Solve it if you can

This is similar to Felipe's approach, but my calculation method differs enough that I thought it was worth posting.

Focus first on the portion of $S$ in the first quadrant lying below the line $y = x$ . (By symmetry this will represent one-eighth of the region $S$ .) Then the boundary line of $S$ for this restricted region will be where

$\sqrt{x^{2} + y^{2}} = 1 - x \Longrightarrow y = \sqrt{1 - 2x}.$

Now this curve intersects with $y = x$ when

$x^{2} = 1 - 2x \Longrightarrow x = \sqrt{2} - 1.$ .

So the region of $S$ in this restricted region will be the integral of $y = \sqrt{1 - 2x}$ from $x = \sqrt{2} - 1$ to $x = \frac{1}{2}$ , plus the area of the triangle with vertices $(0,0), (\sqrt{2} - 1, 0)$ and $(\sqrt{2} - 1, \sqrt{2} - 1)$ .

The triangle has area $(\frac{1}{2})(\sqrt{2} - 1)^{2} = (\frac{1}{2})(3 - 2\sqrt{2})$ , and the integral comes out to

$\displaystyle\int_{\sqrt{2} - 1}^{\frac{1}{2}} \sqrt{1 - 2x} dx = -\frac{1}{3}(1 - 2x)\sqrt{1 - 2x}$ ,

which when evaluated from $x = \sqrt{2} - 1$ to $x = \frac{1}{2}$ gives a value of

$\dfrac{1}{3}(3 - 2\sqrt{2})\sqrt{3 - 2\sqrt{2}}.$

Adding this to the area of the triangle and multiplying the sum by $8$ gives us a total area of

$\dfrac{4}{3}(3 - 2\sqrt{2})(3 + 2\sqrt{3 - 2\sqrt{2}}) = \boxed{0.8758}.$

Our region on the first quadrant is $\begin{cases} \sqrt{x²+y²} \leq 1-x \\ \sqrt{x²+y²} \leq 1-y \\ 0 \leq x \leq 1 \\ 0 \leq y \leq 1 \end{cases}$

Finding the contour of the region we get that $(\sqrt 2-1,\sqrt 2 + 1)$ is the symmetrical point. Using polar coordinates to integrate we have the angle from 0 to $\frac{\pi}{4}$ (symmetrical point) and the radius from 0 to $\frac{1}{1+\cos t}$ . This comes from: $\sqrt{x²+y²}=1-x \\ \Rightarrow r + r \cos t = 1 \\ \Rightarrow r = \frac{1}{1+\cos t}$

Using polar coordinates jacobian r we have

$\displaystyle \int_{0}^{\frac{\pi}{4}} \displaystyle \int_{0}^{\frac{1}{1+\cos t}} r \cdot \text{ dr} \cdot \text{ dt}$

Multipling the result by 8, because there are 8 equal parts, we get $0.875806 .\square$

I used sort of a brute force method. My approach was to parameterize the curve that partitions the square into points closer to the origin than any edge of the square. Because of symmetry, only the first eighth of the curve has to be parameterized. Then, Green’s theorem can be used to determine the area inside of the parameterized curve and multiplied by 8 to give the answer.

The partitioning curve can be parameterized from the unit circle, uc[t]={Cos[t],Sin[t]}; In the first 45 degrees of that unit circle, the length, d, of a ray from the origin to the point on the partitioning curve is equal to 1 - d Cos[t], the horizontal distance from the vertical edge, x=1. Rearranging gives d=1/(1+Cos[t]). Thus, the partitioning curve is: pc[t]:= (1/(1 + Cos[t]) ){Cos[t], Sin[t]}.

The line integral around the partitioning curve over an appropriate vector field, such as F[x, y] = {-y/2, x/2}, gives the area of the wedge (shown in red in the illustration). Multiplying by 8 gives the area inside the whole curve, which is 0.875806. An illustration is below. Since this is a geometry problem, I’m not sure if this approach is ok, but it is one way of looking at the problem.

The curves are just parabolas. Use the symmetry of the picture and its easy to see what we want to integrate. Try to understand why the answer is given by $4\left (\int_{1-\sqrt{2}}^{\sqrt{2}-1} -\dfrac{1}{2}x^2+\dfrac{1}{2} \mathrm{d}x-\left(\sqrt{2}-1\right)^2\right).$

The region $S$ has an obvious symmetry; it can be subdivided into eight congruent sectors. I will calculate the area of the sector between the line $x = 0$ and the line $y = x$ .

The boundary is defined by a parabola (each point is equidistant to a given point and a given line). It is easy to see that the parabola contains the point $(0, \tfrac12)$ and, with a little more work, that it passes through $(u, u)$ with $u = \sqrt 2 - 1$ .

Symmetry requires that the parabola is given by an equation of the form $y = c - ax^2$ . Substituting the known points we get $c = \tfrac12$ and $a = \tfrac 12$ . We must therefore find the area bounded by $x = 0$ , $y = x$ , and $y = \tfrac12(1 - x^2)$ . The corresponding integral is $A = \int_0^u \left( \tfrac12(1 - x^2) - x \right) \, dx \\ = \tfrac12 u - \tfrac12 u^2 - \tfrac16 u^3 \\ = -\tfrac56 + \tfrac23\sqrt 2 \approx 0.10948.$

Multiplying this by eight, we get $\text{area of}\ S = -6\tfrac23+5\tfrac13\sqrt 2 \approx \boxed{0.87581}.$