Saudi Arabia Mathematical Competitions 2012

Assume that $P$ is the midpoint of segment $B_2{C}_2$; that is, $B_{2}P=C_{2}P$. 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 $B{A}_3=A_{3}C$. Let $B_3,C_3$ denote the intersections of pairs of segments $B{B}_1$ and $AC$, $C{C}_1$ and $AB$, respectively. By Ceva's theorem, we have $$\frac{A{C}_3\cdot B{A}_3\cdot C{B}_3}{C_{3}B\cdot A_{3}C\cdot B_{3}A}=1\text{or}\frac{A{C}_3}{C_{3}B}=\frac{A{B}_3}{B_{3}C}$$

Note that $A{B}_3/C{B}_3$ is equal to the ratio between the areas of triangle $AB{B}_1$ and $CB{B}_1$; that is, $$\begin{array}{rl}\frac{A{B}_3}{C{B}_3}& =\frac{[AB{B}_1]}{[CB{B}_1]}=\frac{AB\cdot A{B}_1\cdot \sin \widehat{BA{B}_1}}{BC\cdot C{B}_1\cdot \sin \widehat{BC{B}_1}}\\ & =\frac{AB}{BC}\cdot \frac{A{B}_1}{C{B}_1}\cdot \frac{\sin \widehat{BA{B}_1}}{\sin \widehat{BC{B}_1}}.\end{array}$$ Because line $A{B}_1$ is tangent to $\omega$, $\widehat{C_{1}AB}=\widehat{ACB}=C$. Hence $\sin \widehat{BA{B}_1}=\sin \widehat{C_{1}AB}=\sin C$ and $$\frac{\sin \widehat{BA{B}_1}}{\sin \widehat{BC{B}_1}}=\frac{\sin C}{\sin 2C}=\frac{1}{\cos C}.$$ Because line $A{B}_1$ is tangent to $\omega$, we also have $\widehat{B_{1}AC}=\widehat{ABC}=B$. Thus, triangle $A{B}_{1}C$ is similar to triangle $BAC$. By the Law of sines, we have $$\frac{AB}{BC}=\frac{\sin C}{\sin A}\text{and}\frac{A{B}_1}{C{B}_1}=\frac{\sin C}{\sin B}.$$ It follows that $$\frac{A{B}_3}{C{B}_3}=\frac{AB}{BC}\cdot \frac{A{B}_1}{C{B}_1}\cdot \frac{\sin \widehat{BA{B}_1}}{\sin \widehat{BC{B}_1}}=\frac{{\sin }^{2}C}{\sin A\sin B\cos C}.$$ In exactly the same way, we can show that $$\frac{A{C}_3}{B{C}_3}=\frac{{\sin }^{2}B}{\sin A\sin C\cos B}.$$ It follows that $$\frac{{\sin }^{2}C}{\sin A\sin B\cos C}=\frac{A{B}_3}{C{B}_3}=\frac{A{C}_3}{B{C}_3}=\frac{{\sin }^{2}B}{\sin A\sin C\cos B}.$$ 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.