P is a point on the circumference of a circle radius r, which touches another circle radius 2r, on the inside. The smaller circle rolls, without slipping, around the inner circumference of the larger circle.

Herbert Pang, age 18, Sha Tin College, Hong Kong sent a very good solution to this problem. Well done Herbert. The point P is a fixed point on the smaller circle. The point Po is the position of P when P is at the point of contact between the two circles. Consider the general position where the point of contact is the point C but here we do not assume that P1 is the position of the point P.

By showing that the lengths of the arcs PoC and P1C are equal, we shall prove that P1 must be the position of the point P when the point of contact is at C. Hence we shall show that P must always lie on the diameter of the large circle through OPo.

Let M be the centre of the small circle, then MO = MP1 = MC = r and the triangle OMP1 is isosceles. Hence
∠MOP1 = ∠MP1O = θ

∠P1MC = π− (π− 2θ) = 2θ.
Hence, using the formula " arc length = radius x angle at the centre of the circle":
P0C = (2r)(θ) = 2rθ
and
P1C = (r)(2θ) = 2rθ.
Hence P must be at the point P1 because the circle rolls without slipping, which shows that the locus of P is the diameter of the larger circle.