IMC 2014 Q10

\(A\), \(B\), \(C\), \(D\), \(E\) are the centres of the five semicircles. Note that \(AC\) joins the centres of two touching semicircles and therefore passes through the point of contact of the semicircles. So \(AC\) has length 2 + 2 = 4.
This also applies to all of the other sides of triangles \(ACD\) and \(BED\). Hence both triangles are equilateral. So each of the nine arcs which make up the perimeter of the shaded shape subtends an angle of 60° at the centre of a semicircle.
So the length of the perimeter of the shaded figure is 9 × \(\tfrac{1}{6}\) × 2 × π × 2 = 6π.