Hi, everyone. I'm so sorry that today I forgot to record the live session. So unfortunately, we won't hear the wonderful useful tips that David gives to the class, but I will try my best to recreate experience that we have in class today. So today's our last lesson on our 10 lesson series about 3D geometry. So what are 3D geometry and spatial reasoning? We pretty much divide into two different topics. The first one is spatial reasoning that focuses on the views of an object after it is rotated, folded, unfolded, cut, stacked, or taken apart. We have more of these types of questions in Level 7 and 8, so I'm sure if you've been with Matt Kangaroo for a few years, you are quite familiar with spatial reasoning and visualizations. And then what's new in this Level 7 and 9-10 is the 3D geometry in which we study in-depth solids such as the cube, the rectangular prism, the spheres, cones, and cylinders, the sort of very most common 3D geometric figures. And in terms of problem solving for the visualization and spatial reasoning part, we will solve questions that involve creating a solid from smaller blocks, views of a solid from different directions, and a nest of 3D geometric figures. And for the second part, we'll learn to calculate the area, the surface area, the volumes, and angles of solids. And for other 3D geometry, we will find that it is very closely related to 2D geometry, and what we learned in the last two lessons on 2D geometry will come in very handy. Okay, so pretty much the problem-solving strategies that we had in the last few classes are also very helpful. And imagine, you know, a lot of things when you can't quite draw, imagine how it's rotated and moving in your head would be very helpful. But of course, drawing additional lines and, you know, turning 3D problems into 2D problems would be some of the strategies that we'll be talking about. Okay, so let's go with the first part on visualizations and spatial reasoning. So we have a rectangular prism made out of four blocks, each consisting of four little cubes. Three of these blocks are shown completely in the picture. Which of the blocks below has the same shape as the white block? Okay, so we have two fours, that's 16, right, 16 blocks, and 16 small cubes, and four blocks, four different colors. Three of them are shown completely, and we see all of them here. And what we do not see is the white block. Yeah, so for these types of questions, you know, some of you are naturally good at visualizations, usually you can just figure it out right away. I'm usually not good at that, and for some of the questions that are more complicated, it's really not obvious for me. So I usually rely on diagrams, and I really encourage you to find a way to, you know, do it yourself, whatever is an effective and easy way for you to do it. For example, for me, in this case, I would just draw a diagram of the bottom part, because I know that the white block is all in the bottom layer, so it has two rows, four columns, right. And I would just mark where the little squares are. So for example, this one, the most obvious, has one, so I just put one here, this one, right. And then I would just go, you know, layer by layer, the first layer here. In front, there are two. In the back, I see only one, so there's one at the bottom, mostly the white ones. I have another one here. And then for these layers, the same thing, one in front, and there's one in the back, so another one here. And then for this one, I see three. One is not seen, and another one, and that's four. And I double-check that here, there's a gray one in front, and in this, all four of them are visible. So that's what I have, one, one, one, in this configuration, and they're all in the bottom layer, so that means it's C, right. It could be easy in this case, but this method really helped me to figure out a more complicated case, so by all means, develop your own method. And then, so these are put here for completeness, but most of the time, you would deal with nets of a cube, and if you remember from 1178, a cube has 11 different nets, and we have a method called side-pairing, in which we can sort of rotate the face over here, so that we do it 90 degrees, so that this side, this edge gonna match with this edge, so when you fold it, this one match with this one. Do the easy one first, these two match together, and when you rotate this 90 degrees, this one gonna go over here, this one goes over here, and this one goes over here, so you have 11 nets of a cube. Try to look it up and make sure you are comfortable with this method of side-pairing, and this one goes here, this one goes over here, and this one goes all the way here. And then, okay, so that's one question about nets of a cube. Imagine that the cube in the diagram is cut open along the dotted lines and unfolded. Two of the given configurations are impossible, so they are, so that means we, our five, three of them are the correct nets, and two of them are not. So for this kind of question, sometimes you can, you know, you know, you have to find three, just find the three easy ones, and, you know, the leftover, sometimes it's sort of easier to eliminate the obviously wrong choice. And remember that we can, you know, a face in a net can rotate 90 degrees, not 180, but 90 degrees to have identified pair sharings and the orientation of the pattern. So, also this is a re-recording rather than a live session. In a live session, we wait for the students to have the time to actually try and solve the questions before we discuss the solutions, but over here, you are, I probably will move a little bit faster because then you can always pause the video, you know, if you need more time. So I would go for the easy one first, right? And two is obviously