Boundary-phobia

Notation: Let $x$ and $y$ be the usual horizontal and vertical Cartesian variables. We will make use of the polar coordinate system as well, using radius $r=\sqrt{x^2+y^2}$ and angle $\theta=\tan\left(\frac{y}{x}\right)$ .

Convention: Here, we use the convention that $\theta=0$ on the positive $x$ -axis and increases counter-clockwise. We center the square at the origin (the black dot is at x=0, y=0). Since the box is of side length 1, we are limited to the domains $-\frac{1}{2}\le x\le \frac{1}{2}$ and $-\frac{1}{2}\le y\le \frac{1}{2}$ .

Strategy: Let's start by figuring out how to be lazy. This problem has a lot of symmetry we can exploit; in fact, we can break it up into 8 identical wedges of angle $\pi/4$ . Knowing this, we can focus on finding the area of that single wedge and multiply our result by 8 at the end to get our final answer. Let's work with the wedge atop the positive x-axis, for which $0\le\theta\le\frac{\pi}{4}$ .

Now, how do we find the area of this wedge? We could integrate along $x$ and $y$ , bounded by the x-axis, the line x=y, and the curve from the top right corner to the $x$ -axis. This is unappealing; setting up an integral with a piecewise upper bound (part of which we do not yet know) would be ugly. Let's consider polar coordinates instead. We know from our earlier argument we need only integrate along $0\le\theta\le\frac{\pi}{4}$ . If we can find a function for $r$ (which seems inescapable either way), we can just plug it in and integrate over area.

Step 1 - Find constraints on $r$ : Pick a point on the outer edge of the wedge. By premise, the distance $r$ from the origin to this point is equal to the distance from this point to the nearest edge. Here the nearest edge is located at at $x=1/2$ , so the distance from the edge to the point is $1/2-x$ . Thus $r=1/2-x$ .

Step 2 - Construct $r(\theta)$ : Recall that $x=r\cos\theta$ ; plugging this into our equation from Step 1 gives $r=1/2-r\cos\theta$ , which we can rearrange to find $r(\theta)=\frac{1}{2}\frac{1}{1+\cos\theta}$ .

Step 3 - Integrate over the wedge: Recall that a line drawn from the origin to a point $(r,\theta)$ moved by a very small increment $\delta\theta$ covers area $\delta A=\pi r^2\frac{\delta \theta}{2\pi}$ . Integrating this over the wedge gives us our area:

$A_{wedge}=\int_0^{\pi/4}\pi r^2\frac{d \theta}{2\pi} = \frac{1}{2}\int_0^{\pi/4}\frac{1}{4}\frac{1}{(1+\cos\theta)^2}d\theta = \frac{1}{8}\frac{1}{3}\left(4\sqrt{2}-5\right)$ .

Mathematica was used to compute the integral.

Step 4 - Wrapping up: Multiplying by 8 gives us the area of the entire region: $8A_{wedge}=\frac{4\sqrt{2}-5}{3}$ . We can read out and sum the top three numbers to arrive at our solution: $4+2+5=11$ .

For any point in the triangle $OAB$ , the nearest point on the boundary of the square to that point lies on the line $AB$ Thus the curve $CED$ which marks the boundary of the desired region in the triangle $OAB$ is the locus of points in $OAB$ that are equidistant from $O$ and $AB$ ,

Thus the curve $CDE$ is the intersection with $OAB$ of the parabola that has focus $O$ and directrix $AB$ . Setting up a coordinate system with $O$ at the origin, this parabola has equation $y = x^2 - \tfrac14$ . If $\Delta$ is the area of the sector $ODE$ , then $\Delta \; = \; \int_0^{\frac12(\sqrt{2}-1)} \big[-x - (x^2 - \tfrac14)\big]\,dx \; = \; \tfrac{1}{24}(4\sqrt{2}-5)$ and so the desired area is $8\Delta \; = \; \tfrac13(4\sqrt{2}-5)$ making the answer $4+2+5 = \boxed{11}$ .

Let's be the origin $O$ in the $xy$ -plane the square center. Let be

$\displaystyle P=\bigg\{(x,y) \in \mathbb{R}^2 | -\frac{1}{2} \leq x \leq \frac{1}{2}, -\frac{1}{2} \leq y \leq \frac{1}{2} \bigg\}$

a generic point inside the square. The area $\textbf{A}$ of the blue region is

$\displaystyle \textbf{A}= (2AC)^2+8 \cdot Area(ABC)$

because the $(2AC)^2$ square is 'surrounded' by $8$ $ABC$ small triangoloids.

Now, let's find the locus of point such that the distance of $P$ from $O$ is equal to the distance of $P$ from the left vertical side of the unit square. That is

$\displaystyle \sqrt{x^2+y^2}=\frac{1}{2}-x \hspace{5pt} \Longrightarrow \hspace{5pt} y=\sqrt{\frac{1}{4}-x}$

considering only the positive solution respect to $y$ . This curve follows exactly the shape of the curved-edge $\overset{\frown}{BC}$ , although we need to find its boundaries. In order to do so, we intersect $y=\sqrt{\frac{1}{4}-x}$ with $y=x$ to find the $x$ -coordinate of $A$ and $C$ and with $y=0$ to find the $x$ -coordinate of $B$ .

$\displaystyle \begin{cases} y=\sqrt{\frac{1}{4}-x} \\ y=x \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} x=A_x=\frac{\sqrt{2}-1}{2}$

considering only the positive solution of $x$ .

$\displaystyle \begin{cases} y=\sqrt{\frac{1}{4}-x} \\ y=0 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} x=B_x=\frac{1}{4}$

Now we are able to evaluate $\textbf{A}$

$\displaystyle \textbf{A}=(2AC)^2+8\int_{A_x}^{B_x} \sqrt{\frac{1}{4}-x} dx=\bigg(2 \cdot \frac{\sqrt{2}-1}{2}\bigg)^2+8\int_{\frac{1}{4}}^{\frac{\sqrt{2}-1}{2}} \sqrt{\frac{1}{4}-x} dx={\frac{4\sqrt{2}-5}{3}}$

Eventually $\displaystyle A+B+C=2+5+4=\boxed{11}$