Okay, thanks for joining. I will launch the poll for this warm-up problem now. Tom used six squares with a side length of one to form the shape shown in the picture. What is the perimeter of the shape? Okay, let's take a look at our warm-up problem. Thank you for joining us today. So the results of this poll is looking really good. Over 80% of you think the answer is 12. That is correct, so I'll just really quickly go over this. We know that they're all squares with a side length of 1, so we have 1 here, 2, I'm just counting 3, 4, 5, 6, 7, 8, 9, and 10. But we also notice we have a couple of little pieces. And if I look at the pieces together, I have a piece here and a piece here. Those together are going to sum to 1 because I know that this square is the same size, so I've just shifted it over. So the two pink pieces have to be another one, which takes me to 11. I'm going to switch colors. I guess I will use blue. I have a piece over here that was supposed to be blue. I have a piece here and a piece here. Again, I've shifted the two in the second row over a little bit, so I know that these two pieces on the third, on the top of the third row, have to also add up to 1. So that gives me another 1, and adding that, I have my answer would be 12. So as you can see, today's lesson is going to have some pictures and diagrams. Sometimes our geometry lessons have pictures and diagrams. This one is not specific to geometry, but it is going to involve pictures and diagrams. So let me just give you the title slide so you can see. So we might give you a diagram where you need to interpret it and use that to solve the problem, or it might be that you drawing your own picture is going to help you find a solution to the problem. Okay, so if this is your first webinar, I know we have students joining us in this series all the time, welcome. This is the fourth in the webinar series. If you need to know how to access the recordings, please stay to the end of the webinar and I can show you how to do that through your Math Kangaroo account. Okay, I hope you enjoy the problems today. Our TA is with us and he'll explain some problems as well. If during the webinar you have questions, you can send them in the chat to me or to Jacob, and we'll do our best to try to help you out. Okay, sometimes when a problem is not illustrated, you may want to draw your own illustrations. Sometimes a lot of us are visual learners and actually putting those words into a picture or a small brief diagram can be really helpful. Sometimes you might need a diagram or picture to help you keep track of stages. You might want to cross out steps you've done and steps you need to do. I will caution you to keep your keep your diagrams pretty simple. There may be times where you need to draw to scale. So if, for example, a problem tells you that the you have a rectangle and the length is twice the width, it might really help you if you draw a rectangle that is, let's see if I can, I have all my little, that is about twice as wide as it is long. So that might really help you is if you draw to scale because it might say, then it might say that you have to divide that in half, right? If I hadn't drawn it to scale, it would be pretty hard to divide it in half. So that might be something you want to keep in, keep in mind that sometimes scale can help, sometimes it's not necessary. I have polls for quite a few of the problems today, but not all of them. If there are images and the answer choices, it's not something that I can readily do in a Zoom poll, but you'll see like for this one I do have a poll and you'll still be able to see the whole problem in the top of the poll. By drawing two circles, Mike obtained a figure which consists of three regions. You can see the three regions, one, two, and three. At most, how many regions could he obtain by drawing two squares? So you want to maximize it when it says at most. So in Math Kangaroo, we want to always make sure we're reading the problem and understanding what we need to find very carefully. So you'll find I emphasize words, underlining them, and just with my voice as I'm reading. I will launch the poll so you can answer there. I did not, I guess I did not do this one as a poll. Somehow I wound up with the warm up in number one repeated. Okay, I'm seeing quite a few students give me the answer A, and I can definitely see why you might choose A, but I'm going to give you a hint. The squares do not have to be, they do not have to share like the same base or anything. So I think quite a few of you are drawing maybe squares like this. However, the squares do not have to make right angles to each other. Does that help? So this is what I mean by the right angle to each other. They do not have to be this way. I'm seeing students get close. Some students have, yeah, this is better. Okay, good. So I'm gonna just use my little square tool to draw the first one, cause that's nice and easy. But how will I draw the second square so that I can maximize the number of regions? And I had a student tell me, and I thought it was a good insight that you could make an eight pointed star. So by rotating the second square, you can make an eight pointed star. So quite a few students are giving me the answer eight. And if I look at this very carefully and I wanna count the regions that I have, I think you'll see that there's one, two, three, four, five, six, seven, eight, and don't forget the center. So there are nine if you turn it into like a star shape. So hopefully that helps all of you. For those of you who got it, really good work. And if not, now you'll know, right? So that's one of my important points is if you don't get a problem correctly when you first try it, but after we explain it, you say, ah, I can see why that was the better answer, then you've gone through the learning process. And that's the big goal here, right? So number two, Agatha used 36 identical cubes to build a fence around a square region. The cubes are all connected. How many cubes identical to the cubes she used to build the fence does Agatha need to fill the enclosed area? And I guess I do have a poll for this one because we saw it before. All right, so here are the results of the poll. And this is 52%. More