length segments? EC, if AF is 4 and FB is 3. So very, very little information here. Very, very little information here. We don't even know the side length of the square, but we know that this is a triangle, this is a right angle, and we only know that AF is 4 and FB is 3. So you don't have to drag, the handout we can quickly draw, first you started with A, F, B, 3 and 4, and that means A, B is uniquely determined, and then from there you draw B, C straight up, D, A straight up until, you know, to the point A, D equal to A, B. In a sense it's sort of like construction problem, you can construct everything uniquely knowing this 3 and 4, and you get C, here you get D, you connect them, and that's where the point E is. So actually the question has all the necessary condition information. Yes, so see what we have in our toolbox, right? If you are asked to calculate the length, right? So what should we do? What are the toolbox we have? We have a couple of like, you can look at special triangles, right? Because those tell us like what lengths are the same, the equilateral, isosceles, right triangle, yeah. So we look at special triangles. You can use a Pantheon theorem. And then we look at congruence triangles. And if we don't have a congruent triangles, then we have to look at similar triangles, right? Because where the ratio comes into play. So pretty much that's the kind of toolbox you have to in this case, to calculate length. So over here, three and four, we just started by indicating all the lengths that we know, right? So three and four, and that means it's a right triangle. That means this one has to be five, and I just indicate five right here in the figure. So I put five here. That's the first step. And then the next step is that we are sort of done with length. Oh, that's good. That means the triangle, I'm sorry, the square has a side length five. So that's a big step, right? And the next step we do is geometry. After you do everything with length, you see what angles are the same, and we can mark it on the pictures. So we look at here, what angles are the same? If I name this angle is alpha, right? And this is beta. Then we know that alpha plus beta is equal to 90 degrees, right? They are complementary. And that means this angle over here, because it also complementary to beta, so it has to be alpha, right? So it's more like do angle chasing. And then I will look at the ECB, that triangle. Alpha, this is 90 degrees. So alpha is also complementary to E. So that means E has to be beta, right? So really just mark as much as you can find out right in the pictures, and that really usually is gonna lead you to the solutions. So now I don't have anything special except for the right triangle. I don't have any congruent triangles, but right away, you can see that we have a similar triangles, right? We have AFB, AFB. That triangle is similar to which one? It's similar to, okay. So usually when you write, you want to write the corresponding vertices so that you don't make mistakes. So we notice that AF is the longer side. I know C in the middle because C is right angles. And AF is the longer side next to alpha. So I should start with BCE in that order, right? And now I'm sort of very close to the answer because I can use a ratio of length. I would have AF, which I know, over CB, right? That would be FB over EC, right? And AF is gonna be four over five is equal to FB, which is three over EC, right? So I know that EC is gonna be, what is it? 15, 15 divided by four. Yeah, I think that's correct. 15 divided by four, that's 3.75, right? So usually that's the way to go. I didn't do anything special. I just start doing all the lengths, all the angles that I know. And from there, I can get the solution. Okay, so. That's the next one. Okay, so next we do with circles. So these are, I mean, we're gonna work more next time. I don't think we use a lot of these today, but yes, a little bit when a tangent line to a circle, when it intersect at one point, then right away we know that this angle has to be 90 degrees and this is a radius. And if you have two circles that are tangent to each other, then, I mean, almost all the time, you want to draw a line that connects the two centers and that goes to the tangent point. And we know that the length, the distance between the two centers has to be the sum of the two radii, right? So that is something. And when you have a, we're gonna do a lot more of this next class, right? But so for this, for now we need to know this. And also when you have one figure tangent to the other, you just draw as many lines as possible to kind of verify that relationship right in the pictures. Okay, so let's see if we can. So the next one. Two circles are located inside a square with size of length one, as shown in the diagram, the side length of the square. The circles are tangent to the side of the square and to each other, right? What is the sum of the length of the radii of the circles? Right, so for these pictures, if you don't have the handout, we just draw a big square, draw two circles that are tangential to the square and tangential to each other. And then you want to, you know, put, you want to put label the special points in the pictures and draw as many lines, many line segments as you can that shows a relationship, the tangency. Okay, so that will be the first step in solving the geometry problems. So I can give you a few minutes to, like a minute to draw the pictures and I will start drawing. Okay so hopefully you've finished drawing by now so I can start drawing. So usually we name it O1. You can name it different ways but I can use O1 and O2. If we name it this way then they say that the circle is tangent to the square so that means if I put the line here I can call it maybe M and N. These are perpendicular to the side length of the square and the radius is going to be R and R. So you want to translate a word problem everything into the picture and similarly this is I can call P and Q here. P and Q perpendicular and that is R Okay, so that, and the question asked to calculate which length, A, B. You have an answer question that sent to the chat and I reply you. The question asked to calculate the distance between the two, between the centers, right, the sum of the length. Okay, so actually it's kind of tricky, it asked to sum of the length, it didn't tell us right away, but we know that because they are tangent to each other, so it