Let there be two squares with side length 1 (in th...

We will show that if you start with two unit‐squares sharing a side and then rotate each one by 45° about its own center, the area where their interiors overlap is

$$\frac{3-2\sqrt{2}}{2}\,,$$

which may also be written as $\frac{3}{2}-\sqrt2$ (approximately 0.086).

In what follows we describe one way to arrive at this answer.

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The Setup

1. Initial placement before rotation.

   Suppose we have two squares of side length 1 that share a side. For convenience choose coordinates so that:

   • The “top square” has vertices

     $$A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1)$$

     so that its center is at $(0.5,0.5)$.

   • The “bottom square” has vertices

     $$A=(0,0),\quad B=(1,0),\quad E=(1,-1),\quad Z=(0,-1)$$

     so that its center is at $(0.5,-0.5)$.

2. Rotation by 45° about the centers.

   Each square is rotated by 45° (say counter‐clockwise) about its center. (It does not matter which direction, as long as both are rotated by the same angle.) A standard fact is that a square of side 1 rotated by 45° becomes a “diamond” whose four vertices lie on the lines $x=0.5\pm\frac{\sqrt{2}}{2}$ and $y=0.5\pm\frac{\sqrt{2}}{2}$ (for the top square) and similarly for the bottom square. In other words, one may show that a square of side 1 rotated 45° about its center has the equation

$$|x-x_c|+|y-y_c|\le \frac{\sqrt{2}}{2}\,,$$

where $(x_c,y_c)$ is its center.

Thus after rotation we have:

• The top square becomes $$S_{\rm top}=\{(x,y): |x-0.5|+|y-0.5|\le \sqrt{2}/2\}\,.$$
• The bottom square becomes $$S_{\rm bot}=\{(x,y): |x-0.5|+|y+0.5|\le \sqrt{2}/2\}\,.$$

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Finding the Intersection Area

We wish to find the area of

$$S = S_{\rm top}\cap S_{\rm bot} = \{(x,y) : |x-0.5|+|y-0.5|\le \sqrt{2}/2 \quad\text{and}\quad |x-0.5|+|y+0.5|\le \sqrt{2}/2\}\,.$$

A natural strategy is to “slice” the region horizontally (or vertically) and integrate.

1. Determine the vertical ( $y$–) range for the intersection.

   For a point $(x,y)$ to lie in $S_{\rm top}$ the inequality

   $$|x-0.5|+|y-0.5|\le \frac{\sqrt{2}}{2}$$

   must hold. In particular, even if we choose $x=0.5$ (so that $|x-0.5|=0$), we need

   $$|y-0.5|\le \frac{\sqrt{2}}{2}\,,$$

   i.e.

   $$y\in\left[0.5-\frac{\sqrt{2}}{2},\,0.5+\frac{\sqrt{2}}{2}\right]\,.$$

   For the bottom square $S_{\rm bot}$, setting $x=0.5$ we need

   $$|y+0.5|\le \frac{\sqrt{2}}{2}\,,$$

   i.e.

   $$y\in\left[-0.5-\frac{\sqrt{2}}{2},\,-0.5+\frac{\sqrt{2}}{2}\right]\,.$$

   Notice that

   $$0.5-\frac{\sqrt{2}}{2}=\frac{1-\sqrt{2}}{2}\quad\text{and}\quad -0.5+\frac{\sqrt{2}}{2}=\frac{\sqrt{2}-1}{2}\,.$$

   Hence the two vertical ranges intersect only when

   $$y\in\left[\frac{1-\sqrt{2}}{2},\,\frac{\sqrt{2}-1}{2}\right]\,.$$

   (Since $\sqrt{2}\approx1.414$, these endpoints are approximately $-0.207$ and $0.207$, so the total vertical extent of the intersection is about $0.414$.)

2. For a fixed $y$ in that range, find the allowed $x$.

   For a given $y$ in the common range, the two inequalities become

   $$|x-0.5|\le \frac{\sqrt{2}}{2}-|y-0.5|$$

   and

   $$|x-0.5|\le \frac{\sqrt{2}}{2}-|y+0.5|\,.$$

   When $y$ is between $\frac{1-\sqrt{2}}{2}$ and $\frac{\sqrt{2}-1}{2}$ the expressions simplify (since then $y-0.5\le0$ and $y+0.5\ge0$); one obtains

   $$|x-0.5|\le \frac{\sqrt{2}}{2}-\max\{0.5-y,\;y+0.5\}\,.$$

   In fact one may show that the most restrictive condition is

   $$|x-0.5|\le \frac{\sqrt{2}-1}{2}-|y|\,.$$

   (A short exercise in checking the two cases $y\ge0$ and $y\le0$ confirms this.)

3. Set up the integral.

   It is convenient to change variables by writing

   $$u=x-0.5\,.$$

   Then for a given $y$ the allowed $u$ satisfies

   $$|u|\le d-|y|\quad \text{with}\quad d=\frac{\sqrt{2}-1}{2}\,.$$

   Thus the horizontal “width” at height $y$ is

   $$2\Bigl(d-|y|\Bigr)\,.$$

   And the overall area of the intersection is

   $$A=\int_{y=(1-\sqrt{2})/2}^{(\sqrt{2}-1)/2}2\Bigl(d-|y|\Bigr)\,dy\,.$$

   Since the integrand is even in $y$ (and the limits are symmetric about 0) we may write

   $$A=2\cdot 2\int_{0}^{d}\Bigl(d-y\Bigr)\,dy=4\int_{0}^{d}(d-y)\,dy\,.$$

   Evaluating the integral gives

   $$\int_{0}^{d}(d-y)\,dy = \left[dy-\frac{y^2}{2}\right]_0^d = d^2-\frac{d^2}{2}=\frac{d^2}{2}\,.$$

   Thus,

   $$A=4\cdot\frac{d^2}{2}=2d^2\,.$$

   Now, since

   $$d=\frac{\sqrt{2}-1}{2}\quad\text{so that}\quad d^2=\frac{(\sqrt{2}-1)^2}{4}\,,$$

   we have

   $$A=2\cdot\frac{(\sqrt{2}-1)^2}{4}=\frac{(\sqrt{2}-1)^2}{2}\,.$$

   But

   $$(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\,,$$

   so finally

   $$A=\frac{3-2\sqrt{2}}{2}\,.$$

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Conclusion

The area common to the interiors of the two rotated squares is

$$\boxed{\frac{3-2\sqrt{2}}{2}\quad\text{or equivalently}\quad\frac{3}{2}-\sqrt{2}\,.}$$

This is the answer to the problem.