Saudi Arabian Mathematical Competitions, 2012

Assume that $P$ is the midpoint of segment $B_2C_2$; that is, $B_2P = C_2P$. We will show that $AB = AC$.

Set $\widehat{ABC} = B$, $\widehat{BCA} = C$, and $\widehat{CAB} = A$. Extend segment $AP$ through $P$ to meet segment $BC$ at $A_3$. Then it is clear $A_3$ is the midpoint of side $BC$ or $BA_3 = A_3C$. Let $B_3, C_3$ denote the intersections of pairs of segments $BB_1$ and $AC$, $CC_1$ and $AB$, respectively. By Ceva's theorem, we have
$$\frac{AC_3 \cdot BA_3 \cdot CB_3}{C_3B \cdot A_3C \cdot B_3A} = 1 \quad \text{or} \quad \frac{AC_3}{C_3B} = \frac{AB_3}{B_3C}$$

Note that $AB_3/CB_3$ is equal to the ratio between the areas of triangle $ABB_1$ and $CBB_1$; that is,
$$\begin{aligned} \frac{AB_3}{CB_3} &= \frac{[ABB_1]}{[CBB_1]} = \frac{AB \cdot AB_1 \cdot \sin \widehat{BAB_1}}{BC \cdot CB_1 \cdot \sin \widehat{BCB_1}} \\ &= \frac{AB}{BC} \cdot \frac{AB_1}{CB_1} \cdot \frac{\sin \widehat{BAB_1}}{\sin \widehat{BCB_1}}. \end{aligned}$$
Because line $AB_1$ is tangent to $\omega$, $\widehat{C_1AB} = \widehat{ACB} = C$. Hence $\sin \widehat{BAB_1} = \sin \widehat{C_1AB} = \sin C$ and
$$\frac{\sin \widehat{BAB_1}}{\sin \widehat{BCB_1}} = \frac{\sin C}{\sin 2C} = \frac{1}{\cos C}.$$
Because line $AB_1$ is tangent to $\omega$, we also have $\widehat{B_1AC} = \widehat{ABC} = B$. Thus, triangle $AB_1C$ is similar to triangle $BAC$. By the Law of sines, we have
$$\frac{AB}{BC} = \frac{\sin C}{\sin A} \quad \text{and} \quad \frac{AB_1}{CB_1} = \frac{\sin C}{\sin B}.$$
It follows that
$$\frac{AB_3}{CB_3} = \frac{AB}{BC} \cdot \frac{AB_1}{CB_1} \cdot \frac{\sin \widehat{BAB_1}}{\sin \widehat{BCB_1}} = \frac{\sin^2 C}{\sin A \sin B \cos C}.$$
In exactly the same way, we can show that
$$\frac{AC_3}{BC_3} = \frac{\sin^2 B}{\sin A \sin C \cos B}.$$
It follows that
$$\frac{\sin^2 C}{\sin A \sin B \cos C} = \frac{AB_3}{CB_3} = \frac{AC_3}{BC_3} = \frac{\sin^2 B}{\sin A \sin C \cos B}.$$
or
$$\sin^2 C \tan C = \sin^2 B \tan B.$$
Because we can have at most one of $\tan B$ and $\tan C$ being negative, we must have both of them positive, from which it follows that $B$ and $C$ are acute angles. For $0^\circ < \alpha < 90^\circ$, both $\sin \alpha$ and $\tan \alpha$ are monotonically increasing, thus we must have $B = C$, as desired.