the easiest one. I can, this is the bottom, I can just pull up the four sides and the top. That would give me this face. And then what's next? One actually also pretty obvious, that's the right net, right? You fold these four together, all this make one face, and then the opposite one, you fold this one up. So one and two is good. Let's look at this one. So four, remember the pair, side pairing methods. If we rotate this one over here 90 degrees, then it would go like this. And you can, you notice that the triangle points up this way, right? It becomes this one. And then if you rotate this one over here, it actually, the triangles gonna point this way, right? Which is, which is perfect because now two look exactly like, four look exactly like two, so that is the right net, right? So we know that the answer is D, which I think maybe half of the students who attended got it. And we can see what's wrong with three and five. So five is obviously wrong, right? Because when we fold it up this way, this and this, they are not on the same face, while all four of this are on the same face, right? So five is out. And for three, if we, you know, rotate it over here, then the triangle is facing this way. So that is, that is no good, right? So three is out. And this method of when you can actually rotate 90 degrees is super helpful in solving nets of a cube problem. But some of the questions are quite challenging to look at the previous exam and for level seven. so for our class today, it's good to remember this formula for the pyramid. Okay, let's go next to, okay, surface area. A cube with side length equal to 10 centimeters was covered with a certain number of identical yellow squares here. In such a way that all the sides look identical, what area of the cube in square centimeters is covered by yellow squares? Okay, so we have the side length is 10, so this is 10, and same as 10 here, so basically we, we have to calculate the area of each yellow square, right, and how many of them in total. Yeah, I think this question also everyone in the class got it because, you know, it's 3D but actually it's just 2D that add up all the surface area. So we know that tan, you look at this over here, so that is half, half, and the full diagonal, right? So each diagonal, we can just write right here, is equal to 5. And because this area is isosceles, the right triangle, so according to the Pythagorean theorem, this is 5 square root of 2. That is the side length of the square, right? So that's good, this a equal to 5 square root of 2, so the area is a square, it's going to be just equal to 25 over 2. Right, so that's it. I have a 25 of 2, now I have to calculate how many squares do I have? And then for each face, I have one in the middle, it's half and half and half and half, so that's 3. It makes me the second class is 2, so actually 3 squares, and we have 6 faces of the cube, so 6, and that is 6, you know, cross out with 2, so I have a 9. Times 25, which is 250, but minus 25, so that's 225, the answer is C. Okay, so again, 3D, but actually it's 2D, and special triangles over here is something that's very useful. We saw a lot of these in the last two classes. Okay, so now when we go to real 3D geometry, there's some useful tips that we can apply here. So first, if a line is perpendicular to two non-parallel lines in a plane, then it is perpendicular to any line in that plane. This theorem is kind of, you know, when you think about it, a lot of these are pretty intuitive, right? You have something like this, and standing on the table, then any lines that are on the table is going to be perpendicular to the table, go through this line, and then, you know, whatever intersection with the table, you would see that this line is going to be perpendicular to that line, which means it's perpendicular to the plane. So, to illustrate this point, it's actually very nice to use a cube, because with a cube we have a lot of perpendicular lines, right? So, for example, we look at D-H. D-H here is perpendicular to H-G, and it also, D-H here is perpendicular to H-G, it's also perpendicular to E-H. That means D-H is perpendicular to the entire plane E-F, G-H. That means I take any line here, for example, F-H, right, and F-D is actually 90 degrees, right, and when you think about it, it's kind of pretty intuitive. And then we can even do more than that. So, for example, you can see that C-D, C-D here is perpendicular to D-A and D-H, so C-D is perpendicular to the plane A-D-H-E, right? So, if C-D is perpendicular, if I take like M is the midpoint of B-C and N is the midpoint of A-D, then M-N is, because it's parallel to D-C, or intuitively you can see that M-N is also perpendicular to the plane, to this plane. If I take P is the midpoint of D-H, it can be any, it can be anything, but suppose I take the midpoint, and then this triangle here, M-N-P is actually a right triangle, because M-N is perpendicular to N-P. And more than that, I can actually calculate the length of M-P, and why is that so? Because, you know, M-N is the side length, of course, if I'm given the side length of the cubes and M-N is known, M-P is actually half of A-H, right? You see that? M-N is, yes, it's really 2D geometry, so, and A-H is diagonal, which is square root 2 of the side length, so M-N is known, M-P is known, using the Pythagorean theorem, we can calculate M-P, right? So that is super useful, and then symmetry, a lot of the time for solids, we have symmetry, and then, so that means over here, you can see that, for example, B-H is called the inner diagonal, is equal to C-E, right? B-H-C-E, and also equal to H-E, because these are the three equal, yeah, the inner or spatial diagonal of the cube. So, that theorem, perpendicular lines, use symmetry whenever you