The balancing condition

Recall that $p(x) = \min_{\alpha \in S}(c_\alpha + \alpha \cdot x)$ is dual to the regular subdivision $\Delta_p$ of its Newton polygon $\mathrm{conv}(S)$ obtained by lifting each point $\alpha \in S$ to height $c_\alpha$ and projecting the lower convex hull back down. Every edge $e$ of $V(p)$ is dual to an edge $e^*$ of $\Delta_p$: $e$ is perpendicular to $e^*$ (rotate by $90^\circ$), and its weight $w$ equals the lattice length of $e^*$. Every vertex $v$ of $V(p)$ is dual to a $2$-dimensional cell (a polygon) $\sigma_v$ of $\Delta_p$, and the edges of $V(p)$ incident to $v$ correspond, in matching cyclic order, exactly to the boundary edges of $\sigma_v$. Traversing the boundary of the closed polygon $\sigma_v$ once around, its edge vectors $e_1^*, \dots, e_k^*$ sum to zero — a polygon returns to where it started: $e_1^* + \cdots + e_k^* = 0$. Rotation by $90^\circ$ is a linear map $R$, and each $R(e_j^*)$ equals $w_j v_j$ up to the fixed choice of orientation (the weight $w_j$ is the lattice length of $e_j^*$, and $v_j$ is $e_j^*$ rotated and rescaled to a primitive vector). Applying the linear map $R$ to both sides of $e_1^* + \cdots + e_k^* = 0$ gives $w_1 v_1 + \cdots + w_k v_k = R(0) = 0$, which is exactly the balancing condition.