smc

For all nonnegative integers $n$, let $\angle C_n{A}_n{B}_n=x_n$, $\angle A_n{B}_n{C}_n=y_n$, and $\angle B_n{C}_n{A}_n=z_n$. Note that quadrilateral $A_0{B}_0{A}_1{B}_1$ is cyclic since $\angle A_0{A}_1{B}_0=\angle A_0{B}_1{B}_0=90^{\circ }$; thus, $\angle A_0{A}_1{B}_1=\angle A_0{B}_0{B}_1=90^{\circ }-x_0$. By a similar argument, $\angle A_0{A}_1{C}_1=\angle A_0{C}_0{C}_1=90^{\circ }-x_0$. Thus, $x_1=\angle A_0{A}_1{B}_1+\angle A_0{A}_1{C}_1=180^{\circ }-2x_0$. By a similar argument, $y_1=180^{\circ }-2y_0$ and $z_1=180^{\circ }-2z_0$. Therefore, for any positive integer $n$, we have $x_n=180^{\circ }-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to $n$ (and the coefficient of $x_0$ is $(-2{)}^n$). Hence, we let $x_n=p{q}^n+r+(-2{)}^n{x}_0$. We will solve for $p$, $q$, and $r$ by iterating the recurrence to obtain $x_1=180^{\circ }-2x_0$, $x_2=4x_0-180^{\circ }$, and $x_3=540-8x_0$. Letting $n=1,2,3$ respectively, we have $$\begin{array}{rl}pq+r& =180\\ p{q}^2+r& =-180\\ p{q}^3+r& =540\end{array}$$ Subtracting $(1)$ from $(3)$, we have $pq(q^2-1)=360$, and subtracting $(1)$ from $(2)$ gives $pq(q-1)=-360$. Dividing these two equations gives $q+1=-1$, so $q=-2$. Substituting back, we get $p=-60$ and $r=60$. We will now prove that for all positive integers $n$, $x_n=-60(-2{)}^n+60+(-2{)}^n{x}_0=(-2{)}^n(x_0-60)+60$ via induction. Clearly the base case of $n=1$ holds, so it is left to prove that $x_{n+1}=(-2{)}^{n+1}(x_0-60)+60$ assuming our inductive hypothesis holds for $n$. Using the recurrence relation, we have $$\begin{array}{rl}{x}_{n+1}& =180-2x_n\\ & =180-2((-2{)}^n(x_0-60)+60)\\ & =(-2{)}^{n+1}(x_0-60)+60\end{array}$$ Our induction is complete, so for all positive integers $n$, $x_n=(-2{)}^n(x_0-60)+60$. Identical equalities hold for $y_n$ and $z_n$. The problem asks for the smallest $n$ such that either $x_n$, $y_n$, or $z_n$ is greater than $90^{\circ }$. WLOG, let $x_0=60^{\circ }$, $y_0=59.999^{\circ }$, and $z_0=60.001^{\circ }$. Thus, $x_n=60^{\circ }$ for all $n$, $y_n=-(-2{)}^n(0.001)+60$, and $z_n=(-2{)}^n(0.001)+60$. Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^{\circ }$. Therefore, the answer is $\boxed{15}$.