BMO 2010 Shortlist

Let $BC=a$, $CA=b$, $AB=c$ and $2u=a+b+c$. Then $C{A}_1=C{B}_1=u-c$, $A{C}_1=u-a$, $B{C}_1=u-b$. We have $\sigma (ABC)=\frac{1}{2}ab\sin \angle C$ and $$\sigma (A_1{B}_{1}C)=\frac{1}{2}(u-c)(u-c)\sin \angle C=\frac{1}{8}(a+b-c{)}^2\sin \angle C.$$ Therefore $\sigma (ABC)=2\sigma (A_1{B}_{1}C)$, implies $(a+b-c{)}^2=2ab$. Let $A{A}_0\cap BC=A_2$ and $B{B}_0\cap AC=B_2$. The Law of sines in the triangles $AB{A}_0$ and $AC{A}_0$ gives $$\frac{A{A}_0}{B{A}_0}=\frac{\sin (\angle A+\angle B)}{\sin (\angle BA{A}_0)},\frac{A{A}_0}{C{A}_0}=\frac{\sin (\angle A+\angle C)}{\sin (\angle CA{A}_0)}.$$ Hence $$\frac{\sin (\angle BA{A}_0)}{\sin (\angle CA{A}_0)}=\frac{c}{b}$$ and $$\frac{B{A}_2}{C{A}_2}=\frac{AB\sin (\angle BA{A}_0)}{AC\sin (\angle CA{A}_0)}=\frac{c^2}{b^2}.$$ In particular, $$B{A}_2=\frac{a{c}^2}{b^2+c^2},\frac{C{B}_2}{A{B}_2}=\frac{a^2}{c^2}.$$

Let $C_{1}I\cap BC=D$. Then $$BD=\frac{u-b}{\cos \angle B^{\prime }}$$ $$A_{2}D=BD-B{A}_2=\frac{u-b}{\cos \angle B}-\frac{a{c}^2}{b^2+c^2}$$ Let $A{A}_0\cap B{B}_0=E$ and $A{A}_0\cap C_{1}I=F$. We want to show that $E=F$. It suffices to show that $$\frac{E{A}_2}{AE}=\frac{F{A}_2}{AF}$$ By Menelaus' theorem we have $$\frac{F{A}_2}{AF}=\frac{D{A}_2}{BD}\cdot \frac{B{C}_1}{A{C}_1}$$ and $$\frac{A_{2}E}{AE}=\frac{B{A}_2}{BC}\cdot \frac{C{B}_2}{B_{2}A}$$ Therefore $$\frac{E{A}_2}{AE}=\frac{F{A}_2}{AF}$$ if and only if $$\frac{B{A}_2}{BC}\cdot \frac{C{B}_2}{A{B}_2}=\frac{D{A}_2}{BD}\cdot \frac{B{C}_1}{A{C}_1}$$ Now substituting these lengths and using the Law of cosines $2ac\cos \angle B=a^2+c^2-b^2$, we find that $$\frac{E{A}_2}{AE}=\frac{F{A}_2}{AF}$$ if and only if $$\frac{a^2}{b^2+c^2}=\frac{a+c-b}{b+c-a}-\frac{(a^2+c^2-b^2)c}{(b^2+c^2)(b+c-a)}$$ This equality is equivalent to $(a-b)((a+b-c{)}^2-2ab)=0$, and we are done. $\square$