55rd Ukrainian National Mathematical Olympiad - Fourth Round

Let $C{C}_1$ be the altitude (Fig. 42). Since quadrilateral $A{B}_{1}H{C}_1$ is cyclic, we have $\angle A=180^{\circ }-\angle C_{1}H{B}_1=\angle B_{1}H{C}_1$. Since $\angle B_{1}H{C}_1=\angle NH{C}_1$, we obtain $\angle A=\angle CH{N}_1$.

Taking into account $\angle H{A}_{1}C=90^{\circ }$, $\angle HNC=90^{\circ }$, $\angle A{C}_{1}C=90^{\circ }$, we get that quadrilaterals $H{A}_{1}NC$ and $A{C}_1{A}_{1}C$ are cyclic. Hence $180^{\circ }-\angle C_1{A}_{1}C=\angle BAC=\angle CHN=\angle C{A}_{1}N$ and therefore points $C_1,A_1$ and $N$ are collinear.

Since $B{A}_{1}H{C}_1$ is cyclic, we have $\angle H{C}_1{A}_1=\angle HB{A}_1$. Note that $\angle A_{1}BK=\angle HB{A}_1$. Since $C,C_1,B,K$ are cyclic, it follows that $\angle CBK=\angle K{C}_{1}C$. This means that $C_1,A_1$ and $K$ are collinear. Notice that points $C_1$ and $A_1$ are different. This proves that $C_1,A_1,N$ and $K$ are collinear.