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# Challenging perimeter problem

Perimeter of rectangle covered by 9 non-overlapping squares. From 2000 American Invitational Math Exam. Created by Sal Khan.

## Want to join the conversation?

• Can't you simply use substitution when you have X=(2/5)y? instead os Sal's method?
• yes you could. as i did when i figured this one out. i started with x=1 and that left me with a fraction (not an interger). then i went on to x=2 and thus y=5. which just happens to be where sal started and then proved the solution. sal just thought ahead and made sure he had an interger result.
• Why does y=5 and x=2, shouldn't it be the other way around? Please explain.
• Let's say it was the other way around.

We have the equation x = 2/5 y
If we plug in y as 2, then we get x = 4/5, which is not an integer.
(1 vote)
• Why did Khan decide to plug in 5?
• if you have x= (2/5) y, then you want the smallest possible integer possible, which is when y=5. so x= (2/5) * 5 which means that the denominator and 5 cancels out to get 2 so x=2. You CAN plug in 10, or any multiple of five (which yields integers) but that would break the constraints set by the problem itself. Also, if you plug in 1 for y, then THAT would break the constraint as the sides of the squares are supposed to be integers.
• Is there any other way that you can solve this problem without guess and check?
(1 vote)
• Why did you pick 5 for y at
• X and Y are a ratio so they can be any values you want as long as they are constrained by the ratio. Picking whole numbers makes the rest of the problem easier, and at the end you get integer results so you're done.
• I'm confused i get X=2/5Y but when i worked the rest out i got an extra Y?
• i am totally lost how can you find all this with no measurements

i just dont get it
• Relative sizes. Sal compares the lengths of sides of squares, and since he knows that you can add two lines of lengths P and Q together by superimposing the endpoint of one of them on the endpoint of another, he can calculate the relative sizes of squares by considering their sides against those of his selected 'unit' squares.
(1 vote)
• can you try to explain it more simpole
• Wouldn't the most accurate answer be that the perimeter is equal to 52y?
Because (13x + 7y) + (12x + 9y) + (12x + 9y) + (8x + 9y) = 45x + 34y
And we know that x = 2/5y, so:
45(2/5y) + 34y
90/5y + 34y
18y + 34y
52y

Right? And the numbers he chose in the video to represent x and y were arbitrary aside from complying with the "positive integers with a GCF of 1" distinction given in the problem, correct?