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# Just Rolling Round

##### Age 14 to 16Challenge Level

 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 $P$. 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.