smc

First, we must understand two important functions: $f(x)=b^x$ for $0<b<1$(decreasing exponential function), and $g(x)=x^k$ for $k>0$(increasing power function for positive $x$). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant. We will now examine the first few terms. Comparing $a_1$ and $a_2$, $0<a_1=(0.201{)}^1<(0.201{)}^{a_1}<(0.2011{)}^{a_1}=a_2<1\Rightarrow 0<a_1<a_2<1$. Therefore, $0<a_1<a_2<1$. Comparing $a_2$ and $a_3$, $0<a_3=(0.20101{)}^{a_2}<(0.20101{)}^{a_1}<(0.2011{)}^{a_1}=a_2<1\Rightarrow 0<a_3<a_2<1$. Comparing $a_1$ and $a_3$, $0<a_1=(0.201{)}^1<(0.201{)}^{a_2}<(0.20101{)}^{a_2}=a_3<1\Rightarrow 0<a_1<a_3<1$. Therefore, $0<a_1<a_3<a_2<1$. Comparing $a_3$ and $a_4$, $0<a_3=(0.20101{)}^{a_2}<(0.20101{)}^{a_3}<(0.201011{)}^{a_3}=a_4<1\Rightarrow 0<a_3<a_4<1$. Comparing $a_2$ and $a_4$, $0<a_4=(0.201011{)}^{a_3}<(0.201011{)}^{a_1}<(0.2011{)}^{a_1}=a_2<1\Rightarrow 0<a_4<a_2<1$. Therefore, $0<a_1<a_3<a_4<a_2<1$. Continuing in this manner, it is easy to see a pattern(see Note 1). Therefore, the only $k$ when $a_k=b_k$ is when $2(k-1006)=2011-k$. Solving gives $\boxed{1341}$. Note 1: We claim that $0<a_1<a_3<...<a_{2011}<a_{2010}<...<a_4<a_2<1$. We can use induction to prove this statement. (not necessary for AMC): Base Case: We have already shown the base case above, where $0<a_1<a_2<1$. Inductive Step: Rearranging in decreasing order gives $1>b_1=a_2>b_2=a_4>...>b_{1005}=a_{2010}>b_{1006}=a_{2011}>...>b_{2010}=a_3>b_{2011}=a_1>0$.