Three equal circles of radius $r$ each pass through the centres of the other two.
What is the area of intersection that is common to all the three circles?
1. **Identify the centers and the desired region:**
   Let the centers of the three circles be \( A \), \( B \), and \( C \). Each circle has a radius \( r \) and passes through the centers of the other two circles. The desired region is the area common to all three circles.

2. **Calculate the area of the sectors:**
   Each sector formed by the intersection of two circles at their centers has a central angle of \( 60^\circ \) (since the centers form an equilateral triangle). The area of one such sector is:
   \[
   \text{Area of one sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{\pi r^2}{6}
   \]
   Since there are three such sectors, the total area of these sectors is:
   \[
   3 \times \frac{\pi r^2}{6} = \frac{\pi r^2}{2}
   \]

3. **Calculate the area of the equilateral triangle:**
   The equilateral triangle \( \triangle ABC \) has a side length equal to the radius \( r \) of the circles. The area of an equilateral triangle with side length \( r \) is:
   \[
   \text{Area of } \triangle ABC = \frac{\sqrt{3}}{4} r^2
   \]
   Since this triangle is counted three times in the sum of the sectors, we need to subtract it twice to get the correct area of the desired region.

4. **Combine the areas:**
   The area of the region common to all three circles is:
   \[
   \text{Area of the desired region} = 3 \times \text{Area of one sector} - 2 \times \text{Area of } \triangle ABC
   \]
   Substituting the values calculated:
   \[
   \text{Area of the desired region} = \frac{\pi r^2}{2} - 2 \times \frac{\sqrt{3}}{4} r^2 = \frac{\pi r^2}{2} - \frac{\sqrt{3}}{2} r^2
   \]
   Simplifying, we get:
   \[
   \text{Area of the desired region} = r^2 \left( \frac{\pi - \sqrt{3}}{2} \right)
   \]

The final answer is \(\boxed{r^2 \left( \frac{\pi - \sqrt{3}}{2} \right)}\).