Question

What is the blue area in the graphic at the following link:
drive<dot>google<dot>com/file/d/1w4WF6LfpJB34pkmeRnN5EJsOPuk-ZQUS/view

Answer 1

let O = (0,0)
Then the circle with center at O is x^2+y^2 = 1
The circle with diameter OB is (x - 1/2)^2 + y^2 = 1/4
Now let F be (x,y). Then
C is at (x,√(1-x^2))
F is at (x,√(1/4 - (x - 1/2)^2)
and since CF = 2x, we have
√(1-x^2) - √(1/4 - (x - 1/2)^2 = 2x
Now just solve for x^2, the area of the blue region

Answer 2

I am able to see your diagram
Here would be my approach.

Place the figure on the x-y grid with O at the centre.
Then B is (1,0) and A is (-1,0) , with the equation of the large circle is
x^2 + y^2 = 1
the equation with OB as diameter is
(x - 1/2)^2 + y^2 = 1/4
and the equation with AO as diameter is
(x + 1/2)^2 + y^2 = 1/4
Let the coordinates of D be (a,b) and the width of the square as c
then for the other points on the square we have:
C(a+c, b) , E(a, b-c) , and F(a+c, b-c)

Here comes the fun part:

Both D and C satisfy the equation x^2 + y^2 = 1
a^2 + b^2 = 1 and (a+c)^2 + b^2 = 1
subtract them:
(a+c)^2 - a^2 = 0
a^2 + 2ac + c^2 -a^2 = 0
2ac = -c^2
c = -2a <---- easy so far

E lies on (x+1/2)^2 + y^2 = 1/4
(a+1/2)^2 + (b-c)^2 = 1/4
(a+1/2)^2 + (b + 2a)^2 = 1/4
a^2 + a + 1/4 + b^2 + 4ab + 4a^2 = 1/4 , but a^2 + b^2 = 1
1 + a + 4ab + 4a^2 = 0

So we have
a^2 + b^2 = 1 and
1 + a + 4ab + 4a^2 = 0

Not looking forward to solving this, so let's go to Wolfram:
https://www.wolframalpha.com/input/?i=solve+1+%2B+a+%2B+4ab+%2B+4a%5E2+%3D+0+%2C+a%5E2+%2B+b%5E2+%3D+1

We get multiple answers, e.g. there would be a reflection of the whole
thing in the x-axis, etc

looking at the answers and the diagram, a plausible solution is
a = -0.262478 , b = 0.964938
which would make c = -2a = .524956

and the area is c^2 = appr .2755...


better check those calculations, chances of error are great.

Answer 3

Oops. I haven't checked my solution against Reiny's, but I must note that the area of my square is 4x^2, not x^2.

But I'm sure you caught that.

Answer 4

Thanks for your help, guys. This problem's been nagging me for a while. The area of the square was 4x², once you solved for x............