can to reduce the amount of computation you have, and as much as possible, you reduce a 3D problem to a 2D problem, because when you calculate something that you don't know how to, try to put it in a plane, try to create a plane, so, like in 2D, we do actual lines in here, we do actual plane, much easier when we deal with things in 2D. Okay, so let's use what we've learned here to solve some of the questions in 3D geometry. The first one is angles. So the picture on the left shows a cube with four marked angles. What is the sum of these angles? So, remember these are in space, right? Spatial ones, so, if it's a, if it's a quadrilateral, if all these four planes are quadrilateral, like on a plane, then we know that the sum of these angles is a 360 thing. But, oh, I think also worth just to, you know, remind ourselves that a line, I mean, two points determine uniquely a line, right? And three points uniquely determine a plane. Of course, that point should not be, as we call it, you know, non-collinear, it should not be on the line. So three points define a plane, right? But four points do not define a plane. That's why the tripod has only three points. It would have, I mean, something like this. You can imagine cutting the plane here and go about some points, so, you know, four points never take it for granted that four points define a plane. So let me name it while you are working through the solutions. ABCDEFG. So just find out whatever is easy and good for you. So BCD definitely, right? These are two sides of a square, so definitely this is 90 degrees, no doubt about that. What about one and three? So one, we said earlier that CB is perpendicular to AB, perpendicular to PF. So CB is actually perpendicular to this entire plane ABFE, so that means it's perpendicular to any line on that plane, so that means CBE is also 90 degrees. And the same argument of here, either you can use symmetry or you can use the same argument, CG is perpendicular to GF and GH, so it's perpendicular to this plane GH, and that's why this angle three here is also 90 degrees. And we are left with BE and FEEG, and when some student said that this angle is also 90 and up to 60, then David immediately recognized what's the mistake here. The mistake the student think that these two planes are parallel to each other, I'm sorry, perpendicular to each other. The ABFE and EFGH, when you have two planes that are perpendicular to each other, don't take it for granted that any two lines on that planes will be perpendicular to each other. The theorem says that when you have one line perpendicular to a plane, then it's perpendicular to every line on that plane, but be very careful when you think that the two planes are perpendicular. So in this case, EBEG is not 90 degrees, but we know that EBG is a three-point, so three points always define a plane, so what I can do is that I can connect B and G, and then now this, I know that EBG is a 2D object. But more than that, I know that EBG is actually, have you seen that it's an equilateral triangle, because EP is a diagonal of a face, EG also a diagonal of this face, and BG is also a diagonal of this face. So we sort of break the three of them into three different planes to show that they have the same length, and then combine them together in another plane, and that plane is actually the intersection of EBG cutting the cube at this point. So that C here is, angle 4 is 60 degrees, and we add up 90, 90, 90 with 60, we get 330 BSEL. And the answer. Okay. Go to the next one. This question, where the edge of the cube ABCDEFGH has a length of 2 cm, points PQR, PQR are the midpoints of edges of AD, you know, pretty much clear from the pictures, P is midpoint of AD, Q is midpoint of HE, and R is midpoint of FB, and we are asked to calculate the triangle of, the area of triangle PQR. So PQR, three points, always define a plane, so PQR is, you know, is a triangle, always a triangle. And what, what can we, how can we calculate the area? So as David said, you could either use symmetry, or you can use brute force, right, brute force like what we did in the other two lessons, like length tracing, angle tracing, there are just so many lengths, so many angles, so try to find out as much as you can. Or you can use symmetry to simplify your calculations and do it. You know, you may just want to connect, but it's really hard to make a good picture, so you connect P and Q. And you know, it's a 3D, when you draw it, it's sort of preserved parallel lines, right, but for non-parallels, if everything was just swapped, you really, there's, there's no, you know, the scaling is just not preserved. So after a while, you know, I think half of the class also figured it out, we have P and Q over here, it's, if you look at PQR, if you sort of, you know, use symmetry, actually you can see that the distance from P to Q and P to R are the same, and also C to R are the same. Same distance from Q and R, right, because this is also midpoint of this, you know, A, D, and G, sort of on the other side, 90 degrees from that, and same, this also, you know, 90 degrees on a different plane. So you can, either you can use, you can use symmetry argument to show that this PQR actually are, PQR are equilateral, or you can use brute force, and I mean, we need to calculate the sine length anyway, so let's see if we can use brute force to calculate, let's look at PR, maybe that would be the easiest one. So we look at PR here, this PAB, actually a right triangle, right, and we know that this is 2, this is 1, so using our friend Pythagorean theorem, it's square root of 1 square plus 2 squares, that gives us square root of 5, and that is actually PB, right, PB over here, which is square root of 5, and then, if we look at this triangle PRB, if