line from $\pi((pm)^j p^\infty)$ to $\pi((pm)^{j+1} p^\infty)$ is a support hyperplane for $A_{-{\lambda},\mu}$ for all $j<k$. Consider the line from $\pi((pm)^k p^\infty)$ to $\pi((pm)^{k+1} p^\infty)$. This will have slope $-\frac{\mu^k}{{\lambda}^k}$ Consider any $(x,y)=\pi(a_0a_1\dots) \in A_{-{\lambda}, \mu}$. Note that without loss of generality we can assume that $a_0a_1\dots a_{2k}=(pm)^k$, in view of the fact that the sequence of slopes, $-\frac{\mu^k}{{\lambda}^k}$, is is a decreasing negative sequence, so if $a_0\dots a_{2k}\neq (pm)^k$, then we can apply the inductive hypothesis for some $j<k$.

As before, we see that the slope of this point is $$\frac{ \sum_{i\text{ even}} (1-a_i) \mu^i  + \sum_{i\text{ odd}} (\varepsilon_i-a_i) \mu^i}
     { \sum_{i\text{ even}} (1-a_i) {\lambda}^i -\sum_{i\text{ odd}} (\varepsilon_i-a_i) {\lambda}^i }
<  -\frac{ \sum_{i\text{ odd}} (\varepsilon_i-a_i) \mu^i} { \sum_{i\text{ odd}}(\varepsilon_i-a_i) {\lambda}^i },$$ where $\varepsilon_i = -1$ if $i < 2k$ and $1$ otherwise.

We want $$\frac{ \sum_{i\text{ odd}} (\varepsilon_i-a_i) \mu^i} { \sum_{i\text{ odd}} (\varepsilon_i-a_i) {\lambda}^i }
    < -\frac{\mu^{2k}}{{\lambda}^{2k}}.$$ Cross multiplying, this will occur if $${\lambda}^{2k} \sum_{i\text{ odd}} (\varepsilon_i-a_i) \mu^i
  \geq
  \mu^{2k} \sum_{i\text{ odd}} (\varepsilon_i-a_i) {\lambda}^i$$ or, equivalently, $$\sum_{i\text{ odd}} (\varepsilon_i-a_i) \mu^{i-2k}
  \geq
  \sum_{i\text{ odd}} (\varepsilon_i-a_i) {\lambda}^{i-2k}.$$ We see that ${\lambda}< \mu$ and hence $1/\mu < 1/{\lambda}$, from which it follows that $$\sum_{i\text{ odd}, i<2k} (\varepsilon_i-a_i) \mu^{i-2k} \geq \sum_{i\text{ odd}, i<2k} 