Ukrajina 2008

Let $H$ be the orthocenter of $△ABC$, lines $A{A}_1$ and $BC$ intersect at point $A_3$. Then $△A_{3}B{A}_1\sim △AB{A}_3$, $△A_1{A}_{3}C\sim △AC{A}_3\Rightarrow C{A}_3^2=A_3{A}_1\cdot A{A}_3=B{A}_3^2\Rightarrow C{A}_3=B{A}_3$ (fig. 6).

Since $\angle B{A}_{1}C=180^{\circ }-\angle BAC$, by means of the triangles $B{B}_{0}A$ and $C{C}_{0}A$ we can easily find that $\angle BHC=180^{\circ }-\angle BAC$, i.e. $\angle B{A}_{1}C=\angle BHC$. Therefore,



Fig. 6

$B,H,A_1,C$ are circular points and $\angle H{A}_{1}C=180^{\circ }-\angle HBC=90^{\circ }+\gamma$. Since $\angle A_3{A}_{1}C=\angle A_{1}AC+\angle A_{1}CA=\gamma \Rightarrow \angle H{A}_1{A}_3=\angle H{A}_{1}C-\angle A_3{A}_{1}C=90^{\circ }-\gamma +\gamma =90^{\circ }=\angle H{A}_{1}A$. Hence $A,C_0,H,A_1$ are circular points. Therefore, $\angle A{A}_1{C}_0=\angle AH{C}_0=180^{\circ }-\angle AHC=180^{\circ }-A_{0}H{C}_0=\beta$ since points $B,C_0,H,A_1$ are also circular. Therefore, $△A{A}_1{C}_0\sim △AB{A}_3$.

If we find point $C_3=C{C}_1\cap AB$ similarly to the previous constructions, it will be a midpoint of the side $AB$. Therefore, $C_3$ and $A_2$ are midpoints of the respective sides of the similar triangles. Thus, $\angle C_0{A}_2{A}_1=\angle A_3{C}_{3}B=\alpha$. But $\angle C_0{A}_0{A}_3=90^{\circ }+\angle A{A}_0{C}_0=90^{\circ }+\angle C_{0}BH=180^{\circ }-\alpha$ and therefore $A_0,C_0,A_2,A_3$ are circular points. This implies that point $A_2$ belongs to the circle of nine points of $△ABC$. Similarly points $B_2$ and $C_2$ also belong to this circle.

Then we obtain that $\angle C_0{A}_1{A}_2=\beta$, $\angle C_0{A}_2{A}_1=\alpha$, therefore $△A_1{A}_2{C}_0\sim △ABC\Rightarrow \frac{C_0{A}_2}{AC}=\frac{A_2{A}_1}{AB}$. Similarly $\frac{B_0{A}_2}{AB}=\frac{A_2{A}_1}{AC}\Rightarrow \frac{C_0{A}_2}{B_0{A}_2}=\frac{A{C}^2}{A{B}^2}=\frac{\sin \angle C_0{A}_0{A}_2}{\sin \angle B_0{A}_0{A}_2}$ since all the points $A_0,B_0,C_0,$ and $A_2$ are on one circle. Similarly $\frac{B_0{C}_2}{A_0{C}_2}=\frac{B{C}^2}{A{C}^2}=\frac{\sin \angle B_0{C}_0{C}_2}{\sin \angle A_0{C}_0{C}_2}$ and $\frac{A_0{B}_2}{C_0{B}_2}=\frac{A{B}^2}{C{B}^2}=\frac{\sin \angle A_0{B}_0{B}_2}{\sin \angle C_0{B}_0{B}_2}$.

Let's multiply the last equalities $\frac{\sin \angle C_0{A}_0{A}_2}{\sin \angle B_0{A}_0{A}_2}\cdot \frac{\sin \angle B_0{C}_0{C}_2}{\sin \angle A_0{C}_0{C}_2}\cdot \frac{\sin \angle A_0{B}_0{B}_2}{\sin \angle C_0{B}_0{B}_2}=1$. The Ceva theorem implies that lines $A_0{A}_2$, $B_0{B}_2$, and $C_0{C}_2$ intersect at one point.