is actually the distance between O1 and O2. So this one from here, this point, let's call it t, then O1t is small r and O2t is big R. So this question, as I suspected, you got some, some you got very close answer, but it involves a little bit of algebra, which is very common in geometry. You would see like a lot of geometry problem, you study geometry and have a pages and pages of algebra. So let's see, RR here, right? We have to calculate this distance. So we draw as many lines as we can to kind of, you know, make use of all this, you know, special points, special lines. So for example, if I, if I extend this all the way to here, and I extend it all the way to here, right? Cause I, sometimes the, I have to bring the two segments far away together so that I can write out, you know, some relationship between them. So this one going to be R, the smallest one, and this one, because they're parallel, so it's going to be small R. So that means PQ, maybe I can name this S, that means NS, right? This one's going to be one minus small R minus big R, right? Because this one is one. And that also means that this one is RR, this is a square, so this R, that means this QV, for example, this one's going to be also one minus R minus R, right? So just write as many things you can, write on the pictures. And that means this triangle here, O1, O2, and, okay, so I ran out of space, out of letters. So O1, O2, and W, this is a right isosceles triangle. Right isosceles triangle. And then you see, I actually haven't done anything. I've just do angle chasing, length chasing, and put everything on the picture, so I conclude that this O1, O2, W is a right isosceles triangle, right? And what does it mean? That means O1, O2, O1, O2, that would be square root, right? So hypotenuse is going to be square root of the side, which is one minus R minus R, right? And O1, O2 is R plus R. So if we, this is a variable we are looking for, so I just do it as A, so A is equal to square root of two of one minus A. Is that correct? Yes, yeah. So we just bring A on this side. We have a one. If I make a mistake, let me know. One plus two, because today Liev's not here to correct me. A is equal to square root of two. Is that correct? So I think a lot of you got to this point, and then now we have A is gonna be square root of two divided by square root of two plus one. And you don't have anything like this in the choices, right? You don't have anything like this, so, and then, so usually when you write fraction, you don't want the denominator to be irrational. You want it to be like an integer. So one way to do that, you want to get rid of this. Then we multiply everything with square root of two minus one, right? And this one, remember the identity we talked about earlier? That's why it's good. I mean, we need algebra for everything. And this one, you have A square minus B square. It's gonna be equal to A minus B times A plus B, right? So in this case, you have A plus B. This is a trick to get rid of the square root over here. And that in denominator is become two minus one, which is one. And in the top here, we multiply it out. We have a two minus square root of two. Okay, and the answer is D. So I don't know, we may do a different way, but actually it's a crucial step. You can also like test it, but yeah, really this step is important to be able to circle the correct answer. Okay, so I think that's good. We have one more question is kind of also good. So let me do that. But I think this one is more important. So trigonometry, so I know some of you probably have studied it, studied in high school. Some of you haven't. So let me just quickly do like the sort of the simple theorem in trigonometry is a law of cosines. So if you have A square, you have a triangles, and then that is a relationship that this is a law that relates the length, the three lengths and one angle. You have A square going to be two B square plus C square minus two BC cosine alpha, right? And similarly for beta and gamma, you can write the same law. So why would we study trigonometry? I mean, in the exam, of course, you don't expect to, they test you complicated formula for trigonometry, but it's a very good thing to do because it's so powerful. If we don't use trigonometry, we can just, you know, basically start with very simple triangles, but you see how powerful this formula is. It allows us to, if you know two lengths, C and B and the triangle, and the angle alpha, then we can calculate length A, doesn't matter how weird the angles and the lengths are. But more than that, right? If we know, if we know alpha, for example, if we know alpha, if we know B and C and alpha is straightforward to calculate A, but if you know alpha and beta, I'm sorry, alpha and B and A, for example, you actually can calculate C. Do you see that? Because it's just another variable here. The only thing is that now you have to solve a quadratic equations, but we know how to do that, right? So, so that's the thing. We're going to do a problem with that. And in terms of, you know, calculation, you just only need to remember a few familiar angles. Psi of 30 degrees is a half. I actually only remember this. And then this 60 degree is complimentary. So that's going to be the cosi. And then cosi 30 is going to be square root three plus two, because if you square, psi square plus cosi square has to be one. So you only need to remember this and the rest is going to just follow, right? And then this one is going to be two, one over square root of two for 45 and 45. Okay. So let's do one question. It's more like to illustrate, you know, the power of trigonometry, but a lot of the time, as you can see in next class, you can solve it using classical Euclidean geometry without using trigonometry, but I think it's good to know. For level nine and 10. So let's do one question here. ABCD looks kind of complicated, right? It's a square with psi length one and arc of circle centered ABCD. So we can see that from C, the center, we draw this circle. We have this arc here. From A, you draw a circle and get an arc here. You know, if you don't have the handout, you don't have time to print it, you can draw it roughly like here and here. And then you start from B, you draw an arc from C, A. This arc from D, you draw another arc, C, A. And the intersection point is P and Q. P and Q. And the question asks, what is the length of PQ? you know that it's 2008, it's kind of