IMO Hk TST, 1997

The largest possible value of $BC$ is $3 + \sqrt{3}$.
Since $PD = PE$ and $AP$ bisects $\angle EAD$, the quadrilateral $AEPD$ is cyclic or it is a kite. The latter case is impossible since $\angle B > \angle C$. Thus, we have
$$180^{\circ} = \angle EAD + \angle DPE = A + \left(90^{\circ} + \frac{A}{2}\right) = 90^{\circ} + \frac{3A}{2}.$$
This implies $A = 60^{\circ}$.

As usual, let a, b, c and s be the lengths of $BC$, $CA$, $AB$ and the semiperimeter respectively. Firstly, since the inradius is 1, we have
$$s - a = \cot \frac{A}{2} = \sqrt{3}.$$

By the sine law, we have $\frac{a}{\sin 60^\circ} = \frac{b}{\sin B} = \frac{c}{\sin C}$, and so
$$\frac{2a}{\sqrt{3}} = \frac{b+c-a}{\sin B + \sin C - \sin 60^\circ}.$$
Note that
$$\sin B + \sin C = 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2} = \sqrt{3} \cos (60^\circ - C).$$
Therefore, we have
$$a = \frac{\sqrt{3}(s-a)}{\sin B + \sin C - \sin 60^\circ} = \frac{2\sqrt{3}}{2\cos(60^\circ - C) - 1} \le \frac{2\sqrt{3}}{2\cos(60^\circ - 30^\circ) - 1} = 3 + \sqrt{3}.$$
Equality holds when $C = 30^\circ$ and $B = 90^\circ$.