of $t\lambda$, and $t\mathcal{P}_\lambda^{m} = {\mathrm{Newt}}(s_{t\lambda}(\x))$.

Let $\p$ be a point in the $t$-th dilate $t\mathcal{P}_\lambda^m = {\mathrm{Newt}}(s_{t\lambda}(\x))$. Since $s_{t\lambda}(\x)$ has saturated Newton polytope, then $\p$ is the content vector of a semistandard Young tableaux $T$ of shape $t\lambda$. The tableau $T$ decomposes into $t$ semistandard Young tableaux $T_1,\ldots, T_t$ each of shape $\lambda$ such that $T_i$ consists of the $j$-th columns of $T$ for $j\equiv i\mod t$. Letting $\bv_i$ denote the content vector of $T_i$, then $\p = \bv_1 + \cdots + \bv_t$, so ${\mathrm{Newt}}(s_\lambda(\x))$ has IDP.

Let $m=3$, $\x=(x_1,x_2,x_3)$, and $\lambda = (2,1,0)\vdash 3$. Each lattice point in the dilated polytope $3{\mathrm{Newt}}(s_\lambda(\x)) = {\mathrm{Newt}}(s_{3\lambda}(\x))$ is the content vector of a semistandard Young tableau $T$ of shape $3\lambda = (6,3,0)$, and the lattice point can be decomposed into the sum of three points which are content vectors of semistandard Young tableaux $T_1, T_2, T_3$ of shape $\lambda$ by taking the columns of $T$ mod $3$.

$$\begin{aligned}
\begin{ytableau}
   *(pink1) 1 &*(green2) 1 &*(yellow) 2 &*(pink1) 2 &*(green2) 2 &*(yellow) 3 \\
   *(pink1) 2 &*(green2) 3 &*(yellow) 3 
  \end{ytableau}
&=
\begin{ytableau}
   *(pink1) 1 &*(pink1) 2 \\
   *(pink1) 2 
\end{ytableau}
+
\begin{ytableau}
   *(green2) 1&*(green2) 2 \\
   *(green2) 3
\end{ytableau}
+
\begin{ytableau}
   *(yellow) 