, 2010

Let $BC = a$, $CA = b$, $AB = c$ and $2u = a+b+c$. Then $CA_1 = CB_1 = u-c$, $AC_1 = u-a$, $BC_1 = u-b$.
We have $\sigma(ABC) = \frac{1}{2}ab \sin \angle C$ and
$$\sigma(A_1B_1C) = \frac{1}{2}(u-c)(u-c) \sin \angle C = \frac{1}{8}(a+b-c)^2 \sin \angle C.$$
Therefore $\sigma(ABC) = 2\sigma(A_1B_1C)$, implies $(a+b-c)^2 = 2ab$.
Let $AA_0 \cap BC = A_2$ and $BB_0 \cap AC = B_2$. The Law of sines in the triangles $ABA_0$ and $ACA_0$ gives
$$\frac{AA_0}{BA_0} = \frac{\sin(\angle A + \angle B)}{\sin(\angle BAA_0)}, \quad \frac{AA_0}{CA_0} = \frac{\sin(\angle A + \angle C)}{\sin(\angle CAA_0)}.$$
Hence
$$\frac{\sin(\angle BAA_0)}{\sin(\angle CAA_0)} = \frac{c}{b}$$
and
$$\frac{BA_2}{CA_2} = \frac{AB \sin(\angle BAA_0)}{AC \sin(\angle CAA_0)} = \frac{c^2}{b^2}.$$
In particular,
$$BA_2 = \frac{ac^2}{b^2 + c^2}, \quad \frac{CB_2}{AB_2} = \frac{a^2}{c^2}.$$

Let $C_1I \cap BC = D$. Then
$$BD = \frac{u-b}{\cos \angle B'}$$
$$A_2D = BD - BA_2 = \frac{u-b}{\cos \angle B} - \frac{ac^2}{b^2+c^2}$$
Let $AA_0 \cap BB_0 = E$ and $AA_0 \cap C_1I = F$. We want to show that $E = F$. It suffices to show that
$$\frac{EA_2}{AE} = \frac{FA_2}{AF}$$
By Menelaus' theorem we have
$$\frac{FA_2}{AF} = \frac{DA_2}{BD} \cdot \frac{BC_1}{AC_1}$$
and
$$\frac{A_2E}{AE} = \frac{BA_2}{BC} \cdot \frac{CB_2}{B_2A}$$
Therefore
$$\frac{EA_2}{AE} = \frac{FA_2}{AF}$$
if and only if
$$\frac{BA_2}{BC} \cdot \frac{CB_2}{AB_2} = \frac{DA_2}{BD} \cdot \frac{BC_1}{AC_1}$$
Now substituting these lengths and using the Law of cosines $2ac \cos \angle B = a^2 + c^2 - b^2$, we find that
$$\frac{EA_2}{AE} = \frac{FA_2}{AF}$$
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. $\square$