2015 Ninth STARS OF MATHEMATICS Competition

Let ${\gamma }_i$ cross the lines $A_i{C}_{i+1}$ and $A_i{C}_{i+2}$ again at $X_{i+2}$ and $Y_{i+1}$, respectively.

Since the homothety centred at $A_i$, transforming ${\gamma }_i$ into $\gamma$, sends $X_{i+2}$ to $C_{i+1}$ and $Y_{i+1}$ to $C_{i+2}$, the lines $X_{i+2}{Y}_{i+1}$ and $C_{i+1}{C}_{i+2}$ are parallel, so $A_i{C}_{i+1}/A_i{C}_{i+2}=X_{i+2}{C}_{i+1}/Y_{i+1}{C}_{i+2}$.

On the other hand, since $C_i$ has equal powers relative to ${\gamma }_{i+1}$ and ${\gamma }_{i+2}$, $C_i{X}_{i+1}\cdot C_i{A}_{i+2}=C_i{B}_i^2=C_i{Y}_{i+2}\cdot C_i{A}_{i+1}$, it follows that $C_i{X}_{i+1}=C_i{B}_i^2/C_i{A}_{i+2}$ and $C_i{Y}_{i+2}=C_i{B}_i^2/C_i{A}_{i+1}$.

Hence $A_i{C}_{i+1}/A_i{C}_{i+2}=(B_{i+1}{C}_{i+1}/B_{i+2}{C}_{i+2}{)}^2(A_i{C}_{i+2}/A_i{C}_{i+1})$, so $A_i{C}_{i+1}/A_i{C}_{i+2}=B_{i+1}{C}_{i+1}/B_{i+2}{C}_{i+2}$, and consequently ${\prod }_{i=0}^2(A_i{C}_{i+1}/A_i{C}_{i+2})={\prod }_{i=0}^2(B_{i+1}{C}_{i+1}/B_{i+2}{C}_{i+2})=1$. Since the sines of the angles $C_i{A}_i{C}_j$ are proportional to the lengths of the corresponding chords $A_i{C}_j$, the conclusion follows by Ceva's theorem in trigonometric form in the triangle $C_0{C}_1{C}_2$.





Alternative Solution: Let the tangents to $\gamma$ at $A_i$ and $A_{i+1}$ meet at $D_{i+2}$ (fig 1), and notice that the latter has equal powers relative to ${\gamma }_i$ and ${\gamma }_{i+1}$ to deduce that it lies on their radical axis, $B_{i+2}{C}_{i+2}$. Consequently, the lines $C_i{D}_i$ are concurrent at the radical center of the ${\gamma }_i$, and the conclusion follows by the lemma below (see Figure 2).

Lemma. Let $P_0{P}_1{P}_2$ be a triangle, and let $T_i$ be the touchpoint of the side $P_{i+1}{P}_{i+2}$ and the incircle $\gamma$ of the triangle $P_0{P}_1{P}_2$. Let further $Z_i$ be a point on the arc $T_{i+1}{T}_{i+2}$ of $\gamma$ not containing $T_i$. Then the lines $T_i{Z}_i$ are concurrent if and only if the lines $P_i{Z}_i$ are concurrent.