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Current time:0:00Total duration:9:56

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 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 one so they're talking about the width and the height of the rectangle the reason why they're saying greatest common divisor one is they're saying that you can't that they don't have a common divisor that you can divide them both by to get kind of 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 five and 15 but over here our greatest common divisor isn't one we have they're both divisible by five over here so what you'd want to do is say no instead of five and fifteen they need to be one and three now you have the same ratio of sides but now the greatest common divisor is one you kind of have it in the simplified form or the most simplified form if you divide both of both of this this height and this width by five so that's why they're saying greatest common divisor one and then they say find the perimeter of the rectangle so let's see what we can do here so any card 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 the square right over here the center square and they did tell us that there are all squares so let's say that that square right over here has as 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 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 the width of these squares combined made the width of this larger square so what I'm going to do is I'm actually going to actually this might be an easier way to write 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 why I'm saying it's Y by Y this over here is going to be an X plus y times an X and X plus y is going to be each of its dimension so it's going to be X plus y height and X plus y width then this one over here this one over here well if this dimension if this dimension is x plus y and that 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 2 X plus y 2 X 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 this is 2x plus y and then we can go to 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 then to get the entire dimension of one side of the square so it's going to be 3x 3x plus 2y 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 the 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 5x plus 3y 5x plus 3y is going to be that entire length right over there and we can also go to this this side right over here where we have we have this length let me do that same color this length is 3 X plus 2 y this is X plus y and this is y so if you add 3x plus 2y plus X plus 1 Plus y-you get 4x plus what is that 4y right - I 3 y 4 y + 4 y + 4 y 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 7 X plus X is 8 X 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 it's dimensions are going to be the Y plus the 4x plus 4y so that's 4x plus 5y 4x plus 5y and then if we if we think about the dimensions if we think about the dimensions of this of this perimeter of this actual rectangle over here if we think about this 1 over if we think about its height if we think about its height right over there that's going to be it's going to be 5x plus 3y times a plus 8x plus 4y so 5 plus 8 is 13 so 13 X plus 3 plus 4 is 7 y plus 7 Y 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 + 4 X we get 8x we get 8x and then if we add 4y + 5 y we get 9 y + 9 y so these are going to have to be equal to each other so that's an interesting constraint so we have 13 x + 7 y + 7 Y is going to be is going to have to equal 8 X + 9 y 8x + 9 y 8x + 8 X + 9 Y and we can simplify this if you subtract 8x from both sides you get 5x and if you subtract 7 Y from both sides you get 5x is equal to 2y five X is equal to two Y or you could say X is X is equal to two over five Y in order for these to show up as integers we have to pick integers here but let's see if we have some of 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 + 3 y + 3 X + 2 y + 4 X + 4 y this top dimension 5 plus 3 is 8 plus 4 is 12 you get 12 X 12 X 12 X and then you get 3 y + 2 y + 4 y so that's 5 + 4 that's 9 y + 9 y that's this top dimension and if you go down here you have 8 plus 4 is 12 X let me do that same color 12 X 12 X and then your 4 plus 5 is 9 y + 9 y so these actually ended up to be the same in terms of x and y so they're not giving us any more information no more constraints obviously 12 X + 9 y is going to be equal to 12x + 9 y so our only constraint on this problem is what we got by setting this this left-hand side to this right-hand side is that X is needs to be equal to 2/5 Y so let's just pick some numbers so that we get nice integers 4 x + y and then we figure out the perimeter we want to make sure that the the dimensions have don't have any common divisors so if we pick if we pick Y to be 5 if we pick Y to be 5 let's pick Y to be equal to 5 then looking at this constraint what is X well then X is 2 is 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 let's see 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 - 55 + 6 is 61 and then when you look at the when you look at its width you have 12 X which is 24 plus 9 Y Y is 5 so plus 45 plus 45 24 plus 45 is what that is 69 is equal to 69 and 61 and 69 do not share any any common divisors other than one 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 you can have a drumroll now the perimeter is going to be 61 plus 69 plus 61 plus 69 which is equal to well 61 plus 69 is 130 130 that's another 130 right there 130 plus 130 is 200 260 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 these two smallest squares and then we're able to find that find the perimeter