Challenging perimeter problem - Math, Class 9 Video Lecture - Class 9 | Best Video for Class 9

Here's an interesting
problem involving perimeter
from the 2000 American
Invitational Mathematics Exam.
So it says the diagram
shows a rectangle that
has been dissected into
nine overlapping squares.
So one, two, three, four,
five, six, seven, eight,
nine overlapping squares.
Given that the width and
the height of the rectangle
are positive integers with
greatest common divisor 1-- so
they're talking about
the width and the height
of the rectangle.
And the reason why they're
saying greatest common divisor
1 is they're saying that they
don't have a common divisor
that you can divide them both by
to get a more simplified ratio.
And to think about
that is, we might
be faced with two choices,
one where maybe it's
this side over here is-- let me
draw it like this-- 5 and 15.
But over here, our greatest
common divisor isn't 1.
They're both divisible
by 5 over here.
So what you'd want
to do is say, no.
Instead of 5 and 15,
they need to be 1 and 3.
Now you have the
same ratio of sides,
but now their greatest
common divisor is 1.
You kind of have it in the
simplified form or the most
simplified form if you divide
both this height and this width
by 5.
So that's why they're saying
greatest common divisor 1.
And then they say, find the
perimeter of the rectangle.
So let's see what
we can do here.
And I encourage you to pause
it and try to do it on your own
before I bumble my way
through this problem.
So let's start in the beginning.
Let's start at this square right
over here, the center square.
And they did tell us
that they're all squares.
So let's say that that square
right over here has a length
x and a height x.
It's an x by x square.
So let me write it.
So this is an x,
and that is an x.
So this is an x by x
square right over there.
And then you have this
square right over here.
And we don't know
its measurements.
So let's say that this square
right over here is y by y.
So it has y width, and
it also has y height.
Now, what is this
square over here?
Well, this is an x
plus y by an x plus
y square, because the width
of these two squares combined
made the width of
this larger square.
So what I'm going
to do is-- actually,
this might be an
easier way to write it.
Since these are all
squares, I'm going
to write the dimension of
that square inside the square.
So this is going to
be an x by x square.
Kind of a
non-conventional notation,
but it'll help us keep
things a little bit neat.
This is going to
be a y by y square.
So I'm not saying the area is y.
I'm saying it's y by y.
This over here is going
to be an x plus y times--
an x plus y is going to
be each of its dimensions.
So it's going to be x plus
y height and x plus y width.
Then this one over here-- well,
if this dimension is x plus y
and this dimension
right over here
is x, then this whole side or
any of the sides of the square
are going to be the sum of that.
So x plus x plus y is 2x plus y.
You can imagine that
I'm just labeling
the left side of each
of these squares.
The left side of this
square has length y.
Left side of this one, x.
This one has x plus y.
And then this is 2x plus y.
And then we can go
do this one up here.
Well, if this distance
right over here is 2x plus y
and this distance right
over here is x plus y,
you add them together to
get the entire dimension
of one side of the square.
So it's going to be 3x plus 2y.
I've just added the 2x
plus the x and the y
plus the y to get 3x plus 2y
is the length of one dimension
or one side of this square.
And they're all the same.
Now let's go to
this next square.
Well, if this length is
3x plus 2y and this length
is 2x plus y, then this
entire length right over here
is going to be 5x plus 3y.
5x plus 3y is going to be that
entire length right over there.
And we can also go to
this side right over here
where we have this length--
let me do that same color.
This length is 3x plus 2y.
This is x plus y.
And this is y.
So if you add 3x plus
2y plus x plus y plus y,
you get 4x plus-- what
is that-- 4y, right?
2y, 3y, 4y.
And then we can express
this character's dimensions
in terms of x and y because
this is going to be 5x plus 3y.
Then you're going
to have 2x plus y.
And then you're going to have x.
So you add the x's together.
5x plus 2x is 7x, plus x is 8x.
And then you add the
y's together, 3y plus y,
and then you don't
have a y there.
So that's going to be plus 4y.
That's the dimensions
of this square.
And then finally, we have
this square right over here.
Its dimensions are going to
be the y plus the 4x plus 4y.
So that's 4x plus 5y.
And then if we think
about the dimensions
of this actual
rectangle over here,
if we think about its
height right over there,
that's going to be 5x
plus 3y plus 8x plus 4y.
So 5 plus 8 is 13.
So it's 13x plus 3 plus 4 is 7y.
So that's its height.
But we can also think
about its height
by going on the
other side of it.
And maybe this will give
us some useful constraints
because this is
going to have to be
the same length
as this over here.
And so if we add 4x
plus 4x, we get 8x.
And then if we add 4y
plus 5y, we get 9y.
So these are going to have
to be equal to each other,
so that's an
interesting constraint.
So we have 13x plus 7y is going
to have to equal 8x plus 9y.
And we can simplify this.
If you subtract 8x from
both sides, you get 5x.
And if you subtract
7y from both sides,
you get 5x is equal to 2y.
Or you could say x
is equal to 2/5 y.
In order for these to
show up as integers,
we have to pick integers here.
But let's see if we have any
other interesting constraints
if we look at the bottom
and the top of this,
if this gives us any
more information.
So if we add 5x plus 3y plus
3x plus 2y plus 4x plus 4y,
this top dimension-- 5
plus 3 is 8 plus 4 is 12.
You get 12x.
And then you get
3y plus 2y plus 4y.
So that's 5 plus 4.
That's 9y.
Plus 9y.
That's this top dimension.
And if you go down here,
you have 8 plus 4 is 12x.
Let me do that same color.
And then you have 4
plus 5 is 9y, plus 9y.
So these actually ended up to
be the same in terms of x and ,
so they're not giving
us any more information,
no more constraints.
Obviously, 12x plus 9y is going
to be equal to 12x plus 9y.
So our only constraint
on this problem
is what we got by setting
this left-hand side
to this right-hand side, is that
x needs to be equal to 2/5 y.
So let's just pick
some numbers so
that we get nice
integers for x and y
and then we figure
out the perimeter.
We want to make sure
that the dimensions don't
have any common divisors.
So if we pick y to be 5, so
let's pick y to be equal to 5,
then looking at this
constraint, what is x?
Well, then x is 2.
It's going to be 2/5 times 5.
So then x is equal to 2.
So let's see what we get for
the dimensions of this rectangle
then.
So the height of
this rectangle is
going to be 13 times 2 is
26, plus 7 times 5 is 35.
So 26 plus 35 gets us what.
That gets us to 61.
This is equal to 61.
Did I do that right?
You see the 55 plus 6 is 61.
And then when you
look at its width,
you have 12x, which
is 24, plus 9y.
y Is 5, so plus 45.
24 plus 45 is what?
That is 69.
And 61 and 69 do not share any
common divisors other than 1.
So it looks like we're
done, or we're almost done.
We know the dimensions
of the rectangle.
It is a 61 by 69 rectangle.
And if you want its
actual perimeter,
you just add them all up.
So the perimeter
here is going to be--
we could have a drum roll now.
The perimeter is
going to be 61 plus 69
plus 61 plus 69, which is equal
to-- well, 61 plus 69 is 130.
That's another 130 right there.
130 plus 130 is 260.
So it actually wasn't
too bad of a problem
if we just started at the middle
and just built up from there,
built up the dimensions in terms
of the dimensions of these two
smallest squares.
And then we were able
to find the perimeter.