Today we're going to look at my favourite puzzle. It's involving squares, rectangles - I've actually used this a few times in the Oxford University admissions interview questions so it's got some heritage behind it, bit of prestige
behind this question. I won't be using it anymore because I'm going to tell you all how it's done. We're going to start
with a square, like so. And this can be any size square, it doesn't really matter, and what we want to do is fill the square with rectangles. Now the rectangles can be of any size as well, but they have a rule that they have to be in the ratio 2:1. So this doesn't mean this is length two and this is length one, this could be length - let's say this was two quarters, and then - oh I should have picked a better number - and this one needs to be one quarter. So we could have small ones, and really tiny ones - it doesn't matter the size of the rectangle they just have to be in this ratio; long side is twice the length of the short side. And what we want to do is fill the square with these rectangles. And we're interested in, for what numbers of rectangles is it possible to fill the square? So clearly for 1 it isn't because a rectangle in this ratio 2:1 is not a square so you can't just fill it
with one, so we can say for n equals one that's a big no. But then for n equals 2 it's going to work because I can just do that. If we say the square's one by one then both of these
rectangles are one by a half, twice in the ratio, and we can fill it with two. So n equals two is a tick and that was n equals one was the cross. And we want to know for what other values can we do this? Is it possible for any other values? Can we do it for all of them? That's the question,
can we figure this out. (Brady: So you can use
duplicates obviously?) Yes we can use the- so here we've got two of the same size. I could have- it doesn't have to all be the same size. So that's important, we could have one big one and I don't know six small ones or something. Or maybe like four different size rectangles; there's no limit in the size of the specific rectangles or the numbers of each one, just we're interested in the total number of rectangles that fill the square as long as they're in this ratio.
- (So is) (the puzzle question for which numbers) (can we do it? Or is the question) (can you do it for all
numbers or like - ?) I always thought- I always present it as for which numbers can we do it? But within that we're kind of saying can we do it for all numbers? Are there some numbers other than one where it doesn't work? But we're going to hopefully- we're going to try and consider all possible values, that's that's the goal. We're going to actually try and consider
every single number. So the way I would approach a problem like this is- well there's two options the way I see it. So so we've shown so far it's not going to work for one, because then it would be a square. And we've shown that we can do it for two.
So you could continue in this way and say, well, let's try and do it for three, let's try and do it for four. But the problem is, if you hit a number where it doesn't work you've got to show it doesn't work, you've got to like try everything. Like you have to exhaust all possible ways of arranging rectangles, and it's really
difficult to exhaust every single possible thing that you can do in such an open problem like this. So my suggestion would be to think, well, can I find something else that works that isn't 2? Let's try and get another handle on the problem and see if we can use that to build other cases. If we start with the two that we had, we had a square and we divided it into two. So if I now divide this bottom rectangle into two squares I've now got a square that I can divide into two, and another square that I can divide into two - and lo and behold we've got a solution. So we've got five there. N equals five is a tick. So we took what we had, which was two, and then sort of broke it down - so I call this the breaking down method - into uh one of them, we broke it up first into two squares, and then copied what we did at the start to get four. So we replaced one rectangle by four rectangles,
so in total we added three to our two that we started with. Now there's nothing stopping me from doing this again and again and again. So I can pick one of my five rectangles, divide it into two squares, and then divide each square into two rectangles. And now I should have eight; 1, 2, 3, 4, 5, 6, 7, 8. So now we get n
equals eight is a solution. And I could have of course picked another rectangle and divided that up; I could have picked the top one and then I would have got a solution where we had eight all of the same size. But remember we're not interested in the specific configuration, we just care about the number of rectangles - is it a tick
or is it a cross? I can continue to do the same thing. So if I continue with this bottom left one, two squares, two rectangles. And I've added- I've replaced one rectangle by four so I've added three so that would be eleven. And I can do the same again so this is
the last one I'll show you because I'm running out of space - and I can just continue this. And again I don't have to keep picking this one, we can keep adding three. So we get 14, we get 17, we get 20, 23. To come back to the original question; for which- how many like ways can we do this? We now have an infinite number
of ways to fill a square with these 2x1 rectangles; because we just keep adding three and we can just go on forever. But we're missing quite a few,
there's a lot of gaps. There's almost like twice as many gaps as there are numbers we're hitting, because we're hitting every other third number. So what we want to try and do now is think about, well, how could we hit those numbers? This method I called the the Breaking Down method because we took a solution that we had and then we broke down a rectangle and added three to the total. And then the other approach I'm calling the Building Up.