than half of you say that the answer is 64. And that is correct. There are a couple of ways to think about this. So I've kind of drawn. I appreciate if the students do not annotate while I'm annotating. What's going on? Where are we? So what you'll notice is if I have 36 around the outside, one way I thought about it was I thought, OK, well, I have four groups if I outline them like this in green. So 36 divided by 4 gives me 9. So each one of these is 9 across, each of the greens. And since I have one square on each side, that means inside, I have an 8 by 8. And 8 by 8 is 64. Another way students might think about it is they might try to take away that you could do the 36 and you could subtract 4 for each corner. That would give you 32. And then you could say, OK, so I know I must have 8 along each side because 32 divided by 4 then equals 8. So there's kind of two ways to think about getting an 8 by 8. Doesn't really matter how you get there. The answer would be that you have 8 by 8 squares inside and 64 total. OK, there are three points that form a triangle. In how many ways can a fourth point be chosen in order to create a parallelogram? So we know that you need four points for a parallelogram, but this is specifically asking how many points for a parallelogram, but this is specifically asking how many ways can you do it or how many parallelograms could you make? OK, we have the majority of you have answered the poll. Anybody else want to put an answer in there? It's anonymous, so we don't know if you get it correct or incorrect. Only you will know. All right, I end the poll here. More than half of you think the answer is 3, that there would be three ways to add a fourth point and make a parallelogram. We do know that you need four points to make a parallelogram. And a parallelogram is a very special type of shape because the top two sides have to be parallel, and then the other two sides have to be parallel. So if I have the triangle that I've drawn, and let's say I'm going to focus on this side and this side, then I need to make a parallel line here. That would be parallel to the one on the left, and a parallel here to be parallel to the one on the bottom. So that gives me one point right here, right? But I could choose two other sides. I could choose, for example, this brown side and this brown side. So I would, again, have another line parallel, kind of like the first one I drew. But now I need a line that's parallel to the other brown side. So that gives me a dot over here. Now I can choose a third color yet. I've done those two sides. I have not done this side and this side. I would need to make a line parallel here and a line parallel here. And while I'm almost off the screen, I do have a dot. So the correct answer is that there are three ways you can turn a triangle into a parallelogram. You have to use two sides at a time, and you can take two sides at a time in three different ways. And then you would draw the next point to make a parallelogram. How many more triangles than squares are shown in the picture? So the first thing you might need to do is figure out how many triangles and squares you have. That way, you can figure out which one is more and by how many. Sorry, the picture is a little fuzzy. If it's when it gets enlarged, sometimes they get a little pixelated. Be careful on this particular one. Math Kangaroo did not put the answer choices in ascending or descending order. They're kind of random, so carefully choose your answer. Okay, so I'm getting quite a few answers. Anybody else want to answer the poll? Not everyone put a response there. All right, I'll end the poll now. Jacob, can you see the poll? I don't think I've ever seen a result quite like this one. We have practically a four-way tie. So that's really interesting. I think that means we better go over this problem very carefully together. Certainly there are not four correct answers. There's one correct answer. Math Kangaroo is pretty consistent about that. Okay, I was wondering if I had... There is a version of this slide that has everything already pre-colored, but we can do it together. Okay, so it says more triangles than squares. So the first thing I'm going to do is try to figure out how many triangles I have. So there's one large, right? I think everyone knows the whole shape is a large triangle. So that would be this big large one, right? Pretty easy to find that one. Okay, let's... And then there's... We have some small triangles as well. I'll switch my color. We have these small ones. We have four of these small ones as well, right? Excuse me for a second. There's some background noise. Okay, so we do have the four small triangles. Let's see what else we have for triangles. Oops, four small. We also have some medium-sized triangles. Some triangles, I'm going to put them into some categories, some triangles that are let's call it too high. So these are triangles that are too high, and I see three of those. And then we have some triangles that are, I will call, three high. We'll call those three high. And three high, I see only two of those. So all together, the total number of triangles is five plus five or ten total. Now let's take a look at the number of squares. Squares, we're going to start out, I guess we can start out with small squares. So we'll start out with one by one squares, and I see one, two, three, four, five, six of those. Then we can try to see if we have any two by two squares. I see a two by two square here. There are no three by three squares or any other squares, so that gives us seven total squares. And so then to find our final answer, we have to find the difference, which is ten minus seven equals three. And make sure you select the correct answer. They're out of order. E was three. Okay, number five is a problem that you had wanted to lead, Jacob. Did I get that right? We can't hear you, Jacob. Sorry, can you hear me now? Yes. Okay, perfect. So each cell of a five by five table is filled with zero or one, such that each two by two square of the five by five table contains exactly three equal numbers. You want to find what the largest possible sum of all the numbers in the table is, so I'll give you guys some time to think about this, and