India — Team Selection Test, 2009

We first observe that $AI = r \cos\frac{A}{2} = 2r$. Let $T'$ be point on $AI$ such that $AT' = T'I = r$. Obviously $T'$ lies on the in-circle of $ABC$. We show that $T'$ also lies on the nine-point circle of $ABC$.
Let $D, E, F$ be the midpoints of $BC, CA, AB$ respectively. Note that $T'F$ is parallel to $IB$. Hence $\angle AFT' = \angle ABI = \frac{B}{2}$. We also observe that $DF$ is parallel to $AC$, so that $\angle BFD = \angle BAC = A$. It follows that
$$\angle T'FD = 180^\circ - (A + \frac{B}{2}).$$
Similarly, we obtain
$$\angle T'ED = 180^\circ - (A + \frac{C}{2}).$$
It is easy to check that
$$\angle T'FD + \angle T'ED = 180^\circ.$$
Hence $T', F, D, E$ are concyclic. Since $F, D, E$ are on the nine-point circle, it follows that $T'$ is also on the same circle.
However, there is only one point at which the in-circle and the nine-point circle touch each other (Feuerbach's theorem). It follows that $T = T'$. Hence: $AT = r$.