we cannot do, like, quadrilateral or something, but we can always do triangles, because triangles, 3 points on another plane, so PRB, we know that RB is perpendicular to AB and BC, that means perpendicular to the bottom, here ADCB plane, so RB is perpendicular to PB, so this angle is also 90 degrees, and then we know that RB is equal to 1, right, half of the sine length, so, beautiful, we have a PR, it's going to be actually square root of PB square, which is 5 plus 1, DR square, so that's square root of 6. And similarly for all other ones, so now I have this equilateral triangle with sine length of square root of 6, and I need to calculate this area, and last week, I think last two weeks, we talked about these special triangles, and we know that if this one is, it's a drawing, but we have this equilateral triangle, and we know that if you take half of it, this goes in 90, 30, 60, if this dimension is A, okay, so this is 1, this is 2, this is square root of 3, right, so just remember 1, 2, square root of 3, and then that means the area is going to be square root of 3, half of square root of 3 times 2, so that's the area, S is going to be just equal to square root of 3, right, in this case. But if you have a thing, in general, if you have A, right, A, A, A, so everything here is going to be scaled up by A, then this A is going to be A over 2, and this one is going to be square root of 3 divided by 2A, and the area is going to be square root of 3 and times, this square is going to be square root of 3 over 4A square, so actually, it's a very, very useful formula to remember, but if you do not have it or don't remember it you can derive it at least remember this 2 1 square root of 3 ratios so over here I would have the area equal to square root of 3 over 4 and times a square which is 6 right so I cancel it out 2 by 2 so that's going to be 3 square root 3 over 2 centimeter square and the answer is c and that is the same thing again the strategy is that you break down and calculate each side of the angle of the triangle separately using different planes and from and then put them up together in one plane and calculate the area so lots of 2d geometries in this problem let's combine them to get to 3d okay the next oh yeah here we just type up everything so to give you the solution sort of nicer and neater if you want to look at it at home now it's another one about volume in a pyramid s a b c all the angles with a common vertex s are right angles the area subsides s a b s a c and s b c are respectively equal to 3 4 and 6 what is the volume of pyramid s a b c okay so this is a pyramid right and we talked about earlier but I just wanted to okay remind you of the formula which has the factor of one third here um times the height right it's the height from the apex to the um base and times um area of the base that is like this this formula is general for any pyramid the nice thing about this triangular pyramid is that any size can be a base right if you have if you have a peer if you have a pyramid that you know um the base are a the base is is a you know a quadrilateral then there's only one way to draw the height here from the apex to the base there's no other way right but in our case all the all faces are triangles so we actually have the choice of choosing which one's going to be the apex right and which one's going to be the the base you have that choice so make that choice that makes life easier for you because right now we think that s is the um the apex and a c b is the height then uh we need to calculate a b b c c a and um you know could be tough if you if you use this one it could be tough but is there a different way to do that um notice that s a s b and s c they are all um perpendicular to each other so maybe i could use you know one of them at the height and the other is the base right so um you know you might i think it might sort of yeah it might be helpful to why not right draw a rectangular prism you don't have to draw the whole thing but um yeah it sort of gives it an idea so yeah so um you can you can have a okay s has to be here a you can call it a apex and then um we can have you know b because s s a perpendicular to s b and perpendicular to s c so the a is apex and our base is actually s b c and the height is uh the height is is as did you see that right so a b here a c here right so that is our this thing is our rectangular um our pyramid right so s a let's go with a s b let's go with b and s c is called c so the base um so v gonna be just equal to a third of what um the height is a right so that's great and the area here is b and this is this is a right triangle so it's a half of b times c and which is um a b c which is like take a b c and then divide by six right and that is a very beautiful formula because um there's a lot of symmetry over here the the the units is correct this volume so it's a you need to multiply three lengths together and there's absolutely no reason for a or b or c to stand out they should have an equal row so that's why when you have this formula instead of you know plugging in the numbers you can you can see the big pictures and you sort of more or less sure that um this is the correct answer so now the job is to find a b c and that was not too hard because a b is twice the areas right is twice um area of s a b so that means a b equals six right and what about b c is twice the area of um s b c so that was six so that's going to be 12 and a c is twice the area of um uh s a c so that is four so that's eight right does it does it lots of area lots of algebra here and then a b c of course from here you can calculate for a b c separately but we don't have to because we only need the product of a b c so using as you remember we use you make use of symmetry in