the largest number of boxes that can be added to the $r$-th row of $\lambda$ is $a_r$. Thus $\lambda_r \leq \lambda^{(k)}_r \leq \lambda_r+a_r$ for each $k=0,\ldots, N$. By construction, $$\lambda^{(b_i)} = (\lambda_1 + a_1, \ldots, \lambda_i + a_i, \lambda_{i+1},\ldots, \lambda_m)$$ for each $i=1,\ldots,m$, so that each part of the partition is either at a maximum or a mininum. Thus if $\mu,\nu\in {\mathrm{Newt}}(G_{h,\lambda}(\x))$ are lattice points such that $t\mu+(1-t)\nu = \lambda^{(b_i)}$, then $\mu = \nu = \lambda^{(b_i)}$ necessarily. So $\{\lambda^{(b_1)}, 
\ldots, \lambda^{(b_m)}\}$ is a set of extreme points of ${\mathrm{Newt}}(G_{h,\lambda}(\x))$.

On the other hand, suppose $k\notin \{b_1,\ldots, b_m\}$. Then there exists $j$ such that $$\lambda^{(k)} = (\lambda_1+a_1,\ldots, \lambda_{j-1}+a_{j-1}, \lambda_j + c , \lambda_{j+1},\ldots, \lambda_m)$$ with $0<c<a_j$. In this case, we have $$\begin{aligned}
\lambda^{(k-1)} 
    &= (\lambda_1+a_1,\ldots, \lambda_{j-1}+a_{j-1}, \lambda_j+c-1, \lambda_{j+1},\ldots, \lambda_m), \\
\lambda^{(k+1)} 
    &= (\lambda_1+a_1,\ldots, \lambda_{j-1}+a_{j-1}, \lambda_j+c+1, \lambda_{j+1},
\ldots, \lambda_m),\end{aligned}$$ so $\lambda^{(k)} = \frac12(\lambda^{(k-1)}+\lambda^{(k+1)})$. Thus $\lambda^{(k)}$ is an extreme point of ${\mathrm{Newt}}(G_{h,\lambda}(\x))$ if and only if $k\in \{b_1,\ldots, b_m\}$.

Lastly, $a_1=0$ and $N=a_1+\cdots+a_m$, so $\lambda^{(b_1)} = \lambda^{(0)}=\lambda$ and $\lambda^{(b_m)} = \lambda^{(N)}$.

\[eg.210a\] Let $\lambda =(2,1,0)\vdash 3$ and $m=3$. Then ${\mathrm{Newt}}(G_{\lambda}(\x))$ is the convex hull of the $S_3$-orbit of $$\textcolor{blue}{\lambda^{(0)} = \ydiagram{2,1}}\quad
\textcolor{green}{\lambda^{(1)} = \ydiagram{2,2}} \quad\hbox{and}\quad
\textcolor{orange}{\lambda^{(3)} = \ydiagram{2,2,2}}$$ In Figure \[fig.Grothendieck\], we see 