## Problem A2

Two circles are concentric. A chord AC of the outer circle touches the inner circle at Q. P is the midpoint of AQ. A line through A intersects the inner circle at $\mathrm{R}$ and $\mathrm{S}$. The perpendicular bisectors of PR and CS meet at $\mathrm{T}$ on the line $\mathrm{AC}$. What is the ratio AT/TC?


## Solution

We have $\mathrm{AR} \cdot \mathrm{AS}=\mathrm{AQ}^{2}=\mathrm{AQ} / 22 \mathrm{AQ}=\mathrm{AP} \cdot \mathrm{AC}$, so $\mathrm{ARP}$ and $\mathrm{ACS}$ are similar, so $\angle \mathrm{ACS}=\angle \mathrm{ARP}$, so $\mathrm{PRSC}$ is cyclic. Hence T must be the center of its circumcircle and must also lie on the perpendicular bisector of CP. Hence it must be the midpoint of CP. So CT $=3 / 8 \mathrm{CA}$ and hence $\mathrm{AT} / \mathrm{TC}=5 / 3$.

However, that is not quite all. If CS is parallel to $\mathrm{PR}$, then their perpendicular bisectors coincide and both pass through A. So one could also regard A as a possible position for T.