our problem um so we can just multiply the three equations together such as a b c square right and on the other side i have 6 and 8 is 48 so i have a 12 times 48 but i realize it's 12 times 4 so that's going to be 24 square right so you're definitely like being comfortable with numbers going to helps it a lot um so a b c is going to be equal to 24 and i plug it in that formula v it's going to be equal to 24 over 6 that is 4 um after a while i think i remember most of student got it but um yeah just changing the point of view doing it this way would be so hard because you know after we find out a b c okay so a b c is 24 right so that means c is um 4 c is 4 in this case b c so we have a b c equal to 2 and a c equals to 4 to 3 right so that's another b equal to 3 so suppose that we can find a b c this way you know um we need a formula that's a formula that calculate area of a c b knowing the three sides of you know pretty involved we don't cover uh in math kangaroo uh and then but a problem that we still have to carry is in our life which is just difficult well if we change our point of view like putting you know the apex here instead of s um things got so much easier so all these kind of these techniques and skill it comes to experience but um it really has to do with 3d geometry um this is just just just be so hard okay good so uh yeah also this is the solutions um yeah we can look at it got more neat okay so also in class with time so we continue doing a few questions with visualizations a student who had 2009 cubic blocks of psi length one and two and 2009 square color stickers of size length one view a rectangular prism and completely cover the surface with the stickers making sure that none of them overlapped it turned out that some stickers were not used how many stickers were left over right so we read these questions and we see what's going on um we have a 2009 cubics right versus the year 2009 um and we use all of them we use all of them to be a rectangular prism we also have 2009 stickers but we do not use all of them there's some of them left over that's actually important piece of information some of them are left over and and when you think about the 2009 this is actually the volume right because each of them is one and then 2009 over here whatever the stickers are just the surface area right this is the surface area and that's what we have to figure out we have a rectangular prism we don't know we don't know the dimensions here we can go with abc whatever but we don't know the dimension but we know something very nice we know the the volume and think about it when we talk about algebra when you look at geometry questions like all algebra in this case um it's actually number theory we have abc is that the volume right it's the only thing geometry that we use now is 2009 so and we we talk about it in the lesson about algebra and um number theory we have only so many variables and only one equation that means have a constraint and the constraints here but they are integers right so next step would be to factorize 2009 um as a product of three numbers to see what we can have so you can try two which definitely not um we tried we try okay so um yeah abc which is um which right yeah definitely not a multiple of three right so seven five and then uh it's gonna be um if we do 2009 um 2009 you're gonna see it's equal to seven times um like you guys say two eight seven two or seven so that one is nice because it divided by seven divisible by seven again so seven times seven times only one and we were just joking in class that suppose that we did the wrong division so we didn't quite 2009 suppose that we've divided by seven and and we have a remainder so we skip seven and we have to try really hard because number would you check if um if you check if the number is prime or not from even number and then we have to check all the primes less than equal to square root of n and this 41 here is a very big prime number is very close to square root to the nine so make sure we know how to do long division correctly but once we have this one and then we say oh that's good because now i can have abc equals to a and so i have maybe a equal to 7 b equal to 7 and c equal to 41 right so super helpful um round factorization long divisions make sure you're comfortable with that and and and that that's great because you can use all the 2009 cubit blocks to build this and now the number of circles just in surface area and we know that ab plus bc plus ca and what did we have so it's like two times um seven squares that's 49 and then plus um seven um times 49 but then uh we have times two right right so that would be inside gonna be two also this is 14 14 plus one is 15 15 times two that's 30 30 times 49 which is pretty good 30 centimeters so that is um well this is 41 how could i miss such 41 not 49 okay so it's not i'm sorry i cannot get this okay so we do we do so one three four six four six and then the leftover is going to be we're not used uh going to be 2009 minus one three four six which is seven six three so the the answer is first thing is b yeah b is a candidate um but then you know some some got e and yeah i e so and then i i initially i thought that oh actually you know when we factorize this into this can it be i mean we know that 2009 so these all nice and well but like how about i write 2009 as one times one times 2009 right or i can do it like one times 49 times um two times times 41 for you yes i'm 41 49 and then and then david realized right away that there's no way this could be the solutions right because if abc equals to 2009 and just one of them equal to one if i equal to one then i equal to one then bc gonna be 2009 but if bc equal 2009 then this thing then this two um the number of sticker