CPhill

(a + 2) / ( a + 5)  =  b /4

4 (a + 2)  = b ( a + 5)

4a  + 8 = ab + 5b

b ( a + 5)  = 4a + 8

b =  ( 4a + 8) / (a + 5)

b =  4 ( a + 2)  / (a + 5)

b will never be less than 5 when  a approaches negative infinity

b will never be more than 3 when a approaches positive infinity

a         b

-17     5

-11     6

-8       8

-7       10

-6       16   

-4       -8

-3       -2

-2        0

-1        1

 1        2

 7        3

Here :   https://web2.0calc.com/questions/square-and-triangles#r1

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Here :  https://web2.0calc.com/questions/sector_4

EF  parallel to QB

So triangle DEF is similar to  triangle DQB

 F is the midpoint of DB (the diagonal of the square)

So

DF / DB   =    EF / QB

(1/2)DB  / DB  =  EF / QB

(1/2) = EF / 1

EF =1/2

And EF is the diagonal of the square formed by the overlap

So.....the side of the squre  =   (1/2)(1/sqrt2)  =  1/ sqrt 8

Area of the quadrialeral   =  (1/sqrt8)^2  =   1/8

n = 2

Nothing shown....

y = x (x^2 +1)^(-1)              take the  derivative and set to 0

y'  = (x^2 + 1)^(-1)  - 2x*x (x^2+ 1)^(-2)   =0

( x^2 + 1)(-2)  ( x^2 +  1 - 2x^2)  = 0

-x^2 + 1 = 0

x^2   -1   =  0

(x - 1) ( x + 1)  = 0

x -1  = 0                                                         

 =1   y =  1 / (1^2 + 1) = 1/2  = max  y                       

 x + 1 =  0

x =  -1    y =   -1 / ((-1)^2 + 1)  =  -1/2 = min y

The sum of max + min  y's   =  0

Very crafty, proyaop !!!

Continuing with proyaop's answer

s^2   = (3 -sqrt 5)   / 2  

So  Pn =  sqrt [ (3 -sqrt 5) / 2) ]  = s

But  P1 * P2     is  =  s * s  =   s^2

So....the given sum should just be  10(s*s) = 10s^2   =   10  (3 - sqrt 5)  /  2 =    5 ( 3 - sqrt 5)  = 15 - 5sqrt5