55rd Ukrainian National Mathematical Olympiad - Fourth Round

Consider translation by vector $\overrightarrow{A_{1}A}$ (Fig. 40). Thus $A{A}_{1}C{C}_3$ is a parallelogram, $C{C}_1{C}_2{C}_3$ is a rhombus. Then $\angle C_{1}C{C}_3=\angle C_{1}FA=\angle CF{A}_1=2\alpha$, therefore $$\angle C_{1}A{C}_3=180^{\circ }-\angle ABC=180^{\circ }-\frac{1}{2}\angle CF{A}_1=180^{\circ }-\alpha .$$ If we build a circle where the point $C$ is its centre, then the central angle $\angle C_{1}C{C}_3=2\alpha$, thus the inscribed angle, that subtends the arc $C_1{C}_3$, equals $\alpha$, therefore the point $A$ belongs to the circle. Therefore, $AC=C{C}_3=A{A}_1$, this completes the proof.

Alternative solution. Let $l$ be the bisector of the vertical angles $A_{1}FC$ and $C_{1}FA$. Let the line intersect the segments $A{C}_1$ and $C{A}_1$ at the points $P$ and $Q$ respectively (Fig. 41). Fig. 40 Since from the conditions of the problem $\angle CF{A}_1=2\angle ABC$, then $\angle QF{A}_1=\angle PB{A}_1$. That means that quadrilateral $PB{A}_{1}F$ is inscribed. Similarly the quadrilateral $QB{C}_{1}F$ is also inscribed. $A{A}_1=AC$ if and only if $\angle A{A}_{1}C=\angle AC{A}_1$. Since quadrilateral $PB{A}_{1}F$ is inscribed, then $\angle A{A}_{1}C=\angle QPB$. It is enough to prove that $\angle ACB=\angle BPQ$. The equality of angles is equivalent to the congruence of the triangles $ABC$ and $QBP$, that is equivalent to $\frac{AB}{BC}=\frac{QB}{BP}$. We will use the fact that the quadrilateral $PB{A}_{1}F$ is inscribed. From secant-secant theorem $AP\cdot AB=AF\cdot A{A}_1$. Similarly, $CQ\cdot CB=CF\cdot C{C}_1$. Thus $$\frac{AF}{CF}=\frac{AP}{CQ}\cdot \frac{AB}{BC}\text{ (using }A{A}_1=C{C}_1).$$ Now it is enough to prove that $\frac{AF}{CF}=\frac{AP}{CQ}\cdot \frac{QB}{BP}$.

$\frac{QB}{BP}=\frac{AF}{PA}\cdot \frac{CQ}{CF}$. Let $\angle ABC=\angle QFC=\angle PFA=\beta$, $\angle FQB=\angle F{C}_{1}A=\phi$, $\angle FPB=\angle F{A}_{1}C=\psi$. From the sine rule for the triangle $QBP$ $\frac{QB}{BP}=\frac{\sin \psi }{\sin \phi }$. From the sine rule for the triangle $FQC$ $\frac{CQ}{CF}=\frac{\sin \beta }{\sin (180^{\circ }-\phi )}=\frac{\sin \beta }{\sin \phi }$. From the sine rule for the triangle $FPA$ $\frac{AF}{PA}=\frac{\sin (180^{\circ }-\psi )}{\sin \beta }=\frac{\sin \psi }{\sin \beta }$. Therefore, $\frac{QB}{BP}=\frac{AF}{PA}\cdot \frac{CQ}{CF}\iff \frac{\sin \psi }{\sin \phi }=\frac{\sin \psi }{\sin \beta }\cdot \frac{\sin \beta }{\sin \phi }$, and the last equality, obviously, is correct.