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# Challenging perimeter problem

## Video transcript

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.