that that $\lambda^{(2)}=(2,2,1)$ is not an extreme point of the Newton polytope.

\[prop.dilate\] Let $t$ be a positive integer. Then $$t{\mathrm{Newt}}(G_{h,\lambda}(\x)) = {\mathrm{Newt}}(G_{th,t\lambda}(\x)).$$

Let $\mathcal{P} = {\mathrm{Newt}}(G_{h,\lambda}(\x))$ and $\mathcal{Q} = {\mathrm{Newt}}(G_{th,t\lambda}(\x))$. Also let $\deg G_{h,\lambda}(\x)= |\lambda|+N$ while $\deg G_{th, t\lambda}(\x) = |t\lambda|+N'$. By Proposition \[proposition\_vertices\_of\_sgp\], the vertices of $t\mathcal{P}$ and $\mathcal{Q}$ are determined by the partitions $t\lambda^{(N)}\vdash t|\lambda|+tN$ and $(t\lambda)^{(N')}\vdash t|\lambda|+N'$, respectively, so it suffices to show that $t\lambda^{(N)} = (t\lambda)^{(N')}$. Furthermore by Lemma \[lem.dominance\](b), $(t\lambda)^{(N')}$ is the unique partition $\mu\in A(th,t\lambda)$ such that $b_{th,t\lambda\mu}\neq 0$ in the definition of the inflated symmetric Grothendieck polynomial $G_{th,t\lambda}(\x)$ and $|\mu|\vdash t|\lambda|+N'$, so it suffices to show that $t\lambda^{(N)} \vdash t|\lambda|+N'$. Suppose $$\begin{aligned}
\lambda^{(N)} &= (\lambda_1+a_1,\ldots, \lambda_m+a_m),\\
(t\lambda)^{(N')} &= (t\lambda_1+a_1',\ldots, t\lambda_m+a_m').\end{aligned}$$ We shall show via induction that $ta_r = a_r'$ for $r=1,\ldots, m$. The base case is $r=1$, where by definition, $a_1' = 0 = a_1 = ta_1.$ Assume that $a_j' = ta_j$ for all $j<r$.

If $\lambda$ has a maximum of $a_r$ addable boxes in its $r$-th row, then at least $ta_r$ boxes can be added to the $r$-th row of $t\lambda$, and so $ta_r \leq a_r'$ for $r=1,\ldots, m$.

Suppose $ta_r < a_r' \leq th(r-1)$. Then $a_r < h(r-1)$ implies that 