IMC 2014 Q25

Let each side of the octagon have length \(x\). The octagon may be divided into eight triangles by joining the centre of the circle to the vertices of the octagon. One such triangle is shown. Each of these triangles has one side of length 1 (the radius of the smaller circle), one side of length 2 (the radius of the larger circle) and one side of length \(x\). So all eight triangles are congruent. Therefore \(\angle AOB = 360^\circ \div 8 = 45^\circ\).
Let \(D\) be the foot of the perpendicular from \(B\) to \(AO\). Then triangle \(BDO\) is an isosceles right-angled triangle.
Let \(OD = DB = y\). Applying Pythagoras' Theorem to triangle \(BDO\):
\(y^2 + y^2 = 1\). So \(y = \tfrac{1}{\sqrt{2}}\).
Applying Pythagoras' Theorem to triangle \(ADB\):

So the length of the perimeter is \(8x = 8\sqrt{5 - 2\sqrt{2}}\).
(Note that the length of AB may also be found by applying the Cosine Rule to triangle OAB.)