older. I don't know what's going to come up because you know that the exam questions are selected, you know, in a math kangaroo conference or when 80, when representatives from 80 countries, 100 countries come and select the questions, so we don't we don't know what's coming up. But I think it's good to know geogrammetry in general, and it's just like the concept without even if you don't know the calculations and how to carry out the calculation yet. So for this question, what I suggest is that you should identify an angle, a triangle, in which you already know two sides and one angle, and PQ is going to be the third side. Right, and it doesn't matter where that angle is, it doesn't have to be opposite PQ. It can be anywhere as we discussed earlier, but we need to find one triangle. That has two known sides, we need two known sides, one known angle, and that includes PQ. And wow, I got an answer that's very impressive, that's correct answer. And the third side, that's very impressive, and the third side is PQ. Okay, so let me, let me quickly go to, it's a long computation, so I don't, I already tied up the solution, but let's see, it's a lot of important concepts, so I want to go through this question. You can use symmetry, right, so sort of connect everything we can, we know that CB, Z04, you know, arcs, they are the same. So if you connect PQ, for example, BD and CA, right, and that's going to exactly intersect PQ at its midpoint, right, so that you sort of can take for granted due to symmetry, so that really is going to be midpoint of PQ. And we look at over here, PQ, you can't quite find any triangles that has PQ, but because QO equal to OP, so I can only have to look at half of it, OP, for example. And then again, I connect as many lines as I know, so I will do CP over here, right, and PD over here. So remember that CA is an arc of the triangle, of the circle that center at D, so that means DP, right, is equal to DC, so sine length is also equal to 1. And look at C, BD is an arc of the circle centered at C, so that means CP is also equal to 1. And then we know that CPD is an isosceles triangle, right, this one is a, I'm sorry, equilateral triangle. And this one, also due to symmetry, is right in the middle, so we know that this angle here is actually, this angle here is actually 30 degrees, 30 degrees. And we know that CP is equal to 1, we know that length CO, because this is an isosceles, no, it's a isosceles, right triangles is 1, so CO is 1 over square root of 2, right. So that would be the first step, we know that we're going to look at the triangle CPO. Okay, this is a long question with a lot of computation, so I, let me go through the solution over here. After that we still have to do a lot of algebra unfortunately, so this is 30 degrees, right. So if I write the cosine law for the triangle CPO, I look at here, I have a CO square is, CO square equal to CP square. Okay, so let's see. Oh, great. One student have a very good, very good solution. So just stay after this so we can hear her solution. So if we have a CO squared equal to CP squared. Okay. And then CP squared, PO squared. And I just know that this cosine of 30 is going to be square root 3 over 2. So I plug it in and the only thing I need to do is to solve this quarter equations and remember when we did algebra, we saw it using complimentary, I'm sorry, completing the square. So I have this 2 here. I should leave it as it is. And then I have a square root of 3 over 2. I take over here and I only need 3 over 4 as a constant. I have an extra quarter here. So I put it on the other side. So I have a quarter here and then I just take the square root of a quarter, which is a half. So I have a 2 solution, square root of 3 over 2 plus or minus a half. And PQ is equal to 2PO. So I have a square root of 3 plus minus 1. And we're going to take the small solution, square root of 3 minus 1, right? Because square root of 3 plus 1 is too big, bigger than square root of 2. Okay. So that is my solution. But one student have a very short solution. So would you like to, I know this is a webinar, but it's a small class. But would you like to present the solution? You can speak out if you don't want to be recorded. I'm not sure if this is correct, but this is kind of what I did. So if you drop this point here, E, do you mind if I annotate? Of course, yeah, feel free, yeah. So if you drop this point here, E, and we know that this CPE is a 30-60-90, we can say that PE is root 3 over 2. And we can find EQ, oh, sorry. So because of the overlap, we know that if we just take 1 minus 2 times PE, we'll get PQ because the overlap here. So there's this, and then there's this, and we just want to complimentary count that off. So if we take 2 times this value right here, and then just subtract it off of that 1, you'll get that over count. That is a very smart solution. And I suspect that that's what the person who invented would expect you to do in the exam because it's very short. Okay, so let me go back here. Is this a smart solution? Let me go back to this so I can go slowly so other student can follow. Okay, thank you. It's a very smart solution. So you can extend this one into E here and F here, right? And we know that EQ equal to PF, right? Then you will save QP plus 2EQ. Is that correct? Equal to 1, right? But then because we look at CPD, it's a 330, 60, 90 degree triangle. This is going to be a half. So this is going to be PE, going to be actually equal to square root of 3 times a half, right? Which is square root of 3 divided by 2. And then from here, you can solve this 2 and PE is equal to QP plus EQ. That is equal to square root of 3 divided by 2, right? And then you know that this one equal to 1. And then you know that EQ is going to be equal to 1 minus square root of 3 divided by 2. And you plug it back and you have a QP equal to square root of 3 minus 1. Okay, great. Thank you so much. That's an awesome solution. As you can see, geometry is a beautiful subject. It involves algebra, it involves symmetry, involves like many different things. And yes, a lot of time you could have a many different solutions to the same questions. Okay. Okay, thank you so much. So we'll see you next time when we deal more with angles.

2D geometry

congruent triangles

similar triangles

circle properties

perimeter calculations

area calculations

Pythagorean theorem

isosceles triangles

law of cosines

deductive geometry