Congruent unit circles intersect in such a way that the center of each circle lies on the circumference of the other. Let $R$ be the region in which two circles overlap. Determine the perimeter of $R.$
1. Consider two congruent unit circles, each with radius \( r = 1 \), intersecting such that the center of each circle lies on the circumference of the other. This means the distance between the centers of the two circles is exactly 1 unit.

2. The overlapping region \( R \) is formed by the intersection of these two circles. To find the perimeter of \( R \), we need to understand the geometry of the intersection.

3. The centers of the circles and the points of intersection form an equilateral triangle with side length 1. This is because each side of the triangle is a radius of one of the circles.

4. The boundary of the overlapping region \( R \) consists of two circular arcs. Each arc is part of the circumference of one of the circles.

5. Since the triangle is equilateral, the angle subtended by each arc at the center of the circle is \( 60^\circ \) (or \( \frac{\pi}{3} \) radians).

6. The length of each arc is a fraction of the circumference of the circle. Specifically, each arc is \( \frac{1}{6} \) of the total circumference of a unit circle (since \( 60^\circ \) is \( \frac{1}{6} \) of \( 360^\circ \)).

7. The circumference of a unit circle is \( 2\pi \). Therefore, the length of each arc is:
   \[
   \frac{1}{6} \times 2\pi = \frac{\pi}{3}
   \]

8. Since there are two such arcs forming the boundary of \( R \), the total perimeter of \( R \) is:
   \[
   2 \times \frac{\pi}{3} = \frac{2\pi}{3}
   \]

The final answer is \(\boxed{\frac{2\pi}{3}}\).