# 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...

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 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.

First let's start by finding the perimeter of a single square in terms of Length(which will be denoted as L)

Perimeter(denoted as **P**), length of first square(denoted as **L**)

to find the perimeter of the first square we simply add the four sides together

(1)** `P_1` = `L_1 ` + `L_1 ` + `L_1 ` + `L_1 ` **or `P_1 ` ** = **4**`L_1 ` **

next, let's find and connect the midpoints of our first square(see image), to obtain a new square which has vertices at the midpoints of our first square. More importantly, we need to find the perimeter of this square. To find the perimeter of our second square we will use the equation.

(2) `P_(2)` = 4`L_2 `

**` ` `L_2 ` **can be found by noticing that it is the hypotenuse of a **right triangle** with legs of length (`L_1 `/2). Using Pythagorean's theorem we obtain,

`(3) L_2 ` = `sqrt((L_1/2)^2 + (L_1/2)^2)`

`L_2 ` = `sqrt((L_1^2/4) + (L_1^2/4)) `

`L_2 ` = `sqrt(2L_1^2/4) `

`L_2 ` = `sqrt(L_1^2/2) `

`L_2 ` = `L_1/sqrt(2) `

so to find the total perimeter(P_t) we currently have

`(4)P_t ` = 4`L_1 ` + 4`(L_1/sqrt(2)) `

*notice how we rewrite L_2 in terms of L_1 so our equation is just one variable*

Lets repeat this process for a third square, which has vertices at the midpoints of the second square. To find the perimeter of the third square, we need the length of one of its sides. Once again we notice that the side of the third square is the hypotenuse of a **right triangle** with legs `L_2/2 `

using Pythagorean's theorem we find

`(5)L_3 ` = `sqrt((L_2/2)^2+(L_2/2)^2) `

*this is the same form of the theorem we used to find `L_2 ` * reducing to...

`L_3 ` = `L_2/sqrt(2) `

now lets rewrite our total perimeter to include the perimeter of our third square

` `

`(4)P_t ` = 4`L_1 ` + 4`L_1/sqrt(2) ` + 4 `L_2/sqrt(2) `

lets rewrite in terms of just L_1

`P_t ` = 4`L_1 ` + 4`L_1/sqrt(2) ` +4`L_1/(sqrt(2)sqrt(2)) `

`P_t ` = 4`L_1 ` +4`L_1/sqrt(2) ` +4`L_1/(sqrt(2))^2 `

notice how introducing the perimeter of a new square formed by our midpoint method will have a length that can be written in terms of L_1

`L_(n+1) ` = `L_1/(sqrt(2)^n) ` 0 `<=` n `<= ``oo`

where n is the number of times we have used our midpoint method to find a new square. To find the perimeter of any square

`P_(n+1)` = `4L_(n+1) `

knowing how to write the perimeter of any square, we can simply right a summation that contains the perimeters of an infinite amount of squares.

`P_t = ` `4 sum_(n=0)^oo L_1/sqrt(2)^n`

try it out for yourself! Instead of letting n go to infinity, only allow n to go up to 2. You will notice that the perimeter of that summation will equal the total perimeter that we found for three squares!