is plus bc plus c a that one's going to be way way too big it's much bigger than one than 2009 so i'm sorry um no no sign can be equal to one right and we forget about one and that is the only way the only way we can factorize um 2009 as product of three numbers none of this equal to one so that means um it definitely two solutions and we will say oh this solution is impossible and that would be uh you know a tricky choice here but since i have only one choice then the answer is b so geometry would turn out to be like more like another theory question and then this question is like a viewpoint of you know look at a solid from different directions so all three figures below represent the same pyramid made out of wooden cues as seen from different views from the front from above and from the left of how many cues is this pyramid made so you can choose So I was just trying to draw it. And this is question number 30, the last question. No computation whatsoever. I mean, very little, right? And it could be that you got it right away or it just gets stuck for a long time. And yeah, that was manifested in class that one student just got it right away, because someone which is very good at visualizations. And for the most of us who are not, including me, we just have to find a way to solve this kind of question, like find the representation that you'll feel comfortable with, whether like dissecting them into layers, have a system of to take note of different layers, whatever that help you solve the problem. And after a while, just look at it long enough, you sort of figure out. If you look at it in the front here, so we see that you only see four. Of course, I mean, here is to see four, but maybe behind it is something, right? Four or in front is something. Four is the highest that you can see. And same over here. So here, the highest is one. Here is highest two, but there could be one behind it. But what I know is that I have four rows or four rows in these directions, right? Four rows in that direction and no more. And then when I look at the, and I look from the top down, I see this. So that means I have only two rows. I have two rows. And that makes sense because now this is from the top, right? The top, I also have only two, right? Two rows, or you can say two in the direction, called columns or row, but only two. And that means when I look from here over here, that makes sense. I would see this one has only two blocks. So I can use numbers like what I did in number two and in question number one, the number systems is to label it. So when I look here in the top, I have this one has two blocks on top of each other. And over here, I see four. The four could be any of these. This is the highest, right? But in the front is a four. So that mean this thing has to be four. This thing has to be three. This has to be two, which comes from here, right? This comes from here and just translate the whole thing. This one comes from here and that's one. And now it makes sense that once you have it, you know, it's still hard to, you know, draw all of it, you have to draw it in angle. But once we have the diagram, we sort of can visualize what the construction looks like. And then everything, two and four is eight. Okay, to six and nine, 10, 12, and so it's 12. Okay, and some of the, as you can see, some of the visualization question actually are harder than the truly 3D geometry in which you have to use formulas. So that's pretty much it for the webinar. So we went through 10 lesson and remember the first one, we talked about different problem solving strategies and then all of this, you know, basically algebra, number theory, counting, combinatoric and geometries are big areas that you would do in high school. I mean, after that you do calculus, but these are sort of big separate topics. In math, we also have quite a bit of logical reasoning and sequences and patterns in the lesson in which we are just, you know, have a given a sequence we ask if some configurations are possible or not. And some of the hardest questions that I found will actually fall into these categories in which you have to use, you know, tools from all different topics to solve them. So remember math can go, you know, it's really a lot of times it tests you out of the box thinkings rather than, you know, complicated formulas, but remember things do help. I mean, remember things faster, the faster you are, expedite your, you know, the solutions and also, you know, if you keep thinking long enough about the question and doing enough practicing, it just automatically remembers things like the area of a equilateral triangle is the square root of three over four a square. So practice really, really helps. I know this is only more than one week before the competition, but if you have time, definitely do practice at least a few full exams. And for those of you who, you know, pretty much like comfortable with all the material, definitely look back at the old exam and every exam that could be maybe two, three questions that are quite challenging and interesting and all the way surprises. So good luck with the exam, good luck with everything. And this is the first time we offer courses for level nine and 10. And unfortunately, we don't have a class. We only have the webinars. There are little interactions, but I hope David and I did our best to, you know, we tried and don't know how it goes, but we would really appreciate it if you receive a request from the EdScan group, provide us with any feedbacks, constructive feedback, just can be frank about your opinions that will help us improve the course and I'd be very much appreciated. So,

3D geometry

spatial reasoning

problem-solving

visualization

surface area

volume

pyramids

cubes

symmetry