then we'll go over it. Okay, Jacob, I think we're going to have to get to a solution so that we can solve more problems today. So I'll end the poll and share the results. So can you see those results? Maybe you can give them some feedback on that. Okay, so it seems that most people have answered 21 and some have answered 20 and some other answer choices. So we'll go over it now. The key to notice is that since each 2x2 square contains three equal numbers, that means that you have these two options. You can either have three 1s and one 0, or you can have three 0s and one 1s. And this is important because this basically narrows down the possible number of tables, five by five tables that we have. But notice that we want to find the largest possible sum of all numbers in the table. So that means that we kind of want to take the three 1s and one 0s. So each square should have three 1s and one 0. So let's try to figure out what the possible sum of all numbers in the table can be. So notice that if we consider these four squares, notice that they don't intersect. And notice that they have three 1s. Each square has to have three 1s and one 0. So in order for that to happen, if each square has three 1s and one 0, that means that in total, across all these squares, since they don't intersect, they have four 0s in total. So we can straight off the bat realize that you can all have four 0s, or you have to have at least four 0s in the table. And so that means our sum has to be at most 21. So our answer to this obviously can't be A. So now we know that the sum has to be at most 21. And each of these squares, 2 by 2 squares, have to have a 0. And we want to maximize the sum. So in order to maximize it, we want to have 1s in each of the squares that are not one of the highlighted 2 by 2s. From here, notice that since this is a 2 by 2, and it has three 1s, then that means that this has to be 0. And from here, we can just fill out the rest of the grid. So we have to have 1 here, because in this square, you have three 1s and one 0. And similarly, in this square, you have to have three 1s and one 0. So this has to be a 1. And this square has to have three 1s and one 0, so I mean that this has to be a 1. And you can do it for the rest of the grid. And then you get one 1, one 0 in that square. You can have that in that square. And then finally, in the last square, you have that following labeling. So in total, to maximize this, we have this 0, this 0, this 0, and this 0. So in total, we have 21 1s. So the largest possible sum is what we thought it would be, which is 21. So answer choice should be B. I really like this problem. Jacob, I have students saying that you usually pick the toughest problems. OK, number 6, Wally wrapped a wire around a notched board. The picture to the right shows the front side of the board. Which of the pictures below shows the back side of the board? So this is one of these picture problems that you could not possibly do without having images as your choices. So I could not make a poll with this. But this is something that Math Kangaroo likes to do, is they like to test your spatial reasoning. Can you turn this? Can you rotate it? What can you do? And see the back side. So we're asking about the back. Give you a moment. You can answer into the chat. Make sure you're using your three-dimensional reasoning. A lot of students are giving the incorrect answer because they're showing where the string is on the back side or the wire is on the back side, but they're showing it from the front view. And we want to look at it as if we're looking at the back. So it requires you to think about this in a slightly different way. This board is not see-through, right? So to see the back, we have to actually turn it around. Okay, so I'm going to take the first step, the first step is I'm going to use a nice bright color and I'm going to try to show you what the back side of this looks like, right? So I'm going to connect the dots basically. We have a piece of wire wrapped around and it has to go from the bottom to the top here, right? And it has to go from, so that's what the wrapping would look like on the back side. So that is why a lot of students are saying that the correct answer is A. That is what the wrapping looks like on the back side if this was a see-through board. But we have to imagine that we're looking at it from behind. So if we're looking at it from behind, then which picture makes sense? Well, we know we need two wrappings that are more in the vertical up and down, right? So the only one that really works with vertical up and down is B, right? So let's see how does B actually work. If we take this picture that we have, you'll notice that this is upside down. So we basically have to take this thing and to see it from the back, we have to flip it so that the top is the bottom and the bottom is the top. So if I, I'm not going to use this color, if I flip it around, I'm going to have my board this way. And my original string, when I flip it, is going to be around the back. Original around the back is going to look like this. And then it's going to be from the bottom and from the bottom. And we have a little tail. So those, those are going to be in the back and now the front is the pink. So the front makes this diagonal and makes it up and down and it makes an up and down. So the correct answer is B, but it takes a little bit of creative visualization to imagine what it looks like if you take the whole thing and flip it like, you know, like, so you can think about it, those little tails are like my thumb, my thumb is pointing up and when I flip it, now my thumb is pointing down and you're seeing the palm of my hand versus the back of my hand. So try to visualize it that way. The best I can do is an explanation. Do you have something that helps with it, Jacob? No, I think that was a pretty good explanation. Okay. Jacob, you want to try to lead us through number seven? There is a poll when they're ready. Sure. Yeah, I can lead to this question. So in each cell of a six by six board, there's a lamp. We say that two lamps on this board are neighbors if they lie in cells with the common side. And at the beginning, some lamps are lit and each minute every