An Elegant Intersection of Circles

First, notice that the four line segments form a square ( $45^\circ, 45^\circ, 90^\circ$ triangles). We know that the line joining the intersection points of 2 circles is perpendicular to the line joining their centers. So we can say that it is a diagonal of the square.

Therefore half of red region = (Area of a quarter circle) - ( $\frac {1}{2} \cdot$ area of the square) $= \frac {\pi(1)^{2}}{4} - \frac {1}{2}= \frac {\pi}{4} - \frac {1}{2}$ .

So, area of red region = $2( \frac {\pi}{4} - \frac {1}{2}) = \frac {\pi}{2} - 1$ .

The area of the triangle formed by the circle centers and the upper intersection of the circles is (1/2) sqrt(2) [sqrt(2)/2] = 1/2. Define A as the area formed by the difference between this triangle and the sector forned by either circle inside the triangle. The sector area is pi 45/360 = pi/8, so A has area of 1/2 - pi/8. The area of the upper half of the figure of interest is the area of the triangle - 2 (area ofA) = 1/2 - 2(1/2 - pi/8) = pi/4 - 1/2. The area of the figure of interest is double this, or pi/2 - 1.

Call the centers of the circles $A,B$ and the intersections of the circles $P,Q$ . Let $O$ be the midpoint of $AB$ , which is also the midpoint of $PQ$ .

It is clear that $OA = OB = \tfrac12\sqrt 2$ . With $AP = 1$ and Pythagoras in $\triangle OAP$ we find $OP = \tfrac12\sqrt 2$ . Likewise $OQ = \tfrac12\sqrt 2$ . We see that $\angle OAP = \angle OAQ = 45^\circ$ , making $\angle PAQ = 90^\circ$ .

Arc $\stackrel{\frown}{PQ}$ of circle $A$ and segment $PQ$ enclose half of the red area. It can be viewed as the difference of sector $A\stackrel{\frown}{PQ}$ and triangle $APQ$ .

Sector $A\stackrel{\frown}{PQ}$ is precisely one-quarter of circle $\odot A$ , so its area is $\tfrac14\pi$ . Triangle $APQ$ has basis $PQ = \sqrt 2$ and height $AO = \tfrac12\sqrt 2$ , so its area is $\tfrac12\cdot\sqrt 2\cdot \tfrac12\sqrt 2 = \tfrac12$ . Thus the area of half of the red area is $\tfrac14\pi - \tfrac12$ ; the total red area is twice as much, i.e. $\boxed{\tfrac12\pi - 1}$ .