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Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Tomas of Malmsbury School, Mark of The British School of Manilla and Herbert of Sha Tin College, Hong Kong sent good solutions to this problem. Well done all of you.

$P$ 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.

The point $P$ is a fixed point on the smaller circle which moves as the small circle moves. The point $P_o$ 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 $P_1$ is the position of the point $P$.
circle rolling in circle
By showing that the lengths of the arcs $P_oC$ and $P_1C$ are equal, we shall prove that $P_1$ 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 $OP_o$.

Let $M$ be the centre of the small circle, then $MO = MP_1 = MC = r$ and the triangle $OMP_1$ is isosceles. Hence $$\angle MOP_1 = \angle MP_1O = \theta$$ $$\angle P_1MC = \pi - (\pi - 2\theta) = 2\theta.$$ Hence, using the formula "arc length = radius x angle at the centre of the circle": $$P_0C = (2r)(\theta) = 2r\theta$$ and $$P_1C = (r)(2\theta) = 2r\theta.$$ Hence $P$ must be at the point $P_1$ because the circle rolls without slipping, which shows that the locus of P is the diameter of the larger circle.