lamp having at least two neighboring lamps is lit. If the three cells have a common vertex, what is the minimum number of lamps that need to be lit at the beginning in order to be sure that at some point, sometime all lamps will be lit? So one suggestion I might make for the students is you might try a smaller grid to see if you can figure out if there's like a method or a mechanism to make it work. That might help because filling in a six by six grid over and over again is going to take you a while, but you might try a smaller one. Okay, Jacob, I think we're going to have to go through it because we only have a few minutes left. All right. Perfect. So this is a very difficult problem, and it seems that we're kind of all over the place. So let's discuss it. So yeah, as Ms. McGee said, this is, it's very difficult to play around with the six by six grid. So let's say we just place a few lamps down. So let's say we have this lamp, this lamp, and this lamp, then after a few minutes, so after the first minute, we have this lamp is filled, so this comes on out. This lamp is also filled, and this lamp is also lit. After the first minute, but on the next minute, no other lamps can be filled, no other lamps can be lit, because as the poem states, some lamps are lit, and every minute, every lamp having at least two neighbors lit is, two neighboring lamps is lit if the three cells have a common vertex. And so let's see, no other lamp has, no other lamp has, no other lamp has, another lamp has a common vertex. So yeah. So if you look at any of the squares around the lit lamps, you'll see they only have one side with a lit lamp, no two sides. Yeah, exactly. So obviously, we can't just have three lamps. So if we extend this again, so then we have this lamp here, then we have this lamp, this lamp, and then we get this lamp again, and then we get this lamp, and then we also get this lamp here. So it seems that as though, it's kind of like the smallest rectangle that surrounds all the lamps. So we want to find the number of lamps that would create the smallest rectangle around all the lamps. So if we have a six by six grid, we have to have six lamps in each row and six lamps in each column. So we have to have something like all the diagonals are filled in order to have each lamp lit in the six by six grid. So and you can test that out, and you can see that actually works. You can see that if you have six lamps on the diagonal, then each lamp in the six by six grid is going to be lit at the end. So the answer should be six. Yeah, so students can test that out. They can light the lamps on the diagonals, and then you'll see that you'll be able to fill in all the others. So once you figure out that little trick that you need the diagonals, the problem goes very fast. It's a matter of can you figure that out. You can kind of see it in this part here that when Jacob had the diagonals in this three by three grid, he was able to fill in the rest of the grid. So if you expand on that idea, you can get there. All right, here's a nice bonus problem. What is the greatest number of shapes in the form, it's this little L shape, that can be cut from a five by five square? There's no poll for this one, so you can put your answers in the chat. We're about out of time. But I think, yes, the question is, can they be rotated? They can be rotated, but not flipped. So you can't, for example, try to make a shape that would be like this, right? That's flipped. So we cannot do that one. You could stand it on its edge. So we can start by outlining one here. We can do another one here. You can see we can make it all the way around the outside using these shapes which is really nice. And that leaves us with a 3 by 3 in the center. With the 3 by 3 in the center, you can get one and two more. There you go. So the correct answer is that there are six of them. Another interesting way to do this is you know that the 5 by 5 gives you 25 squares. This shape has four squares. So if you try to divide the 25 into groups of four squares, you're going to get six with a remainder of one. So you can then try to figure out is there a way that I can get the maximum number which is six and leave only one square blank. So that's kind of another way you can approach this problem is to figure out what would be the best case scenario is that I can get six of them in there and then see if you can make them fit. All right so that's going to be the last problem we solve today. I have had some people asking me how we get the recordings. So I just wanted to show you that. You can get the recordings. So this is me logged into my Math Kangaroo account and mine looks a little different because I have a lot more registrations than you do. If you go to the courses tab, you'll see that down here at the bottom this is the webinar that I'm teaching right now which is the webinar set B. And I have these different buttons that I can press at the bottom. If I press the event info or the contents button, I will get different things. When I press the contents button, I get a list of the recordings that are available to view. So this is the recordings for the first three webinars that we've already had. The other thing I wanted to show you is if I come back here, I am a center leader. I have a Math Kangaroo center. So all of you who have a registration for a Math Kangaroo contest, you can look in your contest registration and you will find some promo codes to actually get some of our Math Kangaroo past contests to use for studying and even free videos of some of our teachers and assistants solving problems from Math Kangaroo contests. So this is another way that you can get some practice materials. So I hope that helps you get a little more information and I hope you can now go back and review problems where you had any difficulties or if you've joined this webinar series late, you'll know how to go and get the other recordings. All right, thank you for coming today and I'll see everybody next Sunday. Thank you, Jacob, for your help and have a good week, everyone.

geometry

spatial visualization

math problems

perimeter calculation

parallelogram formation

rotations

diagram interpretation

visual aids

lamp grid problem

Math Kangaroo