Because what we're going to do is, again, start with our square and we want to now add rectangles around the outside but so that the final shape is a square, because remember the
square can be any size. So if we try and build up, you think about this a little bit, and again you should do this you can kind of create an overhang like this on all four sides. You can hopefully see that's possibly going to work, it could maybe be a square. So again if we take the middle one to be 1 and this one to be a half, so the middle square was 1x1. Then what we've got here is this length is two and then this one, to be in the correct ratio, has to be one, so we do indeed have a square. We've got two different sizes of rectangle, two in the middle and then four of this second size, where the square that we fill in total is going to be three squared but it's built up of six rectangles. So now we've
got a different solution, six, that we didn't have before. And just like before we can keep doing this and building up. So you could just do that; and you could figure out the numbers, it's gonna be a bit bigger again and you're gonna keep doing this. And we keep adding four every time. So we've got six, then we can say that we can do this now for ten, and we add another four to get 14, 18, 22, 26, 30. So we have another infinite list of possible solutions. And some of these numbers are the same; we've got 14 in both lists. But also you know we've got 10 which we didn't have, we've got 18, 22. We've got new numbers as well. So there's some overlap but we've got another way of constructing these solutions. Like before in this first case, we said that what we were doing was we were adding three every step. And you can think of this as saying, well if I have a solution for n for some number, then I know from our method here, the breaking down method, I can take one of the rectangles in the solution, divide it up into four smaller rectangles and that added three to our solution. So if we had a solution for n we can always get n plus three from the breaking down. And then here we've seen the same thing; because we started with two and I added four around the edge. But if I started with the five solution here, then I add four around the edge, and then I add another four. So I can also get- for example there I would get n equals nine plus four on that so then I'd get 13, 17, 21 etc. And again we're getting some numbers we didn't have. The breaking down method; if we can do it for n we always get n plus three, break one of them into four pieces. The building up method; if we can do it for n we can always do it for n plus four. And now these two are really powerful rules. So we're already seeing that we've got a lot of solutions; but if we actually think about this mathematically now using the algebra we can begin to see that we actually get almost all of the numbers. So I think the best way to demonstrate this is sort of write out the list of the first 30 numbers and cross off the ones we can and can't do; and then we'll think about can we prove this. So we know that 1 didn't work, I think that's the only one we know that doesn't work. We showed that 2 did, we haven't looked at 3 or 4. We had 5, uh we had 6, we haven't covered 7 yet. I think we had 8, yes 8 was the 5 plus 3. 9 we got down here by doing 5 and then 4 around the outside. Then we had 10 on our list. Now, we've got something really helpful here, because we've got three numbers in a row. So because I've got 8 I know that I can just add 3 to it and get 11. Because I've got 9 I can add 3 to it and get 12. Because I've got 10 I can add 3 to it and get 13. And then this is just going to carry on. So you can look back through our lists as well and see, you know, we had 14, 17 and you could tick them off. But if you spot that because we had these three numbers in a row then - and we have
this n plus three rule - we're actually going to get all of the other numbers. So there's lots of ways of spotting that you're gonna get all the other numbers; but however you do it it relied on the fact that we had- well actually only really this rule, the n plus three. Once we figured out that we could get 8, 9 and 10. So we did need building up to get some of these, right? So 9 and 10 came from the building up method and 8 came from the breaking down. So we did need the second method but really we just needed the n plus three rule plus those three and then we got all of the numbers. So that just leaves us with 3, 4 and 7. So we know that 1's impossible, we know that every single other number is possible now, except 3, 4 and 7. So we've got to figure these ones out. And this is- we've got to do this case by case. I'm gonna need more paper aren't I? (Brady: I have no intuition
as to whether 3,) (4 or 7 are possible.)
