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.
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...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.