If the midpoints of a square are connected, a smaller square results.  Suppose the process of connecting midpoints of sides and drawing new squares is continued indefinitely. Write an infinite geometric series to represent the sum of the perimeters of all the squares.

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Let the sides of the square have length a. Then the perimeter of the outer square is 4a.

The side lengths of the square formed by the midpoints are `(asqrt(2))/2` , and the perimeter of this square is `2asqrt(2)=(4a)/sqrt(2)` .

The perimeter of the next square is 2a. (The side...

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Let the sides of the square have length a. Then the perimeter of the outer square is 4a.

The side lengths of the square formed by the midpoints are `(asqrt(2))/2` , and the perimeter of this square is `2asqrt(2)=(4a)/sqrt(2)` .

The perimeter of the next square is 2a. (The side lengths are `a/2` .) Note that `2a=(2asqrt(2))/2` .

So each perimeter can be found by dividing the previous perimeter by `sqrt(2)` . (Another way to see this is to realize that each square is similar to the previous square with the ratio of longer side to smaller side of `sqrt(2):1` . Since the perimeter is a linear measure, the ratio of perimeters is the same as the ratio of side lengths which is the scale factor.)

Thus the perimeters form a sequence `4a,2asqrt(2),2a,asqrt(2),a,...`

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The series can be written as `sum_0^(oo) 4a(1/sqrt(2))^n` where 4a is the perimeter of the outer square.

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Note that this is a geometric series with first term 4a and common ratio `r=1/sqrt(2)` . The sum of an infinite geometric series with |r|<1 is the first term divided by 1-r, so the sum will be `S=(4a)/(1-1/sqrt(2))=(4asqrt(2))/(sqrt(2)-1)` where 4a is the perimeter of the outer square.

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