- This is how I think about it: we're gonna consider 3. We're gonna say, right, let's try and
find a solution for 3, suppose we can do that. And you want to think about the number of corners in your square: 1, 2, 3, 4 corners. And we have three rectangles. For example let's take one of our rectangles. Suppose it hits all four corners, it's clearly got to be a square and there's clearly no space for the others. So you cannot have one of your rectangles hitting all four corners. Similarly you cannot have one of your rectangles hitting three corners; because if one rectangle hits three corners it kind of has to hit the fourth. And it kind of has to be a square again. So the highest number of corners a single rectangle can hit is 2. And in fact one of them has to hit 2 because there's only three rectangles and four corners. So categorically we can say one of the rectangles must cover two of the corners of our square. And so the only way to fit that is to divide it in half. If we're trying to find a solution for n equals three this is completely fixed, there's nothing else that can happen. So then we've got to fill the remaining space with two rectangles. So now we've got two rectangles and we've got four corners again. So okay, it's a rectangle instead of a square but the corners is a really helpful way to get stuck into this problem because they've all got to be touched. So now you say, right, I obviously can't hit all four because then I'd fill it, it would work, but I've got one rectangle left. I can't hit three for the same reason so I have to hit two; and each of them has to hit two. If you take a 1x1 square this is a half so this one has to be a quarter. It's the only way you can fill a rectangle fitting two corners. So then what we're left with this- has to be a rectangle. And if we figure it out this is three quarters, this is a half - it's certainly a rectangle but it's not in our ratio 1:2 so that doesn't work and so n equals three doesn't work.
Three is getting crossed out. Now four you do in basically the same way; start with your square, you say you've got four corners - we can't we've got four rectangles. Similar argument, one of them has to touch two. Because
if each one touches one, you can play around with it a bit but you'll see that it doesn't work. It kind of has to be four squares, that's
the only way you can get it to work, it doesn't doesn't fit our rules. So again you have to have one of them like this touching those two corners, so this is out. And then you're left now with three rectangles to fill this half; we've got four corners, three rectangles, same argument as before except now we're on a rectangle, we have to touch two so it has to be this one half by a quarter again and then we've got the same space. You can play around with this as much as you want, if you put one horizontally it has to go the same size, a half and a quarter, and that's obviously a square. Or if you go vertically down, this won't work because this is a quarter and this one here is three quarters. That would be 1:3 ratio, not 1:2. So four, for the same reason really as three, ain't gonna work. (Cross it off!)
- So it all comes down to 7. This is the big finale of our infinite list of numbers for solving my favourite rectangle puzzle. Can we do it for n equals 7? And the answer, maybe surprisingly, is yes. We can do it; it is possible for seven. It's quite tricky for those of you who really want to give this a go, this is the time, have a go. It's it's very fun trying to do this. The trick with solving it for seven - or the way we do it -
is we actually need three different size rectangles. We'll say this is a 1x1 square so I can convince you all that this works. It doesn't look like it's gonna work at first. So to start with we have one big rectangle, that's gonna be our half by one. So you'll notice this is the building block of almost all of our ideas actually. And then you have three little guys; so this is gonna have to be a third for each of them. And the height here to be in the ratio 1:2 is one sixth. We have another one here and then we put another two here. This is a third, that would be one-sixth, and that will be a half. So we can check one sixth, that would be 2 sixths, 3 sixths - so that's one, that works. This bit here is one-sixth, this one's one-sixth, so that fits to be one-third. And then this gap here, this top long side is two-thirds and this one's one-third and we're done. So we've got five of these small ones, one middle-sized one, and one big one. And that is the solution for n equals 7. n equals seven - yeah! We can solve it for all possible values of n, we can fill our square with n rectangles in this ratio 2:1 except for 1, 3 and 4. (All right. So now anyone
who watched this video) (can get into Oxford.)
- Yes! I used this question for the last two years actually so so that's why I've had to retire it; because we- I try to come up with new questions but I
liked this one so much that actually this is the only question I've used on repeat. But it's now been retired. But yeah this is this is the kind of thing; because there's lots going on, there's lots to think about, lots of things to check and you could if you wanted go as fancy as you want in your proof with modular arithmetic or spotting the pattern - there's a lot to get stuck into. Today's episode sponsor is
Brilliant; and no matter who you are they're going to have a course for you. I love how Brilliant are really upping their game on interactivity, it always helps with learning and understanding. What you're looking at now are just the opening exercises in their massive course on mathematical fundamentals, it's just great stuff. I love the style, that clean look; I also really like how everything unfolds at my pace, at the speed that I understand things, that's what the
interactivity is all about. I never really think of Brilliant as being like school or study,
it's more like a a series of fun puzzles but each puzzle is secretly giving you a deeper understanding of mathematics or science. To check this stuff out yourself go to brilliant.org/numberphile. Using that /numberphile will tell them you came from here and also give you 20% of a premium subscription. brilliant.org/numberphile - it's well worth your time. ...keep drawing rectangles until I got that one to work. But you know the good- the good students, the good candidates, will be able to to get through and spot the pattern with the adding of three and like that's the key thing really I think in terms of the maths behind this is spotting that pattern.
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